Nối tiếp nội dung phần 1 tài liệu Cẩm nang ôn luyện thi đại học 18 chuyên đề Hóa học, phần 2 giới thiệu các nội dung: Ăn mòn kim loại - Điều chế kim loại, kim loại kiềm - Kim loại kiềm thổ - Nhôm, các lý thuyết cơ bản của hóa học,... Mời các bạn cùng tham khảo nội dung chi tiết.
Trang 1Ca'm nang 6n luyCn thi dai hpc 18 chuy6n dS H6a hpc - Nguygn Van HSi
C h u y e n de 8
AN M(m K I M LOAI - IHltU €Hlt KIM LOAI
1 A N M O N K I M L O A I ^-^^
a Khai ni?m - ' ' grm nsffq oyt qsii 6v lorn S0;0 = tih •
Su pha h u y k i m loai hoac h o p k i m do tac d u n g hoa hoc cua m o i truorig
x u n g quanh goi la su an m o n k i m loai
b Phan loai ^-''("OVi'v-r^ -.vrHf rtsrfr:
An m o n k i m loai dugc chia thanh 2 loai chinh: A n m o n hoa hoc va an m6n
d i ^ n hoa •„-•!•> • ^ i s-: • - f/.'^ - >5 ^ ff'^v*^ 4'^'^
+ An mon hoa hoc:
A n m o n hoa hpc la sy pha h u y k i m loai do k i m loai phan u n g hoa hoc v6i
axit hoac cac chat k h i (hoi) 6 nhiet dp cao
Dac diem ciia an m o n hoa hoc la qua trinh oxi hoa-khu, electron dupe chuyen
tryc tie'p t u chat oxi hoa deh chat k h u len khong phat sinh dong di§n
Vidu\ , ,.s;Vt«nf ^'-S'iffW • £0.0 •<-^-10,0 ;{o:
Fe + H2S04(io3ng) > FeS04 + H2 +€^q
3Fe + 2O2 '" > Fe304 - 'v ,gA^ ^ " ^ ^ ' ' "
+ An mon dim hod:
A n m o n d i ^ n hoa la s u pha h u y k i m loai do k i m loai tie'p xiic v o i d u n g dich
chat dien l i tao nen d o n g electron chuyen d a i t u cue a m sang cue d u o n g va
phat sinh d o n g d i f n
Co che an m o n dien hoa: •.^_.,f^y.y :M _y.^;_,^^ v \ <» i-^ , , ^ ^
V i d u CO 2 tarn k i m loai Z n va C u dup-c n h u n g vao d u n g d i c h H2SO4
Tai cue am (anot), Z n ( k i m loai m a n h hon) b i oxi hoa: ,
Z n - 2e > Zn^*
, Tai cue d u o n g (eatot), ion H"^ chuyen den be m a t C u de nhan electron t u cue
a m chuyen t o i va b i oxi hoa tao thanh k h i H2:
2 H * + 2e > H2
Ba dieu k i ^ n de xay ra an m o n d i f n hoa: (1) Co hai k i m loai c6 ban chat
khac nhau, (2) tie'p xiic tryc tie'p v a i nhau (hoac gian tie'p qua day dan) va (3)
dupe n h i i n g vao cimg m p t d u n g dich chua chat d i ^ n l i
c Cach chong an mon kim loai:
+ Phuomg phdp bdo ve be mat:
D i m g n h u n g chat ben v o i m o i t r u o n g p h u len be mat k i m loai D o la:
D u n g son cho'ng gi, vecni, dau mo, trang men, p h u h p p chat poiime
M a m p t so' k i m loai ben n h u crom, niken, dong, kem, thie'c len be mat kif^
loai can bao ve
Cty TNHH MTV DWH Khang Vijt
+ Phuang phdp dien hoa:
No'i k i m loai can bao ve v o i 1 tarn k i m loai khac c6 t i n h k h u m a n h h o n
V i dv, de bao v? v6 tau bien bang thep, n g u d i ta gan vao v6 tau (phan chim trong nuoc bien) n h i i n g tarn kem K h i tau hoat dong, tarn k e m b i an m o n dan, v6 tau dupe bao v^ Sau m p t thoi gian n g u a i ta thay cac tarn k e m khac
VI D U M A U
Vi dvi 1: K h i dieu che hidro t u Z n va d u n g dich H2SO4 loang, neu them vao
m p t vai gipt d u n g dich CuS04 t h i tha'y k h i H2 thoat ra m g n h hon Ban chat cua h i ^ n t u p n g tren la ^^"t"' ' ^ '
A A n m o n dien hoa ^' • ' ' B A n m o n hoa hpc 'i'i
C u thoat ra b a m tren hat Z n nen thoa man 3 dieu kien ve qua t r i n h an m o n
d i f n hoa: Co hai k i m loai khac nhau (Zn - Cu), tie'p xuc true tie'p v o i nhau (Cu b a m vao Z n ) va cung n h u n g trong d u n g dich chat dien l i (H2SO4 loang)
D o do xay ra qua t r i n h an m o n dien hoa —> Dap an A
Vi 2: T r u o n g h p p nao sau day xay ra an mon di§n hoa: "'^"^ ^- '
A Thep de t r o n g k h o n g k h i kho, nong ,
B K e m de t r o n g d u n g dich H2SO4 loang
C Sat chay trong k h o n g k h i
D V 6 tau bien bSng thep ngam trong nuoc b i e n ^ 8 " " * ^ '
Loai A, C: Thep de trong k h o n g k h i kho nong hoac dot chay s^t trong
k h o n g k h i - » Deu bj an m o n hoa hpc. SOB^E EJJ9>1£ • •
Loai B: K e m de t r o n g axit H2SO4 loang Kem bj an m o n hoa hpc i
—> Dap an D
Luu y: V 6 tau bang thep la h p p k i m Fe - C (chua 2 d i | n cue c6 ban chat khac
nhau va tie'p xiic trye tie'p v o i nhau) va ciing tie'p xiic v a i d u n g djch d i ^ n l i (nuae bien) nen xay ra an m o n dien hoa :,;
V i d v 3: Cho cac h p p k i m sau: Cu-Fe (I); Zn-Fe (II); Fe-C (III); Sn-Fe (IV) K h i
tie'p xiic v o i d u n g djeh chat dien l i t h i cac h p p k i m m a t r o n g do Fe bi an
m o n truoc la:
A I , I I va IV B I , I I v a i n C I , I I I v a I V D I I , I I I va IV
1Q1
Trang 2C^m nang 6n luy^n thi dgi hpc 18 chuySn 66 Hoa hpc - Nguygn Van HSi
Lai gidi:
Nhan xet: Trong cac qua trinh an mon dien hoa, kim loai m ^ h han se bi an
^ mon —> Fe bi an mon truoc khi no la chat khu manh hon (dong vai tro la
anot - cue am)
Trong cac cap (I), (III), (IV) thi Fe deu manh han, rieng cap (II) thi Zn manh
hon
• ^ D a p a n C , , • , :>;;.f:,,,, , ,,•
V i di^ 4: Gang bi an mon dien hoa trong khong khi am Qua trinh kliir xay ra tren
•n- be mat aia gang la: ,fe^,^|,.,B,M>i.#«i: -)1»ft»l^s*&i^.»v:t<,r
A O2 + 2H2O + 4e > 4 0 H - ' B 2H2O + 4e > O2 +
40H-C Fe > Fe2* +2e D Fe > Fe^* +3e
Lai gidi:
Thep la hop kim Fe-C (vdi mot so nguyen to'khac) Khi tie'p xiic voi khong
khi am (c6 chiia chat di^n li) se xuat hi^n qua trinh an mon di?n hoa Fe la
' cue am (bi an mon), C la cue duong se xay ra su khu
O2 + 2H2O + 4e > 4 0 H - ^nfi^YGx6«<)au^frl9.flj^/l>tm-•
—> Dap an A
Luu y: Cac em de chon riham B vi bo qua vai tro cua chat oxi hoa la oxi
khong khi
V i du 5: Co 4 dung dich: a) HCl, b) CuCh, c) FeCb, d) HCl c6 Ian CuCh
Nhiing vao moi dung dich mpt thanh Fe nguyen chat So truong hop xuat
hien an mon dien hoa la: ,
A.O B 1 C.2 D.3
Laigiai, yu,<)(i!M"
Nhan xet:
+ Khi nhiing thanh Fe vao cac dung dich HCl va FeCk, Fe bi an mon hoa hpc
theo cac phan ling:
gr Fe + 2HC1 > FeCk + H2 '
.:o-Fe + 2.:o-FeCl3 > S.:o-FeCk ,r;
+ Khi rihiing Fe vao dung dich CuCk se CO qua trinh: 0
Fe + CuCk > Cu + FeCk
Jfi K i m loai Cu sinh ra bam vao Fe tao thanh cap di#n eye nhiing trong dung
•' dich dien l i la CuCk nen xay ra an mon dien hoa
+ Tuong t u voi dung dich HCl c6 l l n CuCk, Fe se day Cu theo phan ling:
ifl Fe + CuCk > Cu + FeCk
—> Cu bam vao Fe —> Xay ra an mon dien hoa ; W
Chat khu la C: Khu dugc cac oxit cua kim loai tir Zn va tao thanh khi CO
Chat khu la Al: khu dugc cac oxit ciia kim loai tu Cr
Thuy luyen la phuang phap dieu che kim loai bang each dimg kim loai m^nh hon day kim loai yeu hon ra khoi dung dich muo'i
hieu y: Do'i voi cac kim loai kiem va kiem tho (tru Be, Mg) khi tac dung vai
dung dich muoi, chiing se tac dung vai nuoc -> Dung dich baza, bazo tao thanh c6 the tac dung tie'p voi muo'i:
Vi du cho N a kim loai vao dung dich CuS04 xay ra cac phuang trinh:
2Na + 2H2O > 2 N a O H + H2 2NaOH + CuS04 > Na2S04 + Cu(OH)2^
Phttong phap di^n phan
Bien phan nong chay
Day la phuang phap dieu che cac kim loai manh Cac kim loai kiem, kiem tho thuang dieu che bang phuang phap dien phan nong chay muoi halogenua; rieng Al dugc dieu che'bang phuang phap di§n ph?n nong chay AkOs voi duang cue bang than chi o
2Ak03 ''P"" > 4A1 + 3O2
Dien phan dung dich
Day la phuang phap dieu che cac kim loai trung binh va yeu tu cac dung dich muo'i tuang ling ciia chiing
iLuong hap 1: Goc axit tham gia di^n phan:
CuCk _ i P ^ Cu + Ck Iruang hop 2: Goc axit khong tham gia dien phan:
CuS04 + H2O — C u + -O2 + H2SO4 '
2
4AgN03 + 2H2O — ^ 2 ^ 4Ag + O2 + 4HN03
193
Trang 3Ca'm nang 6n luyen thi dgl hoc 18 chuyen d8 H6a hpc - Nguyin Van Hi\
D i n h luat Faraday (tinh lup-ng chat t h u dugc 6 cac d i | n cue): m = A l t
nP
m: K h o i l u g n g chat t h u dug-c a d i ^ n cue (gam)
A: Kho'i Itfong m o l nguyen t u (mol.phan tu) ciia chat t h u dug'c a d i ^ n cue
n: So electron trao d o i m a ion d a nhan de t^o thanh n g u y e n t u (phan tu)
t u o n g u n g 6 dien cue
I : C u o n g d p d o n g dien (ampe) t: thai gian di#n p h a n (giay)
F: H a n g so Faraday (F = 96.500) i ' ^ 1 m,- v
Lieu y: T u cong thue d i n h luat Faraday, so mol electron trao d o i t r o n g qua
t r i n h d i ^ n phan d u o c t i n h theo eong thue: n g = — n = —
Day la cong thue giiip giai nhanh chong nhieu bai tonffcfien phan
V i D V M A U _ • ,^.nn
V i dxf 1: Cho l u o n g k h i H2 (du) qua hon hgp cae oxit CuO, Fe263, A I 2 O 3 , MgO
n u n g a nhiet d o eao Sau phan u n g hon hgp ran con lai la:
A C u , FeO, A I 2 O 3 , M g O B C u , Fe, A l , M g
C C u , Fe, A l , M g O ' D Cu, Fe, AhOa, M g O V ,
rfn-.i,,.,, Lai giai:
Nhan xet: Day la each dieu che k i m loai bang p h u a n g phap n h i f t luyen 6
n h i ^ t d p cao, k h i H 2 C 6 the k h u cac oxit k i m loai d i m g sau Z n
6 bai nay, k h i H2 k h u dupe cac oxit CuO, Fe203 thanh C u , Fe; k h o n g khir
dupe M g O , A I 2 O 3 - > Dap an D
V i d^ 2: Cho l u o n g k h i C O (du) d i qua 12 gam h o n h p p g o m Fe203 v a MgO
n u n g n o n g den k h i phan u n g hoan toan, t h u dupe 9,6 g a m chat ran Khoi
l u p n g Fe203 c6 trong h o n h p p ban dau la: „,,^ , ,^ ^f,;^'
A 8 gam B 16 gam C 24 gam D 12 gam
Nhan xet: K h i C O chi k h u dupe Fe203 theo phan l i n g :
rnpejOa ~ 0/05.160 = 8 gam -> Dap an A
V i d\ 3: K h u hoan toan m o t oxit sat FcxOy a nhi^t dp cao can vira d u V lit kl^'
C O (dktc), sau phan u n g t h u dupe 5,6 gam Fe v a 0,15 m o l k h i CO2 Cong
thiic cua oxit sSt v a gia t r i cua V la
A Fe304 va 3,36 , , ,,, B Fe304 v a 4,48 - m »• EOMS
C FeO v a 2,24 - s ^ D Fe203 va 3,36 ;
Lm giai:
ach 1: Giai theo p h u o n g t r i n h hoa hpc
FexOy + y C O — ^ xFe + yC02
Nhan xet: XXQ ^^xit) = " c o = "002 = 0,15 -> Vco2 = 3,36
M a t khae: n^^: TYQ = 0,5:0,15 = 2:3 -> Fe203 -> Dap an D
V i dv 4: N g a m m o t la k e m t r o n g 100 m l d u n g dich CuS04 a mol/1 Sau k h i
p h a n u n g ke't thue, lay la k e m ra k h o i d u n g dich, rua sach, say kho v a dem can, thay kho'i l u p n g la kem giam 0,05 gam Gia t r i cua a la
Vi dv 5: K h i d i ^ n p h a n d u n g dich CuS04 ( v o i d i ^ n eye tro) t h i cac qua t r i n h
xay ra 6 r y e am v a cue d u o n g Ian l u p t la
A O x i hoa H2O v a k h u ion Cu^* B K h i r i o n Cu^"^ v a oxi hoa H2O
C K h u i o n Cu^* va oxi hoa i o n SO 4 " D K h u H2O va oxi hoa i o n Cu^*
Lai giai: ,r
Cyc a m (eatot) eo i o n Cu^* v a H2O I o n Cu^* bi di$n phan:
Cu2* + 2e > C u {Qua trinh khie) Cye d u o n g (anot) eo H2O va i o n SO 4 —> H2O b i d i ^ n phan:
H2O >-O2 + 4 H * + 2 e
2
(Qua trinh oxi hoa) 80,6 " : l o i
—> Dap an B
Ghi nha cau tha: A n h la A n o t n h u o n g e, Em la Catot n h a n e thoi ma
V i dy 6: D i e n phan 800 m l d u n g dich AgNOa 0,1M v o i dien cue t r a trong t gia,
cuong d p dong di?n k h o n g d o i 1,34A (hi^u sua't qua t r i n h d i ^ n phan la 100%), t h u dupe chat ran X, d u n g dich Y v a k h i Z Cho m o t t h a n h Fe vao Y, sau k h i cac p h a n u n g ke't thiic thay k h o i l u p n g thanh Fe tang them 2,36 gam va c 6 k h i N O (san p h a m k h u d u y nhat ciia N"^"^) bay ra Gia t r i eiia t la
A 0,8 B 1,2 C.1,0 D.0,3
.- " 195
Trang 4dm nang On luy^n thi dgi hpc 18 chuy6n di H6a hpc - NguyJn Van HSi
Lot gidi:
Phan ling dien phan dung dich AgNOs: ' " |
2AgN03 + H2O —^Edd_) 2Ag + ^ 0 2 + 2HN03
Mol: 2a "'^ ' ' ' ' 2a ' 2a ' '^^'
Nhqn xet: Vi khoi lug-ng Fe tang them nen thanh Fe dupe Ag bam vao-^ y
chiia A g N 0 3 - > AgNOs chua bi di^n phan het* l«l^/;v.'
-Cac phan ung khi cho Fe vao Y:
Fe + 4HN03 > Fe(N03)3 + NO + 2 H 2 0 , o y a n i ^ M '
Can luu y Fe tac dung voi Fe(N03)3: / -l^ j '
Fe + 2Fe(N03)3 > 3Fe(N03)2 rVrt',
Mol: 0,25a •(- 0,5a /^^v r, ''^ " - ^ ' '',i»Viii W.KWMi *
Fe + 2AgN03 > Fe(N03)2 + 2Ag ' ' ' ofex*
Mol: b <- 2b - > 'jsufTO-f 2b
Theo bai: 2b.l08 - 56(0,75a + b) = 2,36 w: a'.:
Matkhac: nAgNOa = 2a + 2b = 0,08 a =0,02 mol; b =0,02 moi £ jo;
ne= 2a = 0,04 ^ = 0/04 mol ^ t = 2880 giay = 0,8 gio ^ Dap an A
V i 7: Di?n phan dung dich X gom 0,16 mol NaCl va 0,24 mol CuS04 (dien cue
tra, mang ngan xop) den khi tha'y thoat ra 6 anot 2,688 lit khi (dktc) thi ngung
di|n phan Khoi luong dung dich sau dien phan giam di bao nhieu gam?
A 6,96 gam B 12,08 gam C 17,20 gam D 5,12 gam
LMgidi:
2,688 „ \ ,
n t = = 0,12 mol. , „ , v ^ , , ,
22,4
Phan ling "quy doi" gia dinh: i*-^ v
CuS04 + 2NaCl > CuCh + Na2S04 ^^nwib D
Mol: 0,08 <- 0,16 0,08 0,08
^ X gom cac muoi: CuCh = 0,08 mol; CuS04 = 0,16 mol va Na2S04 = 0,08 mol
Thii tu dien phan nhu sau:
A Phuong phap hoa hoe B Phuong phap difn hoa , ,
C Phuong phap bao ve be mat D Phuong phap tao hpp kim khong gi
Bai 2: Co 4 dung dieh rieng biet: CuS04, ZnCh, FeCb, AgNOa Nhiing vao moi dung dich mpt thanh Ni So truong hpp xuat hif n an mon di^n hoa la
A 2 .B.4 C.3 D 1
Bai 3: Trong pin di^n hoa Zn-Cu, qua trinh khir trong pin la
A Zn^* + 2e > Zn B Cu^* + 2e > Cu ^
C Cu > Cu2* + 2e D Zn > Zn^* + 2e
Bai 4: Tien hanh boh thi nghi^m sau: , (1) Nhiing thanh Fe vao dung djch FeCb; ^
(2) Nhiing thanh Fe vao dung dich CuS04; ^' ' (3) Nhiing thanh Cu vao dung djch FeCb; ^^"^ "'"'^ ' " \
(4) Cho thanh Fe tiep xiic voi thanh Cu roi nhiing vao dung dich HCl
So truong hpp xuat hi^n an mon di^n hoa la
dupe 21,4 gam chat ran Y Cho Y tac dung voi dung dich HNO3 loang, du
thu dupe V lit NO (san pham khu duy nha't, 6 dktc) Gia tri eiia V la:
0: A 2,24 B.4,48 C 6,72 D 8,96
^ 7: Tien hanh phan ung nhi^t nhom voi h6n hpp bpt gom A l va CuO, thu
dupe eha't ran X Chia X thanh hai phan bang nhau ,j i"-
+ Phan 1 tac dung vua dii voi 100ml durg dich NaOH I M
^ Phan 2 tac dyng voi dung djch HNO3 loang, du thu dupe V lit NO (san
; pham khii duy nha't, o dktc)
Gia tri eiia V la: j
A 2,24 B.4,48 C 6,72 D 8,96
^ai 8: Hon hpp Y gom CuO, FeO va Fe203 Hoa tan hoan toan 27,6 gam Y bang
dung dich H2SO4, thu dupe 59,6 gam muol M§t khae, ne'u khii hoan toan
27,6 gam X can vua dii V lit khi CO (dktc) Gia trj eua V la
A 4,48 B.6,72 C 8,96 D 5,60
,.CI J 't.8<^,>
Trang 5CJm nang 6n luy^n thi dgi hpc 18 chuygn dg H6a hqc - Nguygn Van HJi
pai 9: Nhung mot la kim loai R (hoa tri 2) c6 khoi lugng 10 gam vao 200ml dung
dich CuS04 0,1M cho den khi phan ling hoan toan thu duoc chat rAn X va
dung dich Y chua 3,22 gam muoi K i m loai M la
A Mg ^'^ B Be C Fe D Z n
pai 10: Cho hon hgp gom Mg va Fe vao dung djch C u C h deh khi cac phan virig
hoan toan, thu dug-c dung dich X gom hai muo'i va chat ran Y gom hai kim
loai Hai muoi trong X la
A M g C h v a F e C h ^ , B M g C h va CuCh
C F e C h va CuCb D M g C h va FeCh
pai 11: Nhung mot thanh Mg vao 100ml dung dich gom Cu(N03)2 0,3M va
AgNOa 0,2M Sau mpt thoi gian lay thanh kim loai ra, rua sach lam kho,
thay khoi luong thanh Mg tang 2,72 gam Khoi lugng Mg da phan ung la
A 1,20 gam B 0,72 gam C 0,48 gam D 0,96 gam
pai 12: Cho 1,3 gam hot Z n vao 100ml dung dich chua hon hgp gom AgNOa
0,2M va Cu(N03)2 0,5M Sau khi cac phan ung xay ra hoan toan, thu dugc
dung dich X va m gam chat ran Y Gia tri cua m la: ^
A 2,80 B.5,36 C 4,08 D 2,16
pai 13: Cho hon hgp bpt gom 1,3 gam Z n va 3,2 gam C u vao 200ml dung dich
^ AgNOa a mol/1 sau phan ung hoan toan thu dugc 12,08 gam chat ran X gom
hai kim lo^i Gia tri cua a la T'lMj
A 0,20 B.0,30 C.0,15 • D 0,50
pai 14: Cho 4,1 gam hon hgp X gom Z n va Fe vao 200ml dung dich CuS04 0,2M
, Sau khi cac phan ihig xay ra hoan toan, Igc thu dugc 4,24 gam chat ran Y va
dung dich Z chiia hai muoi Khoi lugng cua Z n trong X la:
A 0,65 gam B 1,3 gam C 1,95 gam D 2,6 gam
pai 15: Di^n phan 200ml dung dich gom AgNOs 0,5M va Cu(N03)2 I M bang
li dong dif n mgt chieu vol cuong dg dong dif n 2,68A trong thai gian 4 gia
Kho'i lugng kim loai thoat ra 6 catot la:
A 20,4 gam B 23,6 gam C 10,2 gam D 11,8 gam/'
pai 16: Di#n phan dung dich hon hgp chua 0,15 mol CuS04 va 0,1 mol H C l
trong thai gian 2 gio vai dong difn c6 cuong dg la 1,34A Biet hifu suat ciia
qua trinh di|n phan la 100% Khoi lugng kim loai thoat ra tai catot la:
A 6,4 gam B 3,2 gam C 12,8 gam D 9,6 gam
Bai 17: Di?n phan dung dich X gom 0,2 mol N a C l va 0,2 mol CuS04 (di^n afc
tro, mang ngan xo'p) den khi khoi lugng dung dich giam di 17,5 gam thi
ngung di^n phan The rich khi (dktc) thoat ra 6 anot la
A 1,12 lit B 1,68 lit C 2,24 lit D 2,80 lit
^ Cty TNHH MTV DVVH Khang Vigt
T H I / O ' N G D A N - L C J I G I A I
Rai ^'
Sn trang len Fe se bao phii be mat tranh cho Fe bi an mon Day la phvrang phap bao ve be mat —> Dap an C
luu y: day cac em de chgn nham dap an B Can nho rang phuang phap
dien hoa la de bao v§ kim loai, can ghep no vol kim loai m^nh hon
2: :, , ,,.v.,i if-y •••:> • r:-<i '':vr
-+ Nhiing N i vao cac dung dich CuS04 hoac AgNOa thi N i se day C u va A g ra, cac kim loai nay lai bam tren be mat N i Tao thanh cac cap pin di^n Ni -
C u va N i - A g —> A n mon dien hoa
+ Nhung N i vao dung dich Z n C h thi khong xay ra phan ling - > N i khong bi
Catot (cue duang): Ion Cu^* den nhan electron (qua M n h khii ion Cu^*):
Cu2* + 2e > C u ' i A ^^:tt.: y ^_ i_-_Q|i:>,,iA igrm riE/iq tj.u
Dap an B ••:>iMJ%;' ' • '•
BM4:
Nhan xet: 6 thi nghi^m 2 (Fe day C u va C u bam vao Fe) va 4 (Fe tiep xuc vol Cu) -> Deu sinh ra cap di|n c\fc Fe - C u , tiep xiic tryctiep vai nhau va ciing
nhung trong mgt dung dich di|n li Xuat hi|n an mon di|n hoa -> Dap an B
Lieu y: 6 thi nghi?m 1 va 3, Fe b} an mon hoa hgc theo cac phan ling:
|,, Fe + 2FeCla > 3FeCl2 -fy?^ vt;;/''^
C u + 2FeCla > C u C k + 2FeCh f + Xo^»^ •
nAl= ^ = 0,5 mol; n^^^o, = ^ = ^'^ " H 2 ~ = 0'^
2A1 + Fe203 — ^ AI2O3 + 2Fe
Trang 6Ca'm nang On luy^n thi dgi hpc 18 chuy§n 6i Hoa hgc - Nguyjn VSn Hai
AI2O3 + 2NaOH > 2NaA102 + 2H2O
Al^ FeaOg X ^ " ^ ° 3 > AI3^, pe^^ ; r ^ ,
Nhan thdy: tu tr^ng thai dau tien den trang thai cuoi ciing thi: So'oxi hoa ciia
Fe khong doi (luon la +3), con so oxi hoa ciia A l tang tit 0 len +3 ,, ,^
A p dyng bao toan electron: ng=3nAi =3nNO HC*
-> n ^ o = r»Al= 0/2 mol -> V = 0,2.22,4 = 4,48 lit -> Dap an B
Bai7:
Nhan xet:
Khi cho phan 1 + NaOH, tat ca A l va AI2O3 deu chuyen vao hp-p chtYt
NaA102 -> A p dyng bao toan nguyen to'Na va A l ta c6:
nNaOH="NaA102 = " A l nAi= 0,1.1 = 0,1 mol
So do phan ling: A l , C u O — ^ X ) A l ^ \
Nhan thdy: T u trang thai dau tien den trang thai cuoi cung thi: So oxi hoa
ciia C u khong doi (luon la +2), con so'oxi hoa cua A l tang tu 0 len +3
Ap dyng bao toan electron: = Sn^i = Sn^o
-> n ^ o = " A r 0,1 mol ^ V = 0,1.22,4 = 2,24 lit
-> Dap an A
Bai8:
Nhan xet: ^,v,,Wj./«-i5 «ir/;,
Khi cho X + H2SO4 thi oxit chuyen thanh muoi sunfat va mot ion O^" trong
oxit dupe thay the bang mpt ion SO 4" '
1 mol 02- > 1 mol S O ^ ~ -> Khoi lu(?ng tang 96 - 1 6 = 80 gam '
a mol tang 59,6-27,6 = 32 gam
32
-> a = — =0,4 mol -> no (x) = 0,4 mol
80
Khi cho 27,6 gam X tac dyng voi C O thi: no (x) = nco= 0,4 mol
Vco= 0,4.22,4 = 8,96 lit -> Dap an C '
Nhan xet: Y gom 2 kim l o ^ —> Y chiia Mg hoac Fe Mpt kim loai con du ->
CuCk het Loai B va C
M|t khac: Do tinh khii Mg > Fe —> Mg phan ling truac roi moi den Fe
Gia sii Mg con d u —> Fe con nguyen va C u bi "day" ra het -> Chat rSn X gom 3 kim lo^ii —> Dieu gia sii la sai
Mg phan ling het va Fe phan ling mpt phan —> Y gom Fe du va Cu
-> Hai muoi trong X la MgCh va FeCh -> Dap an A
Bai 11:
"AgN03 = 0,02 mol; ncu(N03)2 = 0,03 mol •? >
Nhan xet: Tinh oxi hoa Ag* > Cu^* AgNOa phan ling trudc roi moi den
Cu(N03)2 »i>UV!iJ C/»?T S S > , • , f TlfWC{" &fo «r v
Phan ling hoa hpc: \ i r i * JOP;
Mg + 2AgN03 > Mg(N03)2 + 2Ag Mol: 0,01 <- 0,02 0,02 -> mtang = 2,16-0,24 = l,92g
Mg +CU(N03)2 > Mg(IJ03)2 + C u
Mol: a <— a a mtang = (64-24)a = 40a
Theo bai: m Mg tang = 2,72 1,92 + 40a = 2,72
-> a = 0,02 mol — > mMg (pu) = 24.(0,01 + a) = 0,72 gam
-> Dap an B
Bai 12: ' ' ^ ' ^ ' ^ '
= 0,02 mol; n^^j^o^ = 0,02 mol; ncu(N03)2 = 0,05 mol
Phuong trinh hoa hpc: ^^^^^^ ^^^^^ ^^
Trang 7dm nang On luyQn thi dgi hgc 18 chuySn dg H6a hpc - Nguygn Van HSi
Bai 13:
n z n = 0,02 m o l ; n c u = 0,05 m o l if, fi m i ) j m q f ^ •
C a c h l : - ^ Or ^
N/iflM xet: X g o m hai k i m loai c6 chua Z n hoac Cu —> AgNOs he't
D o tinh k h u Z n > Cu —>• Z n phan u n g truac roi m a i den Cu X g o m Ag
Cach 2: Gpi: ncu (pu) = x m o l Bao toan electron: 2n2n + 2ncu(ph4n ijng) = "^g+
2.0,02 + 2x = 0,2a -> 0,02 + x = 0,la ,4 m ;^a;j i»«.vi
M a t khac: mx = 64(0,05 - x) + 108.0,2a = 12,08 -> x = 0,03; a = 0,5 ? (BH •
- » Dap an D
, Nhan xet: T i n h k h u Z n > Fe ->• Z n phan u n g truoc roi m a i den Fe D o mx <
mv - > Fe = 56 da phan u n g mpt phan de tao thanh C u = 64 —> Z gom
Nhan xet: Day la bai toan di|n phan d u n g djch chua nhieu chat, ca.c em nen
ap d u n g cong thuc tinh so m o l electron trao doi:
m It 2,68.4.3600 ^ , ,
n „ = — n = — = = 0,4 mol „ „
^ A F 96500
Ta c6: So'mol electron A g * nh^n vao: 0,1.1 = 0,1 m o l
-> So'mol electron Cu^* nh|n vao = 0,3 m o l
Cty TNHH MTV D W H Khang Vijt
So m o i CuS04 phan u n g = 0,15 mol. ::.,,f:-./: J.!A.f./'-:'.,-'
mcntot tang = m A g + m c u = 0,1.108 + 0,15.64 = 20,4 gam
4-M o l : 0,1 < - 0,2 - > 0,1 0,1
- > X gom: C u C h = 0,1 m o l ; CuS04 = 0,1 m o l va K2SO4 = 0,1 m o l " ' '
T h u t u dien phan n h u sau:
C u C l 2 Cu + CI2 "-"•••"•^ ^-^^^^"'^
M o l : 0,1 0,1 0,1 m g i a m = 0,1.(64+ 71)= 13,5 gam
-> K h o i l u g n g d u n g dich con giam tie'p 17,5 -13,5 = 4 gam
cuso4 + H 2 0 - i p ^ Cu + io2 + H 2 S O 4 , *
M o l : a a 0,5a ''•TDfeMS ^ cO:dr
64a + 32.0,5a = 4 -> a = 0,05 mol ,.^14 4 r;(HO)!/
Vc,2 + V02 = (ai + 0,025).22,4 = 2,8 lit LaocJio ioinv
-» Dap a n D w i malBlHqlonyrtq [dmx • , \nnU laix w
Trang 8+ Tac dung vai nude ^
Kim loai kiem khir mxac de dang 6 nhi?t dp thvrong, giai phong k h i hidro:
2Na + 2H2O > 2NaOH + H2t
+ Tac dung v&i dung dich axit ;,
2Na + 2HC1 > 2NaCl + H2t
N e u N a d u : 2Na + 2H2O > 2NaOH + H2t
+ Tac dung vai dung dich muoi
Cho K k i m loai vao dung dich CuS04 xay ra cac phuong trinh:
2K + 2H2O > 2KOH + H2t
2KOH + CuS04 > K2SO4 + Cu(OH)2i
b Hidroxit,, „„,„.„:„.„ rfe:)|'iWomr.O.':^
+ Tac dung vdi oxit axit ;
CO2 + 2NaOH > NaiCQh + H2O
CO2 + NaOH > NaHCOa
+ Tac dung vdi dung dich muoi
Fe(N03)3 + 3NaOH > Fe(OH)3i +3NaN03
+ Tac dung vdi hap chat luang tinh
AI2O3 + 2NaOH > 2NaA102 + H2O " "
Al(OH)3 + N a O H > NaAI02 + 2H2O
c Muoi cacbonat
+ Phan ieng thuy phan (quy tim ~^ xanh; phenolphtalein ~^ hong)
do
>0
col' + H2O HCO3 +
OH-Tac dung v&i dung dich axit
Cho tu tu dung dich axit HCl vao dung djch Na2C03
Na2C03 + H C l > NaHC03 + NaCl
NaHC03 + H C l > NaCl + C 0 2 t + H2O
d Muoi hidrocacbonat
+ Tinh chat ludng tinh
NaHC03 + HCl > NaCl + C02t + H2O
NaHCOs + N a O H ^ NaiCO^ + H2O
phan img nhiet phan
2NaHC03 Na2C03 + CO2T + H2O - * •
Cty T N H H M T V DVVH Khang Vi$t
iff
IVluoi clorua
phan leng dim phan: 2NaCl * 2Na + CI2T
^ 2NaCl + 2H2O " P " " ' ' " " > 2NaOH + C h t + H2t
+ TflcdwM ^^<^2''^'*W2S04(fflc(dieuche'khiHCl) f • of 1,j NaCl (ran) + H2S04(^flc) ^ NaHS04 + H C l t
f Muoi nitrat ^
; Phan ung nhiet phan / ^ ^^^^^^^^^ , ^ j , ^ ^ , , , ,
2KN03 2KNO2 + O2 , J _^ •
+ Tinh oxi hod (khi c6 mat axit)
Cho Cu vao dung dich hon h(?p KNO3 va H2SO4 loang, Cu tan dan tao
thanh dung dich mau xanh theo phuong trinh: ^ 3Cu +2NO3 + 8H* ^ 3Cu2- + 2 N O t + 4 H 2 0
V I D U M A U • >'•' '
Vi dii 1: Hoa tan hoan toan 1,7 gam hon hop gom kim loai kiem M va oxit ciia
no vao nuoc, thu dugc 300ml dung dich chiia mQt chat tan c6 nong dp 0,2M
va 0,224 lit khi H2 (dktc) Kim loai M la ^,^^1^ j-j ^ , , ,
Mat khac: 0,02M + (2M + 16).0,02 = 1,7 M = 23 (Na) - > Dap an A
^1 du 2: Hoa tan hoan toan 1,794 gam kim loai kiem M vao 200 ml dung dich
H2SO4 0,1M Co can dung dich sau phan ung thu dugc 4,36 chat ran khan Y
Kim loai kiem M la ifiui >•
A.Rb , B K C.Na | o e D L i
Lcrigidi: , f^Q
Cac phan ung hoa hoc:
2M + H2SO4 ^ M2SO4 +H2 va M +H2O ^ M O H + ^ H 2
205
Trang 9Hoa TiQc - NQuyen van Hai
Nhdn xet: Bao toan kho'i lugng -> m y = I T I M + " ^ c o Z - + ^
V i d u 3: Hoa tan hoan toan 4,6 gam N a vao SOOjnl H2SO4 0,1M, thu dugc khi
H2 va dung dich Y Co can Y thu dugc m gam chat ran khan Gia tri ciia m la
A 11,1 B.7,1 C.14,2 D 18,2
Laigidi:
H f g a = 0-2 mol; nH2 = 0/1 mol; nH2S04 = 0,05 mol
Luu y: N a con d u se tiep tuc phan ung voi nuoc: ^^^^ ^
- N a + H2O ^ N a O H + -Hi
2 Mol: 0,1 '•^•'•'••al
2Na + H2SO4 > Na2S04 + H 2 t
Mol: 0,1 0,05 0,05
^ m = mNa2S04 + rnNaOH= 142.0,1 + 40.0,1 = 11,1 gam —> Dap an A
V i d^ 4: Cho m gam hon hgp gom N a va Ba (ti 1# mol 2:1) tac dyng voi nuoc
(du), thu dugc dung dich X va 4,48 lit H2 (dktc) Cho 300ml dung dich
Al2(S04)3 0,2M vao X, thu dugc m gam ket tua Gia tri cua m la
A 29,54 B 6,24 C 24,86 D 9,34
Voigidi: •••M^^^' -wriq
N a + H2O > N a O H + i H 2 ^
1 Mol: 2a 2a a ' : ' : ^ ^ J '
Ba + 2H2O > Ba(OH)2 + H2 OsH O i M
Mol: a I ' : a a > h ! • ( •
= 0,2 mol -> 2a = 0,2 ^ a = 0,1 mol - aO,0 = SAc.*- •*>:•
= HNaOH + 2nBa(OH)2 = 0,2 + 2.0,1 = 0,4 mol
n^^2 = aimol; n^^ = 0,2mol Q ^ ^ ,
Phuong trinh ion: ' ^ fef M m'jc>{ ifiol f
Ba2- + SO^- >BaS04^ ' 8 d>I
Mol: 0,1 0,1 0,1 - - ^ J ^ i ^ J
A13^ + 3 0 H - > Al(OH)3 4 , g r m nB/!q Di'
Mol: 0,12 0,36 0,12 , :'}H.%<3eJ4'-^^—- w(D&H + :
Hhan xet: V i Y chua hon hgp muo'i —>^ chac chan c6 chua muoi axit —>
K O H phan ung he't
Trang 10Ca'm nang fin luy$n thi dgi hpc 18 chuy§n de 116a hqc - Nguygn Van H i !
V i d\ 8: Cho day cac m u o i : K M n 0 4 , NaNCh, Cu(N03)2, CaCOs, KCIO3, AgNOs
So m u o i t r o n g day k h i nhi^t phan hoan toan t h i tao ra so m o l k h i Ion h o n so
m o l m u o i p h a n l i n g la:
A 2 B 3 C 4 ^ D 5 :d
.(••A:Ab) cp.tid IciA J;'Cd,i' Laigidi: ' i'^-^ -mddo
Nhan xet: Day l a cau h o i n h a m k i e m tra kien thiic cua hpc sinh ve d p ben
n h i | t ciia cac muo'i t r o n g c h u o n g t r i n h t u lop 10 den Idp 12
Cac p h a n ving nhiet phan:
V i d\ 9: T r p n 300ml d u n g dich X gom Ba(OH)2 0,1M va N a O H 0,1M v 6 i lOOml
d u n g d i c h Y g o m Al2(S04)3 0,1M va H2SO4 0,1M, t h u dupe a gam ket tua
G i a t r i c u a a l a ' ptj^si fMsnij! miiq-.:
Cty TNhH 1TV DWH Khang Vigt
> BaS04i 0,03
- > Dap an A ••' ;.(,{:••'• ^ 0~ ,.(y-)n'^ m.im.n
V i d\ 11: Cho 5,65 gam hon h p p X gom L i , N a va K vao 100 m l H2SO4 I M , thoat ra 3,92 l i t k h i H2 (dktc) v a d u n g dich Y Co can Y t h u dupe m gam chat ran khan Gia t r i ciia m la
A 19,80 B 15,25 C 22,45 D 17,80 j
Laigidi: i ?;[
3 92
n H , = - ^ ^ — = 0,175 • ni c n III 1.,
Bao toan electron: = 2nH2 = 0,35 -> n x = = 0,35
A p d u n g d i n h luat t r u n g hoa dien: 1.0,35 = 2n 2 - + l ^ i r ^ u - ^ •
Trang 11CSim nang 6n luy$n thi dgi hpc 18 chuyen dg H6a hgc - NguySn v a n HSi
V i dii 12: Hoa tan hoan toan 6,65 gam hon hop muo'i clorua cua hai kim lo^ij
kiem thuoc hai chu ki ke tiep nhau vao nuoc duoc dung dich X Cho X tg^
• dung voi dung dich AgNOs (du), thu dugc 14,35 gam ket tua Hai kim \Q^^
kiem tren la
A R b v a C s B N a v a K C L i v a N a D K v a R b
"AgCl ^ mo\ Goi cong thijfc chung ciia hai muo'i la MCI ,
Phan ung hoa h o c : + •!('>)!/
MCI + AgNCh > MNO3 + A g C l i 4i =K;;n
MCI = ^ = 66,5 M = 31
mi 0,1
r^i -> Hai kim loai la N a va K —>^DapanB 5'MV<t)h1 rrfi f d HI'i'
V i du 13: Cho 4,86 gam hon hop X gom muoi cacbonat va hidrocacbonat ciia
kim loai kiem M tac dung het voi dung dich H C l (du), sinh ra 1,12 lit khi
(dktc) K i m loai M la
A N a B K C R b D L i
Laigiai: ^^^^
n c o 2 = - ^ = 0 , 0 5 mol Cac phan ling hoa hQc: > >J,QQ •
M2CO3 + 2HC1 > 2MC1 + CO2 + H2O ''^^
MHCO3 + H C l > MCI + CO2 + H2O - A , : , , ^ ^ y i - m
Mian xef:nx=nco2 = 0,05 mol kmqk<^
— 486 ' ' ' '
^ Mx = 7 7 ^ = 97,2 ^ MHCO3 < 97,2 < M2CO3
- > M + 61< 97,2 < 2M + 60 ^ 18,6 < M < 36,2 ^ M la N a ^ Dap an A
V i dy 14: Cho 11,1 gam hon hgp X gom Na, K va Ba vao 200ml H C l I M , thoat
ra 2,8 lit khi H2 (dktc) va dung djch Y Co can Y thu dugc m gam chat ran
khan Gia tri cua m la c
A 27,75 B 19,05 C 17,20 D 18,05 ^
Laigidi:
nH2 = 0,125 Bao toan electron: ne=2nH2 = 0,25 -> nx = ne=0,25
A p dung dinh luat trung hoa di#n: 0,25 = ln^,_ + I n
4 Tac dung vai phi kim
2Mg + O2 ^ 2MgO ^ , 3Mg + N2 ^ MgaNa
+ Tac dung vai nuoc
Cf nhiet do thuong Be khong phan ung voi nuoc, Mg phan ling cham Cac
kim loai con lai k h u manh nuoc tao thanh hidro:
Ca + 2H2O ^ Ca(OH)2 + H2 • r C D i o /
Ba + 2H2O > Ba(OH)2 + H2 ' ^
+ Tac dung vai dung dich axit ' *
Mg + H2SO4 (/oflMg) ^ MgS04 + H 2 t 4Mg + IOHNO3 ^ 4Mg(N03)2 + N 2 0 t + 5H2O , i b q g H : l >
b Hidroxit Tjjnoi
+ Tac dung vai exit axit v inV ,0 <rv-H »Oe«D) ^nun oea tb^li >
CO2 + Ca(OH)2 > CaC03 4 + H2O (rCOfiD) idv KC*
2CO2 + Ca(OH)2 * Ca(HC03)2
+ Tac dung vai hap chat luang tinh
2Al(OH)3 + Ba(OH)2 > Ba(AlC)2)2 + 4H2O
c Canxi cacbonat • ' ' ' ' ' ' H P ' ' T f i O - i f - - - T ^ f ^ r i V v o r i A t ^
+ Tac dung vai dung dich axit K
CaC03 + 2HCl > C a C h + C 0 2 t + 2H2O CaCOs + CO2+ H2O ^ Ca(HC03)2 {Nuac chay da man)
+ Phan ung nhiet phan O.'-v ^ ; :''-: 't!,,?
„ 1200"C , • \ CaCOs ^ C a O + CO2 (Phan ung nung voi)
Tinh chat luang tinh -M + — - iDHS (
Ca(HC03)2 + 2 N a O H > CaCOs + Na2C03 + 2H2O '^S'-^^''
Phan ung phan huy S^"'.yy&"^'
Ca(HC03)2 '"""'"'^ > C a C 0 3 i + CO2 + H2O - ' i •
{Tao thqch nhU a hang dong, can da voi a noi hai, am dun nude) ( ;
Nuoccung H
Khai niem:
Nuoc cung la nuoc c6 chiia nhieu ion kim loai hoa trj 2 nhu Mg^*, Ca^^
Nuoc mem la nuoc khong chua ho^c c6 chua it cac ion tren
211
Trang 12d m nang On luy^n thi dgi hpc 18 chuy6n dg H6a hqc - NguySn V5n Hi\
Phan loai nude cieng
Nuoc cxing tarn thoi: Chua ion goc axit HCO3, khi dun soi se mat tinh
Cling do cac muo'i hidrocacbonat bi phan hiiy: ,
Ca(HC03)2 CaCO.U + CO2 + H2O
Nuoc Cling vTnh cuu: chua ion goc axit CI", SO 4" De lam mem nuoc ciing
vTnh cuu, thuong cho ket tiia cac ion kim loai: j^,, j ^^^
CaCh + NaaCOs ^ CaCOa^ + 2NaCl f
Nuoc Cling toan phan: Co ca tinh ciing tarn thoi va tinh ciing vinh ciiu (c6
chiia ion goc axit HCO 3 , Ch, SO 4")
V I D V M A U
V i du 1: Hop chat nao ciia canxi dugc dung de due tugng, bo bgt khi gay
xuong?
A Thach cao nung (CaS04.H20) C Voi song (CaO)
B Da voi (CaCOs). 5 D Thach cao song (CaS04.2H20)
Laigidi:
Nhan xet: Nhac den thach cao, cac em c6 the loai ngay B va C
De CO the diing due tugng, bo bot khi gay xuong, nguoi ta can nung thach
cao song de tao thanh thach cao nung Thach cao nung moi c6 kha nang hiit
nuoc va dong Cling—> Dap an C
V i du 2: Hoa tan hoan toan hon hgp X gom Be va Mg bang mot lugng vua dii
dung dich HCl C%, thu dugc dung dich Y Nong do phan tram ciia BeCh va
MgCh trong Y Ian lugt la 7,18% va 4,27% Gia trj cua C la
Xet voi 100 gam dung dich Y: , r ^ ; s ) u / l
mBeCl2 = 7,18 gam va m^gciz = 4,27 gam
^ n^HjO (Y) = 100 - 7,18 - 4,27 = 88,55 gam
Nhan xet: Lugng H2O trong Y ciing ehinh la lugng H2O c6 trong dung dich
HCl ban dau
Mat khae, do nong do HCl bang C% nen
m H c i C C 88,55.C
CtyTNHH MTV DWH Khang Vigt
Bao toan nguyen to CI:
Yi du 3: Hoa tan hoan toan 3,84 gam hon hgp gom Fe va kim loai M (hoa tri 2) trong dung dich H2SO4 loang du, thu dugc 2,24 lit khi H2 (dktc) Mat khae, 1,8 gam M tan hoan toan trong 160ml dung dich HCl IM Kim loai M la A.Mg B Ca C Zn
= 22,5 Loai D Dap an A fit'
Vi dii 4: Hai chat dugc dung de lam mem nuoc ciing vTnh cuu la , ||
A Na2C03 va HCl B Na2C03 va Na3P04
C Na2C03va Ca(OH)2 D NaCl va Ca(OH)2
LM giai:
Nhan xet: De lam mem nuoc ciing vTnh eiiu bang phuong phap hoa hgc,
nguoi ta them vao nuoe ciing cac muo'i tan chtia goc axit c6 the ket tiia voi ion kim loai hoa tri 2 (Mg^*, Ca^*, )
Cac goc axit do thuong la goc cabonat, goc photphat -> Dap an B
Vi d^ 5: Cho 2,74 gam Ba vao 100 gam dung dich CuS041,6 %, thu dugc khi X
va ket tua Y Nung Y 6 nhiet do cao den khoi lugng khong doi thu dugc m gam chat ran Gia tri eiia m la
A 2,33 B.3,31 C 5,64 ' ''* D 3,13
Lai giai:
nB.-, = 0,02mol; ncuS04 =0,01 mol "<) l i ' H
Ba + 2H2O -> Ba(OH)2 + H2 CuS04 + Ba(OH)2 > Cu(OH)2i + BaS04i Phan ling nhi^t phan:
BaS04 -» BaS04 {khong bi nhiet phan)
CuO + H2O Cu(OH)2 -
^ = mBaso4 + i " C u O = 0,01.80 + 233.0,01 = 3,13 gam ^ Dap an D
213
Trang 13dm nang 6n luy$n thi dgi hpc 18 chuy6n d6 H6a hpc - Nguyjn Van HSi
Vi du 6: Tron 200ml dung dich X gom NaOH 0,1M va Ba(OH)2 0,1M voi lOOrril
dung dich Y gom H2SO4 0,1M va MgS04 0,2M, thu dugc m gam ket tua Gia
tri cua m la 1
A 5,24 B 5,82 C 4,66 D 6,99
' Lai gidi:
TrongX: n„ ,+= 0^02 mol; n^,+ = 0,02 mol; n^^_ = 0,06 mol
TrongY: n 2+ = 0,02mol; n o_ = 0,03 mol; n^+= 0,02mol , t
Cac phuong trinh phan ling khi pha trpn: ;, , , ^
Mol: 0,02 ^ 0 , 0 4 -> 0,02 v S^VAK (GiC':
Vay: a = 0,02.58 + 0,02.233 = 5,82 gam - > Dap an B M :5Bi:M ;•
Vi dv 7: Hap thu hoan toan 2,688 lit khi CO2 (dktc) vao 2,5 lit dung djch
Ba(OH)2 a mol/1, thu dugc 15,76 gam ket tua Gia tri cua a la
A 0,032 B 0,04 C 0,048 D 0,06
Lai giai:
nco2 =0,12mol; nBaC03 = ^'^^mol." ; , ^ „ , r , - v , , ^
Cach 1: Giai theo phuong trinh
Ba(OH)2 + CO2 ^ BaCOs + H2O
Bao toan nguyen to C: nco2 = ^BaCO^ + 2nBa(HC03)2 nBa(HC03)2 =
Bao toan nguyen to Ba: nBa(0H)2 = "BaCOg + nBa(HC03)2 = 04 +
-> a = 0,04M Dap anB ;())sJj +
cO-V i dii 8: Hoa tan hoan toan 2,42 gam hon hop gom hai muoi cacbonat cua kim
loai kiem M va kim loai kiem tho M ' bang dung dich HCl du Sau phan
ling, thu dugc dung djch X va 0,448 lit khi (dktc) Co can X thu dugc m gam
muoi khan Gia tri cua m la
A 2,20 B.2,64 C 2,30 D 2,34
Cty TMHH M i V DVVI! Khang Vigt
Lai gidi:
j^'COs + 2HC1 > M'Ch + CO2 + H2O i
M2CO3 +.2HC1 > 2MC1 + CO2 + H2O " '••fc.J'j
Vi dy 10: De hoa tan hoan toan 3,2 gam hon hgp X gom kim loai R (thugc
nhom IIA) va oxit cua no can vua du 200ml dung dich HCl I M Kim loai R
A Ba B Be C Mg D Ca ^ •
Lbigidi: 5 ^ ^ , ,
-Cac phuong trinh phan ung: , ,
R + 2HC1 > RCI2 + H2 5^^^!,, ^^^g^ ^i^^^^^ :
RO + 2HC1 > RCI2 + H2O ri'K
^han xet: nx = - nHci = 0,1 mol -» Mx = — = 32 Kit-ai ^
2 0,1
- ^ R < 3 2 < R O - > R < 3 2 < R + 16
^ ~* 16<R<32 -> R = 24(Mg) -> Dap an C
11: Sue 0,448 lit khi CO2 (dktc) vao 100 ml dung djch X g o m Ba(OH)2
O'lM va NaOH 0,1 M Sau khi cac phan ung hoan toan thu dugc m gam ket tua Gia tri cua m la
^•1,970 B 2,364 ' C 0,985 D 1,379 "
215
Trang 14Ca'm nang On luy^n thi dgi hge 18 chuy6n d § H6a hgc - Mytiyeii Vin HJi
+ Tdc dung voi phi kirn i K O f U x o ' ;
4A1 + 3O2 — ^ 2AI2O3 2A1 + 3Cl2 — ^ 2AICI3
+ Tdc dung v&i dung dich ax it nuwjw* • ^ 1 ^
2A1 + 6HC1 > 2AICI3 + 3 H 2 t UiM'~BO
,•.„.„ — f _
A l + 4HN03(loang) > A1(N03)3+ N O t + 2 H 2 O
2A1 + 6 H 2 S 0 4 ( d a c ) > Al2(S04)3 + 3SO21 + 6H2O
Luu y: Nhom bi thu dpng hoa (khong tan) trong dung dich HNO3 dac,
nguQi va H2SO4 dac, ngupi /vne
A l + N a O H + H2O > NaA102 + - H 2 t
2
+ Phan ung nhiet nhom « , " , ; U H : V J '-^ :
O nhift dp cao, nhom khu dugc nhieu oxit kim logi nhu Fe203, Cr203,
thanh kim loai: ^ ^ , ^ ,, *
2A1 + Fe203 — ^ AI2O3 + 2Fe , '-tfj^ftx
+ Sdnxuainhom , ^.nif/jVi H,' ' ^ • \ r
Di$n phan nong chay AI2O3 voi criolit (3NaF.AlF3 hay Na3AlF6) trong binh
di^n phan voi hai di?n eye bang than chi, thu dugc nhom:
2AI2O3 '^""^ > 4A1 + 3O2
Chu f Khong k h u dupe ion AP* trong AI2O3 bang nhung chat k h u thong
+ Tinh chat luang tinh:
M2O3 + 2 N a O H ^ 2NaA102 + H2O J ' t - ^'"'^^ '• " ' '
4A1 + 3O2 — ^ 2 A I 2 O 3 " 2Al(OH)3 — ^ AI2O3 + 3H2O Nhom hidroxit
Tinh chat luang tinh:
Al(OH)3 + N a O H > NaAXOi + 2H2O
+ Phan ung nhiet phan £(|i05:)r;A i 2Al(OH)3 — ^ — A I 2 O 3 + 3H2O +
:,;;.•,>•',.;-+ Dieu che •, • •
T u m u o i A l ( I I I ) '-'"-^^ '''"\^&8;,4;^'n -'
AICI3 + 3NH3 + 3H2O > A l ( O H) 3 ® + 3NH4CI
AlCl3 + 3 N a O H Al(OH)3@ + 3NaCl '*
Luu y: Ket tua A l ( O H ) 3 « se tan dan neu cho N a O H dvr: x :% q^ , g «
Al(OH)3 + N a O H ^ N a A 1 0 2 + 2 H 2 0 , / >• + i?(JiO)!:saf
T u muoi aluminat KC NaA102 + C O 2 + 2 H 2 O ^ Al(OH)3@ + NaHC03.^£ + i(HO)fiS
N a A 1 0 2 + H C l + H 2 O > Al(OH)3© + NaCl ^ f.^"' Ltfw 1/: Ke't tua Al(OH)3^ se tan dan neu cho H C l du:
Al(OH)3 + 3HC1 > AICI3 + 3H2O i = f;
d Phen chua Phen nhom la mot loai muoi kep c6 cong thuc K2S04.Al2(S04)3.24H20 hay Viet gon la KA1(S04)2.12H20 4 r-f*.? • ,X cr':v
Phen chua c6 ung dung lam trong nuoc, cam mau vai sgi, j.; j ,,
V I D V M A U
Vi dy 1: De dieu che dupe 84 gam Fe tu Fe203 (du) bang phuong phap nhi?t nhom v a i hi^u suat cua phan ung la 90% thi khoi lug-ng bpt nhom can dung toithieu la: ' " *
A 81,0 gam B 54,0 gam C 40,5 gam D 45,9 gam
• Lcrigidi: ^ ^ , ;,„
Phuong trinh phan ling: 5,, <,
2A1 + Fe203 — ^ AI2O3 + 2Fe ^rW>>;|lftX'''S,0 =• , i r '
217
Trang 15Cam riaiig On luy§n thi dai hgc 18 chuygn dg Hoa hgc - Nguyin van H5i
„, , , 84 ^ _ , 100 5 ,
Theo phan ung: np^ = — = 1,5 mol n^l = np^ = — mol
- > m A i = 45 gam Dap an D
V i d u 2: Cho 700 ml dung dich Ba(OH)2 0,1M vao V lit dung dich Al2(S04)3
0,1M; sau khi cac phan ung ke't thiic thu dugc 15,54 gam ke't tiia Gia tri cua
WW' ' •
• A.aSO B 0,10 C.0,20 * D 0 , 1 5 / " "
Lcngtat:
Cac phan ung hoa hoc: + ;(}{';/),
•3Ba(OH)2 + Al2(S04)3 > 3BaS04i + 2Al(OH)34 (1) " ^ i " ' ^ " ^ '
Ba(OH)2 + 2Al(OH)3 > Ba(A102)2 + 4H2O (2) " f,(HO)l
Truong hop 1: C h i xay ra phan ung (1) Ba(OH)2 het "Sfo s;
Ket tua gom: BaS04 = 0,07 mol va Al(OH)3 = ^ mol '
' > Khoi lugng ket tua = 0,07.233 + ^ 78 = 19,95 gam Loai j, ^ j
-3 Truong hop 2: Xay ra phan ling (1) va (2) * ^''^ "
3Ba(OH)2 + Al2(S04)3 > 3BaS04l + 2Al(OH)3i (1)
Mol: 3a a 3a 2a :I F - ' o i - T f i
Ba(OH)2 + 2Al(OH)3 > •Ba(A102)2 + 4H2O (2)
Mol: 0,07-3a 0,14-6a 1,1 : , ^,
Ke't tiia gom: BaS04 = 3a mol; Al(OH)3 = 2a - (0,14-6a) = (8a-0,14) mol
Khoi iugng ket tua = 3a.233 + 78(8a - 0,14) = 15,54 gam
- > a = 0,02 mol V = 0,20 lit -> Dap an C
V i d u 3: Cho mot mau hop kim Na-Ba (ti le mol 2:1) tac dung voi nude (du),
thu dugc dung djch X va 4,48 lit H2 (dktc) Cho 300ml dung dich Al2(S04)3
0,2M vao X, sau khi phan ung hoan toan, thu dugc m gam ket tiia Gia tri
p u n g dich X gom: Na* = 0,2 mol; Ba^* = 0,1 mol; O H - = 0,4 mol
fOii tron X voi dung dich gom: AP+ = 0,12 mol; SO 4" = 0,18 mol: ^ Ba2- + S O 4 ' > BaS04i s<,4i T t ; uh ' S^h ,r Mol: 0,1 ' u,n M - > 0,1 R;; ii: <~
AP* + 3 0 H - > Al(OH)3>L Mol: 0,12 - > 0,36 0,12 » r l fioH ^>mi nc
OH Ba A I O 2 '^f fii Trong 100 ml Y : n , = 0,07 mol; n o _ = 0,02 mol; n„, = 0,03 mol
Trang 16Ca'm nang 6n luygn thi dai hpc 18 chuySn 6i H6a hpc - Nguygn Van Hai
V i d\ 5: H o a tan hoan toan m gam hon hop gom N a va A l vao nude, thq
dugc k h i H2 v a dung dich X trong suo't Them tu tu dung dich H C l IJvj
vao X, khi het 100ml thi bat dau xuat hien ket tua; khi het 150ml hoap
350ml thi deu thu dugc a gam ket tua Gia tri cua m va a Ian lugt la
A 7,3 va 3,9 B 7,3va7,8 C 5 , 0 v a 3 , 9 D 5,0 va 7,8
1 Laigidi: '
i^i.-Cac phan ung hoa hoc: J • i j ; <
N a + H2O > N a O H + H 2 ' ' , f j w > l
-A l + N a O H + H2O > Na-A102 + - H 2 ^.(i/Ji
NMn xet: Dung dich X trong suo't —> A l tan het j.'! , ' y
N a O H + H C l > N a C l + H2O , , /
Mol: 0,1 < - 0,1
NaA102 + H C l + H2O > A l ( O H ) 3 i + N a C l
Mol: 0,05 0,05
- > a = 0,05.78 = 3,9 gam - > Loai phuang an B va D >
Nhqn thdy: K h i cho 350ml dung dich H C l vao X, toan bp NaA102 se chuyen
he't thanh ke't tua Al(OH)3, va sau do bi hoa tan mot phan trong H C l :
NaA102 + H C l + H2O > A l ( O H ) 3 i + NaCI
V i du 6: C h i dung dung dich N a O H de phan bi^t dugc cac chat rieng bif t trong
nhom nao sau day? ' 1, i
A L i , Na,Be B AI2O3, A l , Mg C Zn, A l , Na D Fe, C u , Ag
Laigidi: A , f
Loai A: L i , Na, Be deu tac dung v6i H2O trong dung dich N a O H va sui bpt
khi
Loai C : Zn, A l , N a deu tan trong dung dich N a O H va sui bgt khi
Loai D: F3, C u , A g deu khong tac dung voi dung djch N a O H
- > Dap an B Hien tuong nhu sau:
AhOa tan: AI2O3 + 2 N a O H > 2NaA102 + H2O
220
Cty TNHH MTV DVVH Khang Vi§t
Al tan va sui bgt khi: A l + N a O H + H2O -> NaA102 + - H 2 t ]y[g khong phan ung
Vi d|i 7: Nho tu tit den d u dung dich N a O H vao dung dich AICI3 Hien tugng xay ra la
A Co ket tua keo trang va c6 khi bay len. m
B Co ket tiia keo trang, sau do ket tua tan
C Chi CO ket tua keo trang •!" i H 1 ,4 * ' M 1 > '
D Khong CO ke't tiia, c6 khi bay len
Lai gidi:
AICI3 + 3 N a O H > A l ( O H ) 3 i {keo trang) + 3NaCl '
Al{OH)3 + N a O H ( d M ' ) > NaA102 + 2H2O • i tCtet^: + f A
-> D a p an B \'-''f.H
Vi dv 8: Cho hon hgp gom N a va A l c6 ti 1| so mol tuang ung la 1 : 2 vao nuoc (du) Sau khi cac phan ung xay ra hoan toan, thu dugc 8,96 lit khi H2 (dktc) va m gam chat ran khong tan Gia tri cua m la
A 10,8 ^ B.5,4 C.7,8 D 43,2 , ,
Led gidi:
Cac phan ung hoa hgc: * \ '
N a + H 2 O — > N a O H + 1H2 ,' „ ^ui.<imt Mol: a - > a b,5a
Vi dv 9: Dot chay hoan toan 8,7 gam hon hgp Mg, A l va Z n trong khi O2 (du) thu dugc 15,1 gam hon hgp oxit The tich khi O2 (dktc) da tham gia phan
Trang 17ca'm nang 6n luyQn Ihi dai hoc 18 chuySn dS H6a hpc - Nguyin Van H5i
Vi d\ 10: H 6 n h o p X g o m a m o l Na va b m o l A l Cho X vao m o t l u o n g
nuac t h i thoat ra V l i t khi M a t khac, neu cho X vao d u n g dich N a O H (d^.^
thi duoc 1,75V l i t k h i (cac the tich do 6 cimg dieu kien nhiet do, ap sua't^
M o i quan he giua a va b la
A a = 2b B 3a = b C a = b D 2a = b
+ K h i cho X tac d u n g v o i nude (du): ^,, , , ^ • ^.^ ^^^^^ ,
N a + H2O > N a O H + - H 2 g t i k f o a i fiu) B>l <o i
n H = 2a mol * o lA f / • ' ,\ry '-»: ojr;
+ K h i cho X tac d u n g d u n g dich N a O H (du): ' ' """^ '"
Luu y: N a tan he't do tac d u n g v o i nuoc trong d u n g dich, dong thoi A l ciing
tan he't trong kiem d u (theo hai phan u n g n h u viet a tren) ,v • : f
1 3
Theo bai ta c6: ^ = — = - > 2a = b —> D a p an D
0,5a + 1,5b 1,75V 1,75
V i d\ 11: Cho m g a m h o n h g p X g o m A l , C u vao d u n g dich H C l (du), sinh ra
3,36 l i t k h i (dktc) N e u cho m gam X vao m o t l u g n g d u axit HNO3 (dac,
ngupi), t h u dugc 6,72 l i t k h i NO2 (san pham k h u d u y nhat, a dktc) Gia trj
ciia m la
A 11,5 B laS C 12,3 fti |.,Ci D 15,6
, , Lotgtat:
Nhan xet:
+ K h i cho X tac d u n g v o i axit H C l (du) - > C h i c6 A l phan u n g (do C u dung
sau H trong day dien hoa): ^ ^ ,
A l + 3HC1 > A I C I 3 + f H 2 , , p , y
M o l : 0,1 • 0,15 '
+ K h i cho X tac d u n g v o i axit HNO3 (dac, nguoi) - > Chi c6 C u phan i m g i^'^
A l b i thu dpng, khong tan):
Cu + 4 H N 0 3 > Cu(Na)2 + 2NO2 + 2H2O
M o l : 0,15 < - 0,3
- > m = ai.27 + ai5.64 = 12,3 gam - > Dap an C " '^^ ''^
Cty TNHH MTV DWH Khang Vigt
V i dv 12: Cho 1,62 g a m A l tac d u n g v o i d u n g dich HNO3 (loang, d u ) Sau k h i
phan l i n g xay ra hoan toan t h u dugc 0,224 lit k h i N2 (dktc) va d u n g dich Y chua m g a m m u o i Gia t r i ciia m la
Chat oxi hoa: 2N*^ + lOe > N2; 2N*5 + Se > NH4NO3
Bao toan electron: 3 n ^ i = 10 n + 8 nNH^NOi
3.0,06-10.0,01 , ^ , , • •
-> nNH4N03 ^ =0,01 mol Ei,n'f h ' u b / w r *i4 v.:' =
m= mAi(iM03)3 "^NH4N03 ^ 0,06.213 + 0,01.80 = 13,58 gam -)• Dap an C
4 B A I T A P 6 N L U Y E N Bai 1 : Cho cac chat: NaHCOs, CO, Al(OH)3, Fe(OH)3, H F , CI2, N H 4 C I So chat
tac d u n g dugc v o i d u n g djch N a O H loang 6 dieu kien t h u a n g la
A 4 B 5 C 3 D 6 ^
Bai 2: H a p t h y hoan toan 2,24 lit k h i CO2 (dktc) vao 100ml d u n g dich X g o m
N a O H 2 M va K O H 0,1M thu dugc d u n g dich Y Co can Y t h u dugc a gam chat ran khan Gia t r i cua a la u , , ,
Bai 4: H o a tan hoan toan 1,794 gam k i m loai kiem M vao 400ml d u n g dich H C l
0,1M Co can d u n g dich sau phan u n g thu dugc 3,316 chat ran khan Y K i m loai kiem M la
A Na B K C Rb D L i
Bai 5: H a p th\ hoan toan 0,672 l i t k h i CO2 (dktc) vao 1 l i t d u n g dich X g o m
N a O H 0,03M va Ca(OH)2 0,01 M , thu dugc m g a m ket tiia Gia t r i cua m la
A 2,00 B 1,00 C.1,25 D 0,75
Bai 6: M o t coc nuac c6 chiia cac ion: Na* (0,02 mol), Mg^* (0,02 mol), Ca^* (0,04
mol), C I - (0,02 mol), S04~' (0,01 m o l ) va H C O J (x mol) D u n soi coc nuac tren cho den k h i cac phan u n g xay ra hoan toan t h i nuoc con lai trong coc
A La nuoc m e m B Co tinh cung vTnh cuu
C Co tinh cung toan phan D Co tinh cung t ^ thai
223
Trang 18cam nang on luy^ii thi dai lipc 18 chuyen de Hoa hgc - Nguyjn Van Hki
Bai 7: Mot coc dung dich chua mot lugng nhieu cac ion: Ca^*, Mg^*, HCO3
C\-, SO l" Nude trong coc tren thuoc loai gi?
A Nuoc mem B Nuoc cung vTnh cuu
C Nuoc C l i n g tam thoi D Nuoc cxing toan phan
Bai 8J Tien hanh phan ung nhiet nhom hon hop gom A l va 16 gam Fe203 de'n
khi phan ung hoan toan, thu dupe hon hop ran X Cho X tac dung vua dii
voi V lit dung djch NaOH I M , sinh ra 3,36 lit H2 (dktc) Gia tri cua V la
A. 0,30 B.0,10 C.0,20 D 0,15
Bai 9: Cho khi C O (du) di vao ong su nung nong dung hon hop X gom
AI2O3, Fe304, CuO thu dupe chat ran Y Cho Y vao dung dich NaOH (du),
khuay ki, tha'y eon lai phan khong tan Z Phan khong tan Z gom
A AI2O3, Fe, Cu B Fe304,Cu C Fe, Cu D A l , Fe, Cu
Bai 10: Day gom cae oxit deu bi A l khu 6 nhi^t dp cao la:
A FeO, CuO, Cr203 B PbO, K2O, SnO J ' I A T J
C FeO, MgO, CuO D Fe304, SnO, BaO
Bai 11: Nhi^t nhom hon hpp gom a gam bpt A l va b gam Fe203, thu dupe hon
hop X Hoa tan X trong H N O 3 du, thu dupe 2,24 lit khi NO (dktc) Gia tri
fr cua a la: i'-nOi'' ^ •''a ^/'rrf^oi ff«;,n,-ii rv!+nf;! j
A 2,70 w n i V f f i - B 5,40 C 1,35 D 8 , 1 0 i G :
Bai 12: Tien hanh phan ung nhif t nhom voi m gam hon hpp X gom A l va
Fe304 den phan ung hoan toan, thu dupe eha't ran Y Cho Y tac dung vai
i dung dich NaOH (du) thu dupe dung dich Z, phan khong tan va.3,36 lit khi
0 1 H2 (dktc) Sue khi CO2 (du) vao Z, thu dupe 39 gam ket tua Gia tri ciia m la
A 45,6 B.48,3 C 36,7 D 57,0
Bai 13: Tien hanh phan ung nhif t nhom vai 24,1 gam hon hop gom A l va
= ' Fe203 den phan ung hoan toan, thu dupe chat ran X Chia X thanh hai phan
bangnhau
+ Phan mpt phan ung vira du vai 150ml dung dich NaOH I M
+ De hoa tan het phan hai can vua du dung dich chua a mol HCl
Gia tri cua a la
A. 0,40 > B 0,65 C 0,35 D 0,55
Bai 14: Hoa tan hoan toan 0,3 mol hon hpp gom A l va AI4C3 vao dung dich
KOH (du), thu dupe a mol hon hpp khi va dung dich X Sue khi CO2 (dif)
! 1 vao X, lupng ket tua thu dupe la 46,8 gam Gia tri eua a la
A 0,55 B.0,60 C.0,40 D 0,45
Bai 15: Hoa tan hoan toan m gam hon hpp X gom Na20 va AI2O3 vao H 2 O thi'
dupe 200ml dung dich Y chi chua chat tan duy nhat c6 nong dp 0,5M Tho'
Cty TNHH MTV DWH Khang Vl^t
C O 2 (du) vao Y thu duac a gam ket tua /
Gia tri cua m va a Ian lupt la , • dnhi i"r •••i
A 8 ,3 va 7,2 B 11,3 va 7,8 C 13,3 va 3,9 D 8,2 va 7,8
pai 1^* ^''^ ^^"^ 400ml dung dich HCl 2M, thu
(Jupc dung djch X Cho tung gipt den he't V lit dung djch NaOH 2M vao X,
thu dupe 7,8 gam ket tiia Gia trj Ian nhat cua V la '
A 0,45 B.a35 0 ^ C 0,25 J ; = D 0,05 '
Bai 17: Hoa tan hoan toan 3,78 gam A l bang dung dich H N O 3 loang (du), thu dupe dung dich X va 0,448 lit (dktc) hon hpp khi Y gom N2O va N2 Ti khoi
ctia Y so vai hidro la 18 Co can dung dich X, thu dupe m gam chat tin khan
Gia tri cua m la
A 29,82 B 32,22 C 31,42 D 33,02
5 H I / C ! K N G D A N - L 6 I G I A I ^ ij/t ' - •
B i i l : NaHCOs + NaOH ^ Na2C03 + H 2 O
Al(OH)3 + NaOH > NaAlCh + 2H2O '
Ch + 2 N a O H > NaCl + NaClO + H2O ,., r
N H 4 C I + NaOH > NaCl + N H 3 + H 2 O , f,
-HF + NaOH > NaF + H 2 O ' H0^| ,
~> Dap anB .{, >fn l O O - j ' BAi2: • • ,
^'•'f) Tie-"
• -Kt;
" O H - " " N a O H + nKOH= 0'2 + 0,1 = 0,3 mol n^, + = 0,2 mol; n^+ = 0,1 mol
el: ^!^2H1 = M = 3 > 2 ~^ OH- du, phan ung chi ra tao muoi
nroo 0,1
Trong X:
Nhqn xel:
"CO2 cacbonat. o -r -! ro n - « * Pfvn «Jf • PfiO » 4 + B :<>
Phuang trinh ion: C O 2 + 20H- ^ C O 3 " + H 2 O gfj'y r
Mol: 0,1 0,2 0,1
V gom: Na* = 0,2 mol; K* = 0,1 mol; O H " = 0,1 mol va CO 3" = 0,1 mol
Bao toan khoi lupng: a = m^^+ + + m^^ + m^^2- ^n-/' '
S-^t^ = 0,2.23+ ai.39 + ai.l7 +0,1.60 = 16,2 gam /
->DapanC *
^ a i 3 : nc„= — = 0,05 mol .nartko ^ 'Mr^'l ^
64 ry:, ^
225
Trang 19C^m nang 6n luygn thi dji hqc 18 chuy6n ii H6a hqc - Nguygn VSn H&\
Nhdn xet: D u n g d i c h X c h i i a m u o i K N O 3 va H2SO4 l o a n g - > Can giai theo
Nhqn xet: Bai nay neu cac e m v i e t p h u o n g t r i n h p h a n t u se gap n h i e u kho
k h a n D o vay, n e n giai d u a theo cac p h u o n g t r i n h i o n
Cty TNHH MTV DVVH Khang Vigt
So' m o l di^n t i c h d u o n g = So' m o l d i e n tich a m
1.0,02 + 2.0,02 + 2.0,04 = 1.0,02 + 2.0,01 + 1.x ^ x = 0,1 i;> i/\
Den day, cac em luU y: Khi dun nong, muoi hidrocacbonat cua k i m loai hoa
t r i 2 b i phan huy:
Mg2^ + 2 H C O ; > MgCOs^ + CO2 + H2O ' '
dun nong , " f V>ri ? A i
Ca^- + 2HCO3 CaCOsi + CO2 + H2O ' ' '
N h a n thay: n^^2+ + \^2+ = 0,06 Mg^ Ca^* con du, H C O 3 he't ' ' '
N u o c t r o n g co'c sau k h i d u n c6 chua cac i o n : Mg^^ Ca^*, C h v a SO 4 "
-> ThuQC lo^i n u o c c u n g v i n h c u u - > D a p an D
Bai 7: ' ' ^"''"^'''''' ' ^
Nhqn xet: D61 chieu v o i d i n h nghia, m a u nuoc trong co'c c6 chua ca tinh
Cling tarn t h a i (chua H C O 3 ) va c6 ca fa'nh cung vTng c u u (chua CI", SO^")
-> ThuQC loqii n u o c c u n g toan p h a n - » D a p an D ,
B a i 8 : ''
nFe203 = ^ = 0 , l m o l ^ npe = 0 , 2 m o l
Nhdn xet: K h i cho X + d u n g d i c h N a O H —> Sinh ra H2 nen t r o n g p h a n u n g
nhiet n h o m t h i A l con d u va FezOa het ' " ' " "' ' '
2A1 + Fe203 > 2Fe + AI2O3 ^
3
A l + N a O H + H2O > NaA102 + - H 2 dAm mmi^finml <::'.::
^N.OH= | n H 2 + 2nAi203 = ^ + 2.ai = 0 , 3 m o l V = 0,31it ^ ,
- > D a p ^ n A „ ^^p i^f^
Bai 9:
Luu y: K h i C O k h o n g k h u d u g c AI2O3 nen k h i cho k h i C O d u tac d u n g v o i
X, xay ra 2 phan u n g : Fe304 + 4 C O — ^ 3Fe + 4C02 " ^ • r f g l ^ ( v , , , ^ :
- > Chalt An Y g o m : Fe, C u , AI2O3 ' ' ' * /
K h i cho Y tac d u n g v o i N a O H d u t h i AI2O3 b i hoa tan hoan toan: f AI2O3 + 2 N a O H > 2NaA102 + H2O
Vay Z g o m : Fe, C u D a p an C ^
227
Trang 20Ca'm nang On luy?n thi djii hpc 18 chuyfin dg H6a hpc - Nguygn Van Hii
Bai 10:
Luu y: Al chi khu dug-c cac oxit kim loai dung sau Al trong day di§n hoa
, -> Loai B, C, D vi Al khong khu dugc K2O, MgO va BaO
—> Dap an A .,
Cac phan ung hoa hoc: ^.^^^j^ ^ ; e ' L ^ , ^ ^ 1 ,
2A1 + 3?eO — ^ AI2O3 + 3Fe,^^.,^^.^;, s *, ^
Bai 11: ^
Cac qua trinh hoa hoc xay ra theo so do sauf'^J''' f '
J Al, FejOj — ^ X ) Fe(N03 )3 + AKNOg )3 + NO
N/zfln xet: So oxi hoa ciia Fe khong thay doi a trang thai dau va cuoi (deu la
Fe*^), con so'oxi hoa ciia Al tang tir 0 len +3
Bao toan electron: 3nAi = 3 n N o —>• nAi = nwo = 0,1 mol
a = 0,1.27 = 2,70 gam -> Dap an A '
Bai 12:
Nhqn xet: Bai nay ra't nhieu thi nghiem nen neu cac em giai theo phuong
trinh phan ung se mat nhieu thai gian j^^,
So do duong di ciia Al trong cac qua trinh hoa hpc nhu sau:
^Fe304,t° ^ ^j^o^ ^ J^^^JQ^ ^ ^ ^ ^ ^ ^ ^ ^
39 Bao toan nguyen to Al: n^i (x) = nAi(OH)3 mol t ri*^ij4:'>,
N/zfln xet: Khi cho phan 1 + dung dich NaOH, tat ca Al va AI2O3 deu tac
dung va chuyen thanh NaA102
Do vay, bao toan nguyen to Al, ta c6: nAi= nj^jaOH = O'^S mol
^ Ban dau: n^r 2.0,15 = 0,3 mol -> nFe203= " ^ ^ ^ ^ ^ ^ ^ ^
9 9 S
Cty 1NHH MTV DVVH Kliang Vi$t
Fe203 + 2A1 — ^ AI2O3 + 2Fe Mol: 0,1 -> 0,2 0,1 0,2 -> Phan 2 gom: n^i = 0,05 mol; nFe= 0,1 mol; n^ijOa = 0/05 mol
- > HHCI = 3nAi + Znpe + 6nAi203 = 0,65 mol • ' i '
Dap an B. > r r r 1- w
-Bai 14:
So do phan ung: ^
Al, A I 4 C 3 KAIO2 ^ ^ ° ? ^ " 2 0 > Al(OH)3 '
Bao toan nguyen to Al: n^, + 4nAi4C3 = nAi(OH)3 ^ "Al + 4nAi4C3 = OA Theo bai: nAi +nAi4C3 = 0,3 -> nAi =0,2; nAi4C3 = 0,l ^
3 '
-V$y: a = - nAi + 3nAI4C3 "0,6 ,
-> Dap an B
Bai 15:
rO,0,8^-M -Nhan xet: Khi cho X vao nuoc, Na20 se tac dung ngay vai nu6c:
AI2O3 CO tinh luong tinh nen tan trong dung dich NaOH vvra tao ra
Al + NaOH + H2O > NaA102 + - H 2 t
2
Mol: 0,1 0,1 0,1
T h 6 i k h i C 0 2 : NaA102 + CO2 + 2H2O > Al(OH)3 +NaHC03
Mol: 0,1 0,1 -> a = mAi(OH)3 = 0,1.78 = 7,8 gam; ,
va m = mNa20 + mAI2O3 = 0,05.62 + 0,05.102 = 8,2 gam
-> Dap an D
Bai 16: >
nAi = 0,2 mol; nnci = 0,8 mol; nAi(OH)3 = 0,10 mol
Dung dich X gom: AlCh = 0,2 mol; HCl = a2 mol ^ ; 1, ,5
£ach 1: Giai theo phuang trinh hoa hpc:
NaOH + HCl > NaCl + H2O , ;v,: , ;
AICI3 + 3 N a O H > Al(OH)3i + 3NaCl ^
Al(OH)3 + NaOH > NaA102 + H2O
-> n,^^ = 0,20+ 0,60+ 0,10 = 0,90 mol OH -> nNaOH= nQ^ V = 0,45 lit Dap an A = 0,9moI
Trang 21dm nang On luySn thi dgi hpc 18 chuyfin dg H6a hpc - Nguyjn Van Hi\
Cach 2:
Tinh nhanh: riNgOH max = 4n^,3+" "A1(OH)3 = 4.0,20 - 0,10 = 0,70 mol
nNaOH=nHCl +nNaOHmax =0/2 + 0,7 = 0,9 mol ^ V = 0,45lit , ,
Chat oxi hoa: 2 N ^ + 8e -> N2O 2N*5 + lOe N2
va CO the xay ra qua trrnh: 2N*5 + 8e > NH4NC)3(amol) ' P ; V
Bao toan electron: 3 n = 8 n ^ j o + 10nfj2 + ^ nNH4N03 "
-> 3.0,14 = 8.0,01 +10.0,01 + 8a a = 0,03 mol
V^y: m = mAi(N03)3 + mNH4N03 = ai4.213 + a03.80 = 32,22 gam
Dap an B fiO«jl4£ 4—- O t H + Oa>\
Lim y: A l c6 the tac dung voi axit HNO3 tao thanh muoi NH4NO3! •
+ lac dung vm phi Kim
3Fe + 2O2 — ^ Fe304 2Fe + S C k — ^ 2FeCl3 ^ ,
Fe + S '° > FeS -•'''^••'••-••(•'./•n.J' ^ n:iA<i !t;^vv
+ Tac dung vai dung dich axit t'^-<M
Sat k h u ion trong dung dich H C l va H2SO4 loang thanh khi hidro:
Fe + H2SO4(/0flMg) > FeS04 + H 2 t v -vmPXint
Fe + 4HN03iloang) > Fe(N03)3+ N O t +2H2O ,|- -,\Q,.y:
NeuFedi^: Fe + 2Fe(N03)3 > 3Fe(N03)2 v«s\a J\TU 2Fe + 6H2S04(ifflc) — ^ Fe2(S04 )3+ 3 S 0 2 t + 6H2O
New Fe du: Fe + Fe2(S04)3 > 3FeS04
b San xuat gang ' U A M y G '
Manhetit: Fe304 Xiderit: FeCOs Hbw;, Hematit do: Fe203 Hematit nau: Fe203.nH20
+ Nguyen tac san xuat
K h u oxit sat bang C O 6 nhif t dp cao (phuong phap nhiet luy?n) Trong 16 cao, sat c6 so' oxi hoa cao bi khu dan xuong so' oxi hoa thap theo so do:
Fe203 Fe304 ) FeO ) Fe • ''•-^
c Hgrpchatsitdl)
Oxit:3FeO+ IOHNO3 > 3Fe(N03)3 + N O + 5H2O m orD ;S ^ i ' Hidroxit: 4Fe(OH)2 + O2 + 2H2O > 4Fe(OH)3 m ( u b ;^noii x Mu6i:2FeS04 + 2H2SO4 (dac) > Fe2(S04)3 + SO2 + 2H2O (Ff ^
3Fe2^ + NO3 + 4H* > 3Fe3^ + N O + 2H2O b ; 10FeSO4 + 2KMn04 + 8H2SO4 > 5Fe2(S04)3 + K2SO4 + 2MnS04 + 8H2O FeS2 +I8HNO3 > Fe(N03)3 + H2SO4 + ISNCh + 7H2O
4FeS2 + I I O 2 — ^ 2Fe203 + 8SO2
Tinh oxi hoa
PeO + H2 — ^ 2Fe + 2 C 0 i ' , s ^
Trang 22Vi";.'
Ca'm nang On luygn thi dgi hgc 18 chuy6n dg H6a hpc - Nguyen van Hai
c Hgp chat Fe(III)
+ Tinhbaza rfV.«»? v-i^ ^k.nn.f •
Fe203 + 6HNO3 > 2Fe(N03)3 + 3H2O
+ Tinh oxi hoa
Fe + 2Fe(N03)3 > 3Fe(N03)2
Cu + 2Fe(N03)3 > 2Fe(N03)2 + Cu(N03)2
d Oxitsattir ^
+ Tinhbaza < ~
-Fe304 + 8HC1 > FeCh + 2FeCl3 + 4H2O ''^ '^''^ « ' '
Fe304 + 4H2SO4(/oan^) > FeS04 + Fe2(S04)3 + 4H2O '
+ Tinhkhie- :, „ '.X.':/ V *
Fe304 + IOHNO3 ^ 3Fe(N03)3 + NO2+ 5H2O '
+ Tinh oxi hod
Cty T M H H MfV DVVil Khang Vi§t
Fe304 + 4CO -> 3Fe + 4CO2
V I D U M A U
V i du 1: Hoa tan hoan toan 3,16 gam hon hgp gom Fe va Z n vao m p t lupng
vua d u d u n g dich H2SO4 loang, sau phan u n g thu dupe 1,12 h't H2 (dktc) va
dung dich X chiia m gam muoi Gia tri cua m la i:*, •.si :of • J U K I
A 4,83 gam B 5,83 gam C 7,96 D 7,23 gam
Lai gidi:
6 bai nay, cac em c6 the giai chi tiet dya theo phan l i n g hoa hpc ,;
Nhan xet: 11^2504 "H2 ^'^^
So do phan ung: K i m l o a i + H2SO4 > Muoisunfat + H2 "
Bao toan khoi luong: ^0 i^fb
-> m = 3,16 + 0,05.98 - 0,05.2 = 7,96 g a m - > Dap an C .Mii''
V i d\ 2: Cho m g a m hon hpp X gom Fe, FeS2 va FeS tac d u n g he't v o i HN03
(dac nong, d u ) t h u dupe V lit k h i chi c6 NO2 (6 dktc, san p h a m k h u du)
nhat) va d u n g dich Y Cho toan bp Y vao mpt l u p n g d u d u n g dich BaCl2'
thu dupe 46,6 gam ket tua; con k h i cho toan bp Y tac d y n g v o i d u n g dich
N H 3 d u t h u dupe 21,4 gam ket tiia Gia trj cua m va V Ian l u p t la
,2-46,6
BaotoannguyentoS: ns(x)=nBaS04 = - ^ = 0 ' 2 m o l , ,
Khi cho Y + dung dich NH3 d u : , i / < , / v i !
Fe3* + 3NH3 + 3H2O > Fe(OH)3 i • • : •
Bao toan nguyen to Fe: npe (X)= nFe(OH)3 = 0'2 m o l " ''
Bao toan khoi lupng: m = mp^ + mg = 0,2.56 + 0,2.32 = 17,6 gam
Qui doi X ve hon hpp cac don cha't: Fe = 0,2 mol; S = 0,2 mol »„ j (X) = 3npe + 6ns = 1'8 m o l -> nN02 = "e (X) = 1/7 m o l _
V = 1,8.22,4 = 40,32 lit -> Dap an C
V i dv 3: Trong 16 cao, sat c6 so'oxi hoa cao bi k h u dan xuo'ng so'oxi hoa thap
theo ba giai doan:
Fe203 — ^ Fe304 — ^ FeO Fe r- v,| ^^.^
Giai do^n (3) dien ra 0 khoang n h i f t dp nao? <• ?
A.300-4000C B 500-6000C C 700-8000C D 900-1000"C
Lai gidi:
- > Dap an C
V i dy 4: D a n luong k h i CO d i qua hon hpp gom C u O va Fe203 nung
nong, sau m p t thai gian thu dupe chat rSn X va k h i Y Cho Y hap thy hoan toan vao d u n g dich Ca(OH)2 d u , thu dupe 3 gam ket tua Chat ran
X phan u n g v o l d u n g dich H N O 3 d u thu dupe V l i t k h i N O (san pham
k h u d u y nhat, a dktc) Gia trj cua V la
Nhan xet: K h i cho hon hpp oxit tac dung voi khi CO thi:
ng traodoi = 2nco • M a n ^ o = nco2 " ^'^3 mol -> n^ traodoi = 0,06 mol >
Bao toan electron: n^ traodoi = 3 n ^ o " N O =0,02mol ' ' VNO =a4481it , ~.vn^
-> Dap an B
V i du 5: Cho 0,56 gam Fe tac d u n g v o i 400ml d u n g dich A g N 0 3 0,1M, khuay
k i de phan u n g xay ra hoan toan So gam A g thu dupe sau phan u n g 1^
A 4,32 B.3,24 C 1,08 D 2,16
233
Trang 23Ca'm nang 6n luyjn thi dai hpc 18 chuygn d l Hoa hgc - Nguygn VSn Hii
Lcngidi:
npe =0,01 mol; nAgNO3 = 0'04 mol „.,,' ' ^ ' - - ' - f O
Fe + 2AgN03 > Fe(N03)2 + 2Ag 1 f N f c V oM,
Mol: aOl - > 0,02 - > 0,01 • (MX dl/li: •
LwM y: AgNOa CO phan ling ke tiep vdi Fe(N03)2: •f^i^':!-^^! f^ii: •^•ft^.••••^
Fe(N03)2 + AgN03 > Fe(N03)3 + Ag
M o l : 0,01 - > 0,01 - > ''^.^r^'' 0,01 ' «if^'^* ^''firaf
^ mAg= 0,03.108 = 3,24 gam Dap an B ih':mqc/dimdsvX loli.;
V i d v 6: Cho day cac chat sau: KBr, S, Si02, FeO, Cu va FezOa So'chattrong day
CO the bi oxi hoa boi dung dich axit H2SO4 (dac, nong) la
A 4 B.5 C 3 D 2 '
Lai gidi:
Nhan xet: Axit H2SO4 dac, nong the hien tinh oxi hoa khi tac dung voi chat
CO tinh khu (chua nguyen to dang 6 muc oxi hoa thap)
2KBr + 2H2SO4 2KHS04 + Br2 + SO2 + H 2 O '' *
S + 2H2SO4 — ^ 3 S 0 2 + 2 H 2 O
8 2FeO + 4H2SO4 — ^ Fe2(S04)3 + SO2 + 4H2O
Cu + 2H2SO4 — ^ CuS04 + SO2 + 2 H 2 O
ITS _ ^ £)^p ^ j ^ ^
*' Luu y: VeiOs the hien tinh bazo khi tac dung voi H2SO4 dac:
Fe203 + 3 H 2 S O 4 Fe2(S04)3 + 3H2O
V i dv 7: Hoa tan hoan toan 12,8 g a m hon hgp gom Fe304 va FeS2 trong dung
dich axit HNO3 (dac, du), thu dugc 4,48 lit khi NO2 (dktc) va dung dich X
Cho X tac dung v o i dung dich Ba(OH)2 du, IQC ket tua va nung trong khong
khi den khoi lugng khong doi thu dugc m gam chat ran Gia tri cua m la
Ggi so mol: Fe304 (a mol) va FeS2 (b mol) Ta c6:232a + 120b = 12,8
Cac phan ung khu:
F e 3 0 4 - l e > 3Fe*3 /
FeS2-15e > Fe"3 + 2S^
Bao toan electron:
"NOz = '^Fe304 +15 npeSj -> a + 15b =' 0,2 a = 0,05; b = 0,01
234
Cty TIMHH MTV DWH Khang Vi$t
gao toan nguyen to'Fe va S:
_^ m = 160.0,08 + 0,02.233 = 17,46 ^ Dap an C
y i dv 8: Hoa tan hoan toan hon hgp m gam gom Fe va Fe304 bang dung dich
H 2 S O 4 loang d u thu dugc 896ml khi H2 (dktc) va dung dich Y Dung dich Y lam mat mau vua d i i 1,58 gam KMn04 Gia tri m la:
Vi d\ 9: Cho 19,3 gam hon hgp bgt gom Z n va Cu c6 t i le mol tuong ung la 1:2
vao dung dich chua 0,2 mol Fe2(S04)3 Sau khi cac phan ung xay ra hoan
toan, thu dugc m gam k i m loai Gia tri cua m la
A 12,80 B 12,00 C.6,40 D 16,53
Lai gidi: "
Ggi so mol: Z n = a; Cu = 2a Theo bai: 65a + 64.2a = 19,3 a = 0,1
Cac phan ung hoa hgc:
d^ 10: Hoa tan hoan toan 8,24 gam hon hgp bgt Fe304 va Cu trong 120 ml
dung dich H2SO4 I M (loang) Sau khi cac phan ung xay ra hoan toan, thu
dugc dung djch X (khong chua axit du) Co can X thu dugc m gam muoi khan Gia tri cua m la
A 9,28 B 21,28 C 17,84 D 13,28 •
235
Trang 24dm nang 6n luyjn thi dji hpc 18 chuy6n dg H6a hpc - Nguygn Van HSi
Laigidi: -Si^'
G p i so m o l : Fe304 = a; C u = b Theo bai: 232a + 64b = 8,24
NMn xet: C h i c6 Fe304 phan u n g true tiep v o i axit
Trong X k h o n g con axit d u nen Fe304 phan l i n g vira d u v o i H2SO4:
Cac p h a n l i n g hoa hoc:
Fe304 + 4H2SO4 > FeS04 + Fe2(S04)3 + 2H2O
V i dv 11: Cho 6,72 gam Fe vao d u n g dich chiia 0,3 m o l H2SO4 dac, nong Sau
k h i cac phan l i n g xay ra hoan toan, t h u d u g c d u n g djch X va k h i SO2 (san
p h a m k h u d u y nhat cua S**) Co can d u n g dich X t h u d u g c m gam muoi
khan Gia t r i ciia m la
V i dy 12: Cho cac cgp chat v o l t i 1? m o l t u a n g u n g n h u sau: Fe304 va C u {V.I)',
Sn va Z n (2:1); Z n va C u (1:1); Fe2(S04)3 va C u (1:1); So cap chat tan hoan
toan t r o n g m g t luqrng d u d u n g djch H C l loang nong la
Cu + Fe2(S04)3 > CuS04 + 2FeS04 :Ht -Ac/i >X i-.-^.H f •/ hA-iJj ^•^\y'J •
^ 2 n va C u (ti le 1:1): chi c6 Z n tan, C u khong tan JWJ t q >r ; i n Dap an A
Vi dv 13= ^'22 gam FexOy bang d u n g dich H2SO4 dac, nong (du),
thu d u o c d u n g dich X va 0,812 lit k h i SO2 (san p h a m k h u d u y nhat, 6 dktc)
Co can X, t h u dug^c m gam m u o i sunfat khan Gia t r i cua m la
A 14,50 B 29,00 C 21,75 D 17,40
, ,,,, Laigidi: Nhan xet: O x i t sat FexOy tac d u n g voi H2SO4 dac, nong —>• SO2 t h i oxit la
FeO hoac Fe304
Cac em lu'u y rang 1 m o l FeO hoac Fe304 deu chua 1 m o l Fe*^ nen deu c6 kha nang n h u o n g 1 m o l electron, do do:
He = riPexOy = 2nso2 '^Ve^Oy = ^'^^f^ ^'^^^^ ' * "
(3) Cho Fe vao d u n g djch Fe2(S04)3;
(4) Cho Fe vao d u n g djch H2SO4 (bang, du); , (5) D o t nong h o n hgp bpt Fe va S (khong c6 oxi) ,A f is qfcO «
Co bao nhieu t h i n g h i ^ m t?o ra m u o i Fe (II)? HI " " ^ ^ ' ' A'/ M O M ' ^
Cac t h i nghi?m (3); (4) va (5) t^io ra m u o i Fe (II)
D a p an A
237
Trang 25dm nang 6n luyQn thi d<ii hpc 18 chuy6n dg H6a hpc - Nguygii van Hi\
Vi d\ 16: Hoa tan hoan toan 1,92 gam bpt C u trong 100ml dung dich goVf,
HNOs 0,6M va H 2 S O 4 0,5M thu dupe dung dich X Cho mpt thanh Fe vao x
sau khi cac phan ling xay ra hoan toan, thay kho'i lupng thanh Fe giam 1-^
gam Biet khi N O la san pham khu duy nhat cua N * ' trong cac qua trinj^
tren Gia tri cua m la
A 1,44 B.3,36 C 0,88 D 2,80 ,
' •H •'.'•» ; hi kiZ) {: Loigidi: m :n}fjh
n c , = 1 ^ = 0,03 mol ' ''^"^VfC D QOfiS M r J
64
NMn xet: K h i axit HNO3 c6 mat dong thoi voi axit H2SO4 loang thi lupng
trong dung djch la do 2 axit phan li ra - > Giai theo phuong trinh ion
nj^+ = nHN03 + 2nH2S04 = 0'06 + 2.0,05 = 0,16 mol j , Trong X: n^,^_ = 0,06 mol; n 2- = 0/05 mol
NO3 Phuang trinh ion rut gpn: jsiO '
3Cu + 8H* + 2NO3 > 3Cu2* + 2 N O + 4 H 2 O
Mol: 0,03 < - 0,08 0,02 0,03 ' ,{rx(),o ' '"^^
Khi cho X tac dung voi Fe: •< 'u^n ctkA
' ' Fe + 4H* + N O 3 > ¥e^* + N O + 2H2O
Mol: 0,02 < - 0,08 0,02 0,02
- J Fe + 2Fe3* >• 3Fe2* i-jiirig/t idi ^
Mol: 0,01 <-0,02 ^ob iii-^ gnoiHsis'(fib ^o*
Fe + Cu^'' > ¥e^* + C u '"H/foibgnufcocv "ii^oii
1- Crom khong tan trong dung dich HNOs dac, ngupi va H2SO4 dac ngup''
2- Crom khong tac dyng voi dung dich kiem
Cty INHH IVIIV DVVH Khang Vigt
J,, H<?p chat Cr (III)
+ Crom (III) hidroxit :|''y*';>: -^my-a \-:}tT'^ -y-' ^
Tinh chat luong tinh: Cr(OH)3 + NaOH > N a C r 0 2 + 2 H 2 O
Tinh khu: 2Cr(OH)3 + 3Br2 + lONaOH > 2Na2Cr04 + 6NaBr + 8H2O
+ Mudi crom (III)
Tinh khu: 2CrCl3 + 3Cl2 + 16NaOH > 2Na2Cr04 + 12NaCl + 8H2O
c, Hpp chat crom (VI)
+ Crom (VI)exit - '-s *
CrOs + H 2 O > H2Cr04 (Axitcromic) 2Cr03 + H 2 O > H2Cr207 (Axit dicromic) , '
+ Mudi crom (VI) *
Muoi crom (VI) gom 2 loai: Cromat (CrO^") va dicromat ( C r 2 0 ^ - ) Can bang thuy phan: Trong dung dich, ion dicromat (mau da cam) ton tai cimg vai ion cromat (mau vang) theo can bang:
Cr207' + H 2 O i = i 2Cr04~ + 2H* J'^iv^
Phan ling chuyen doi ^^^^^
2K2Cr04 + H2SO4 > K2Cr207 + K2SO4 + H2O rtrrr:;
A 4,05% B 82,30% C 85,25% D 13,65%
239
Trang 26Ca'm nang 6n luygn thi dgi hgc 18 chuy6n dg H6a hgc - Nguygn Van Hai
Lai gidi:
Ggi so'mol trong X: Fe (x mol); Cr (x mol) va Al (z mol)
Theo bai: 56x + 52y + 27z = 8,54
Khi cho X tac dung voi NaOH loang, nong, chi c6 Al phan ung:
A l + N a O H+ H 2 0 — ^ NaA102 + - H 2 t
2 Mol: 0,02 V > 0,03
Luu y: Cr khong tac dung vdi dung djch kiem. -s: ^
V i dv 3: Nhan xet nao sau day la sai?
A Vat dung lam bang nhom va crom deu ben trong khong khi va nuoc
B Crom la kim loai cung nhat trong tat ca cac kim loai
C Nhom va crom deu bj thu dpng hoa boi H N O 3 dac, ngugi
D Nhom va crom deu phan ung voi axit HCl theo ciing ti If so'mol
Lcn gidi:
A dung vi cac vat dung lam bang nhom va crom deu c6 lop mang oxit ben
viing bao ve
B diing vi crom rat ciing, do cung bang 8,5 (so voi kim cuong bang 10)
• C dung
D sai vi A l va Cr tac dyng voi axit HCl va tao thanh cac muoi clorua voi
hoa tri khac nhau:
Cr + 2HC1 — ^ CrCh + H2t -> T i l | m o l l : 2
Al + 3HC1 — ^ CrCb + - H 2 t
2
Vi dv 4: Nguyen to Cr CO Z = 24 Cauhinh electron cuaCr la
A [Ar] 4s2 3d^ B [Ar] 3d^4s\ [Ar] Sd^ 4si D [Ar] Sd^ ?
f Lai gidi:
Le ra cau hinh electron cua nguyen tu Cr la: [Ar] 3d *4s2 JrfD Y sv
Tuy nhien, do c6 sv chuyen le tir phan lop 4s sang 3d de tao cau hinh nira
bao hoa som, ben han -» Cau hinh thuc te la [ArlSd"* 4s' -> Dap an C "^i
Vi dv 6: Cho 1,58 gam hon hgp bgt X gom Al, Cr tac dving voi lugng du H2SO4
loang, nong, thu dugc 1,120 lit khi H2 (dktc) Mgt khac, cho 1,58 gam X vao
lugng du dung djch NaOH loang, nong, thu dvigc V lit khi khi H2 (dktc)
Gia tri cua V i a
A 1 ,344 B 0,672 C 0,896 D 1,120
Lmgtat:
Ggi so mol trong X: A l = x; Cr = y Theo bai: 27x + 52y = 1,58 4 '-M
2A1 + 3H2SO4 — ^ Al2(S04)3 + 3H2t ' ia«t>W,:!!I)a<iv'-^'
Cr + H2SO4 CrS04 + H2t * *
nH2 = 0,05 l,5x + y = 0,05 x = 0,02; y = 0,02
+ Khi cho X tac dung voi NaOH loang, nong, chi CO Al phan ling:
A l + N a O H + H 2 0 — ^ NaA102 + | H 2 t
Luu y: Cr khong tac dung voi dung dich kiem
Vi dv 7: Phat bieu nao sai khi so sanh tinh cha't hoa hgc cua nhom va crom?
A Nhom va crom deu khong tan trong dung djch HNO3 d$c, ngugi
B Nhom va crom deu ben trong khong khi va trong nuoc
C Nhom va crom deu phan ung voi dung dich axit HCl theo ciing ti 1? ve so'mol
D Nhom C O tinh khvr mgnh hon crom ,
Lm gtat: , i^han xet: Al tac d^ing voi axit HCl - > Muoi AlCb
Cr tac dvng voi axit H C l - > Muoi CrCk
- » Al va Cr phan utig voi HCl theo ti If so mol khac nhau
241
Trang 27elm nang 6n luygn thi dgi hpc 18 chuy6n H6a hpc - Nguyin Van Hii
+ Tac dung v&i phi kim ihrimA
2Cu +02 — ^ 2CuO l a M : ^ ^l^ix:^ _ t i ^ C ^ Q ^
Cu tac d y n g v o i axit HNO3 d^c va loang, axit H2SO4 dac, nong: fi.'
3Cu + 8HNO3(/0flM^) > 3Cu(N03)2 + 2 N O + 4H2O '-^
Cu + mNCh{dqc,nguoi) > Cu(N03)2 + 2NO2 + 2H2O
+ Tac dung v&i dung dich muoi ' *'
Cu + 2AgN03 >• Cu(Na)2 + 2 A g
C u + 2FeCl3 — > C u C h + 2FeCl2
b Dong (II) oxit
+ Tinhoxihoa: C u O + CO '° > Cu + CO2 » 5,^ •
aCuO + 2NH3 — ^ 3Cu + N2 + 3H2O
c Dong (II) hidroxit
Tan trong dung dich NHs tao dung dich mau xanh tham:
Cu(OH)2 + 4NH3 > [Cu(NH3)4](OH)2
d Dong (II) sunfat
+ Phan ung trao doi: CuSOi + > CuS>l' + H2SO4
+ Phan ung voi dung dich amoniac
CuS04 + 4NH3 > [Cu(NH3)4]S04 {mau xanh tham)
+ Phan ung dien phan: CuS04 + H2O - J s ^ Cu + ^02 +H2SO4
e Dong (II) nitrat ? ° ; I ' l - '
Cu(N03)2 CuO + 2NO2+ i 0 2
V I D V M A U
V i d^ 1: Cho 3,2 gam bpt Cu tac d y n g v o i 100ml dung dich X hon hpp gon^
HNO3 0,6M va H2SO4 0,1M Sau k h i cac phan u n g xay ra hoan toan, sinh
V lit k h i N O (san pham k h u duy nhat, 6 dktc) Gia tri cua V la
f\!han xet: K h i axit HNO3 c6 mat dong thoi v o i axit H2SO4 loang t h i l u p n g H+
trong d u n g dich la do 2 axit phan l i ra —> Giai theo p h u o n g trinh ion
r = nHNOg + 2 n H 2 S 0 4 = 0,06 + 2.0,01 = 0,08 m o l
n , , ^ _ = 0,06 mol; n^ o_ = 0,01 mol NO3 SO| _
phuong trinh ion n i t gpn: iV^^t , , ,
Trong X:
3Cu + 8H^ + 2NO3" >• 3Cu2* + 2NO + 4H2O ^
M o l : 0,03 < - 0 , 0 8 0 , 0 2 -> n ""^UnAi.OsS^ A Jsi +;lT:.">(,"'
-> V = 0,02.22,4 = 0,448 lit Dap an B
V i 2: Cho 12,4 gam hon hgp X gom Cu, CuS va FeS tac d\ing het v o i
HNO3 (d§c nong, du), thu duac a m o l k h i chi c6 NO2 (san pham k h u duy nha't) va d u n g djch Y Cho toan bp Y vao mpt l u p n g d u dung djch BaCl2/ t h u dupe 23,3 gam ket tiia; con k h i cho toan bp Y tac d\ing v o i dung dich NH3 d u thu dupe 5,35 gam ket tiia Gia trj cua a la
A 0,80 B.0,95 C.1,05 ,^ , D 1,20 ,
Lin giai:
Nhan xet: D u n g dich Y chtra cac ion: Fe^^ C u ^ SO4', va N O , '
Khi cho Y + d u n g dich BaCkBa^* + S04' > BaS04>l
23 3
Bao toan nguyen toS: ng(x)= n^aso^ =—rr = 0,1 m o l 233
Khi cho Y + dung dich NH3 du: ,
Fe^ + 3NH3 + 3H2O > Fe(OH)3 i
Cu2* + 2NH3 + 2H2O > Cu(OH)2 i ' • * ^ Den day cac em l u u y: Cu(OH)2 tan trong NHs d u t^o thanh phuc chat:
(X) = 2ncu + 3nFe + 6ns = 0,95 mol -» nN02 = (X) = 0,95 m o l
-> a = 0,95 ^ D a p a n B 3: Thi^c h i f n cac thi nghi^m sau (6 dieu kien thuong): ;»s r; ,!
(a) Cho dong k i m loai vao dung dich sat (III) clorua
(b) Sue k h i hidro sunfua vao dung dich dong (II) sunfat ,,, (c) Cho d u n g dich b^c nitrat vao dung djch sat (III) clorua
243
Trang 28ca'm nang On luyQn thi djii hpc 18 chuy6n dg H6a hgc - Nguyin Van Hi\
(d) Cho bpt luu huynh vao thiiy ngan
So thi nghi^m xay ra phan ung la l,>fi o ,jl i ! U,; f^mtb ;
A 3 B l C.4 •;?4-^,o,nf^-
^•2-Lcrigidi:
Ta't ca cac thi nghi^m (a), (b), (c), (d) deu xay ra phan ung: i
Cu + 2FeCl3 > CuCb + 2¥eCh f i l f H o ^ ' j
-H2S + C u s a — > CuSvl + H2sa :^ "^-^^
3AgN03+ FeCl3 > SAgClJ' + Fe(N03)3 r-* £G.O /
—>DapanC ^ i ^ / : , / ^ ^ ( t^^^^ J
V i dv 4: Nhi?t phan 18,8 gam Cu(N03)2 mot thoi gian, thu dug-c 10,16 gam
chat ran Hi?u sua't aia phan ung nhif t phan la
n A 40% B 60% C 80% D 50%
Loigidi: •/nhuiliisl
n c u, N O 3 „ - i ^ - 0 1 m o l > - ^ O * « , 0
Nhan xet: Khoi lu^ng chat ran giam = Khoi lugng khi bay ra r,-^^^
j^-Phuang trinh hoa hpc:
' Cu(N03)2 — > CuO + 2NO2 + - O 2
^ m N 0 2 + m o 2 = 18,8-10,16 = 8 , 6 4 4 6 2 x + 32.0,5x = 8,64 '
-> x = 0,08 ^ H = ^ ^ 1 0 0 % = 80% Dap an C
0,1
V i dv 5: Nung 9,40 gam Cu(N03)2 trong binh kin khong chiia khong khi, sau
mpt thai gian thu dupfc 6,16 gam chat ran va hon hg-p khi X Map thy hoan
toan X vao nuac de dugc 600ntU dung dich Y Dung dich Y c6 p H bang
A I B.2 C.3 D.4
Loigidi: 'i NMn xet: Khoi lugng chat ran giam = Khoi lug-ng khi bay ra
Phuong trinh hoa hpc:
Cu(N03)2 — ^ CuO + 2NO2 +
y i dv 6: Cho 35,6 gam hon hgp Zn, Cu va Ag tac dung vira du voi 950ml
dung dich HNO3 I M , thu dugc dung dich chua m gam muoi va 2,24 lit hon h<7p khi X (dktc) gom NO va N2O Ti khoi ciia X so voi H2 la 18,5 Gia tri ciia
mla
A 69,6 B 82,1 C 86,3 ^ D 84,1 ^ '
Loigidi: -'^iti& + 5?': ^
0 bai nay, trudc het cac em can tim so mol moi khi trong X de thu dupe ket
qua: nNo= 0,05 mol; nN20= mol
Z n , C u , A g > Muoinitrat + NO + N2O V
Chat oxi hoa: N^"^ + 3e > NO; 2N*s + ge > N20'~"^^
va CO the xay ra qua trinh: 2N*5 + ge > NH4NC)3 (amol)
Cac phan ung hoa hgc:
Fe + CuS04 > FeS04 + Cu Mol: 0,15 0,15 0,15
iinj fifed ,(-:':«t; • ^
Fe + 2HC1 > FeCk + H2
Mol: ai 0,2 Tac6:m-0,25.56 + 0,15.64 = 0,725m - > m = 16gam '['\
—> Dap an A
245
Trang 29dm nang 6n luygn thi d^i hpc 18 chuy§n 6i H6a hgc - NguySn van Hai
V i d v 8: H o a tan hoan toan 0,1 m o l FeS2 trong 200ml d u n g dich H N O 3 41vj
san p h a m t h u dugc g o m d u n g d i c h X v a m g t chat k h i thoat ra D u n g dich
fits X CO the hoa t a n to'i da m g a m C u Biet t r o n g cac q u a t r i n h tren, san pham
fil k h i i d u y n h a t ciia N*^ deu la N O Gia t r i ciia m la
V i d u 9: C h o 7,84 g a m h o n h o p g o m Fe304 v a C u vao d u n g d i c h H2SO4 loang
d u Sau k h i cac p h a n u n g xay ra hoan toan, t h u dugc d u n g d i c h X c h i i a m
<, i Chat ran Y la C u d u Phan l i n g hoa tan C u bang HNO3 loang:
3Cu + 8HNO3 > 3Cu(N03)2 + 2 N O + 4H2O
M o l : 0,03 < - 0,02
232a +• 64(a + 0,03) = 7,84 g a m ^ a = 0,02 m o l
Vay: m = 152.3.0,02 +160.0,02 = 12,32 gam ^ D a p an A
V i d^ 10: H o a t a n h o a n t o a n m g a m C u t r o n g 160ml d u n g d i c h H N O a I M , thu
d u g c d u n g d i c h X C h o m g t t h a n h Fe vao X, sau k h i cac p h a n u n g ket thiic
t h a y k h o i l u g n g t h a n h Fe g i a m d i 1,44 g a m Biet t r o n g cac qua t r i n h tren,
k h i N O la san p h a m k h u d u y nhat cua N*^ Gia t r i ciia m la
Fe + 2Fe(N03)3 > 3Fe(N03)2 * , , , , ' M i l ( 1 1 1 ; HI ISAf f jj ,.n>
M o l : b <— Z D
Fe + Cu(N03)2 > Fe(N03)2 + C u
M o l : 3a < - 3a - > 3a Theo b a i : 56(3a + 3 b ) = 64.3a = 1,44 , " , ,
M a t khac: nHNOa = 8a + 8b = 0,16 , ,
- > a = 0,01 m o l ; b = 0,01 m o l f, , ,\ J * <
- > m = 0,03.64 = 1,92 g a m D a p a n B j »u -^m
4 B A I T A P O N L U Y E N Bai 1: D a n i u o n g k h i C O d i qua h o n h g p g o m C u O v a F e 2 0 3 n u n g n o n g , sau m g t t h a i g i a n t h u d u g c chat r a n X v a k h i Y C h o Y h a p t h u h o a n toan vao d u n g d i c h B a ( O H ) 2 d u , t h u d u g c 17,73 g a m ket tua Chat r a n X p h a n
u n g v o i d u n g d j c h H N O 3 d u t h u dugc V l i t k h i N O (san p h a m k h u d u y nhat, 6 d k t c ) Gia t r i ciia V la
A 1,792 B 2,016 C 1,120 D 1,344
Bai 2: C h o 6,72 g a m Fe vao 4 0 0 m l d u n g d i c h H N O 3 I M , den k h i phan i m g xay
ra h o a n toan, t h u d u g c k h i N O (san p h a m k h i i d u y nhat) v a d u n g d i c h X
D u n g d i c h X c6 the hoa t a n to'i da m g a m C u Gia t r i ciia m la: ?^
A 0,64 B.3,20 C 1,92 D 3,84
Bai 3: H o a tan hoan toan h o n h g p g o m 0,01 m o l FeCh va 0,02 m o l N a C l vao mgt l u g n g n u o c ( d u ) , t h u d u g c d u n g djch X C h o d u n g d i c h A g N O s ( d u ) vao, t h u d u g c m g a m chat ran G i a t r i ciia m l a (;
Trang 30dm nang On luyQn thi d^i hgc 18 chuySn dg H6a hgc - Nguygn Van HSi
Bai 5: N u n g hon hg-p X g o m 0,14 m o l A l va a m o l Cu trong k h o n g k h i mQt thoj
gian, t h u dxigc 8,58 gam cha't rSn Y Hoa tan hoan toan Y bang d u n g djc}^
H N O s loang (du), t h u dugc 0,896 l i t k h i N O (san p h a m k h u d u y nha't 5
dktc) Gia t r i a i a a la ^ ,
A 0,03 B.0,10 C.0,05 D 0,08
Bai 6: Cho 2,24 gam Fe vao 200ml d u n g dich gom NaNOs 0,3M va H2SO4 0,6Iy[
tao thanh V m l k h i N O (san p h a m khtr d u y nha't, 6 dktc) v a d u n g dich X
D u n g dich X c6 the hoa tan toi da m gam Cu Gia t r i cua m la
A 3,2 B 6,4 C.4,8 D 8,0
Bai 7: N u n g h o n hqip hot g o m 16 gam FezOa v a m gam A l 6 nhiet dp cao Sau
k h i phan l i n g hoan toan, t h u dugc 24,1 gam hon h g p ran X Cho X phan
l i n g v o i axit H C l (du) thoat ra V lit k h i H2 (6 dktc) Gia t r i cua V la
A 3,36 B.4,48 C 7,84 D 10,08
Bai 8: D o t 2,8 gam Fe trong khong k h i , thu dugc hon hgp chat ran X Cho X
tac d u n g v o i d u n g dich HNO3 loang (du), thu dugc k h i N O (san pham
k h u d u y nha't) va d u n g dich chiia m gam m u o i Gia t r i cua m la
A 12,1 B.11,2 C.7,8 D.9,0
Bai 9: Hoa tan hoan toan m gam oxit FexOy bang H2SO4 dac nong, thu duoc
d u n g dich chiia 6 gam mgt loai m u o i sat d u y nha't v a 0,112 lit SO2 (dktc)
Cong thuc ciia oxit sMt va gia trj m Ian l u g t la
A Fe304 va 4,64 B Fe203 va 3,20
C Fe304 va 2,32 D FeO va 1,44
Bai 10: N h i i n g m g t thanh Fe vao 100ml d u n g dich g o m C u ( N 0 3 ) 2 0,5M va
AgNOa 0,2M Sau m g t thoi gian lay thanh k i m loai ra, r u a sach lam kho
tha'y khoi lugng thanh sat tang them 1,92 gam K h o i l u g n g sat da phan u n g la
A 1,40 gam B 2,16 gam C 2,80 gam D 1,72 gam
Bai 11: C h o 13,5 g a m h o n h g p A l , Cr, Fe tac d u n g v o i l u g n g d u d u n g djch
H2SO4 (loang, nong) thu dugc d u n g dich X va 7,84 lit k h i H2 (dktc) Cac thi
n g h i f m dugc tien hanh trong dieu k i f n khong c6 khong k h i Co can X, thu
d u g c m gam m u o i khan Gia t r i ciia m la
A 48,8 B.47,1 C 45,5 D 42,6 •'•'^^•'^
Bai 12: De o x i hoa hoan toan 0,02 m o l CrCb thanh K2Cr04 bang CI2 k h i co
mat K O H d u , so'mol Chto'i thieu can d u n g la
A 0,01 B 0,05 C.0,03 D 0,02 '
Bai 13: C h o 0,14 m o l Fe tac d u n g v a i 0,4 m o l axit HNO3 toi phan u n g hoaii
toan, thu d u g c d u n g dich Y chua m gam muo'i va k h i N O (san p h a m kh^'
d u y nha't) Gia trj cua m la (bie't cap Fe^VFe d u n g truoc cap Fe^VFe^*)
d day, cac em can su dung so do phan ung:
C u O , F e 2 a > X > Muoinih-at + N O CO2 + Ba(OH)2 > B a C a + H2O
-N^flM xet: K h i cho hon hgp oxit tac dung v o i khi CO thi:
ne t»ao doi = 2 n c o • "CO = " C O 2 = ~^ «"° = 0,18 m o l
Bao toan electron: ngiraod6i = 3ni^Q - > n^Q =0,06 mol '
_> V N O =1/344 lit ^ Dap an D v; ;v j ; ^ r.e:^ • , nFe= 0,12 m o l ; nHNOs = 0,4.1 = 0,4 m o l
Phan u n g hoa hgc: , , ,OHH • „
Nhqn xet: X gom: Na^ = 0,02 m o l ; Fe^^ = 0,01 m o l ; C h = 0,04 m o l )f>
Cac phan u n g hoa hgc: ==="''<J,1M,0T; ' H f«Go!
L u u y rang, trong phan u n g nhi?t nhom, thuong ap d u n g dinh luat bao toan khoi lugng, cy the:
249
Trang 31Ca'm nang a n luygn thi hpc 18 chuy6n tSJ H6a hpc - Nguygn VSn H5i
mcr203 + i^Ai=mx -> mAi= 13-7,6 = 5,4 g a m - » 11^1= 0,2 mol
Phuong trinh phan ung: Cr203 + 2A1 > 2Cr + AkOa
Mol: 0,05 0,1 0,1 0,05 ''''
X gom: A l du = 0,1 mol; Cr = 0,1 mol t» I M i
3 Khi X tac dung voi axit: A l -> — H2 va Cr H2
nen n ^ j = 0/25 m o l - > = 5,60 lit -> Dap an C 1!^ i , ,
BaiS: l>l S T
n N O = - ^ ^ = 0 , 0 4 m o l
Cach 1:
Bao toan khoi lupng:
mo2 = m y - mx = 8,58 - 0.14.27 - 64a = 4,8 - 64a
So do phan ling:
A l , C u (1) ) Y (2) ) A F , Cu-2 (3)
Xet su trao d o i electron 6 cac giai doan:
(1) - > (3): A l - 3 e > A F _ ^ nenhimng = 3nAi =0,42
C u - 2 e > Cu*^ —> nenhu«ng = 2ncu = 2 a
(1) - » (2): O2 +4e > 2CH - > nenhan = 4no2 = ' ^
= 0,6-8a
(2) - > (3): +3e N Q - > nenh5n = 3nNo =0/12
Bao toan electron: 0,42 + 2a = 0,6 - 8a + 0,12 a = 0,03 m o l -> Dap an A *
Cach 2: i
Q u i d o i Y thanh: A l = 0,14 m o l ; C u = a m o l va O = b m o l
Bao toan khoi l u g n g : 64a + 16b = 8,58 - 0,14.27 = 4,8
Bao toan electron: 3.0,14 + 2.a = 2b + 0,12 - > a = 0,03; b = 0,18
X CO chua: n^^g^ = 0,04; n^+ = 0,08; n^^^, = 0,02; Na* va
SO|-3Cu + 8H^ + 2 N O ~ > SO|-3Cu2- + 2 N O + 4H2O
C u + 2Fe3^ > Cu2* + 2Fe2^
Cty T N H H M T V D V V H Khang V i j t
ncu = 0,03 + 0,02 = 0,05 mol - > mcu = 0,05.64 = 3,2 gam v'i
Bai nay cac em se gap kho khan neu khong tirJi dugc so'mol A l ban dau
Lini y rang, trong phan ung nhiet nhom, thuong ap dung dinh luat bao toan l<h6i luong, cu the:
mpeaOa + ^AI = ^ m ^ i = 24,1 - 1 6 = 8,1 gam n^i = 0,3 mol l i •
phuong tiinh phan ling: Fe203 + 2A1 > 2Fe + AI2O3
Mol: 0,1 0,2 0,2 0,1 • _» X gom: A l = 0,1 mol; Fe = 0,2 mol
Nhan xet: Bai nay neu dua theo phuong trinh phan ung se rat da: dong va ton
nhieu thai gian O day, cac em can su dung so do phan ung:
Fe > X (Fe, FeaFe203,Fesa) ) Fe(N03)3 :kM
va ap dung dinh luat bao toan nguyen to'Fe: ''^
m = 232.0,012 = 2,32 gam-> Dap an C
Bai 10:
'^AgN03 = 0,02 mol; ncu(N03)2 =
"^ol-' \: ; • V "^ol-' : ^ "^ol-' % 251
Trang 32Ccim nang 6n luygn thi a^i hgc 18 chuySn dg H6a hpc - Nguyin Van HSi
Nhan xet:
Tinh oxi hoa Ag* > Cu^* AgNCh phan ling truac roi moi den Cu(N03)2
Phan ling hoa hpc:
Fe + 2AgN03 > Fe(N03)2 + 2Ag
Mol: 0,01 < - 0,02 0,02 -> mtsng = 2,16-0,56 = 1,6 gam
Fe + Cu(N03)2 > Fe(N03)2 + C u
Mol: a •<— a a mtsng = (64-56)a = 8a gam
Theo bai: mFeting = 1,92 gam 1,6 + 8a = 1,92 - » a = 0,04 mol
mFe(pir) = 56.(0,01 + a) = 56.0,05 = 2,8 gam -> Dap an C
N/ifl«xet: nH2SO4 = nH2=0'05mol s: , ,
-So do phan ling: K i m loai + H2SO4 > Muoi sunfat + H2
Bao toan khoi lugng cho so do tren:
J C A U T A G N G U Y E N T l /
J H?t nhan nguyen t u
+ E)ien tich hat nhan : ' P r v ^ V/.^'^.'^ l,l\A»i/; ;
Proton mang di^n tfch 1+, neu hat nhan c6 Z proton thi dien tich hat nhan bang Z+ va so'don vi dien tich hat nhan bang Z
So don vi di?n tich hat nhan (Z) = so proton = so'electron I 'i '
+ Sokho'i
So khoi (A) la tong so'hat proton (P) va tong so'h^t natron (N) ciia h^t nhan
do: A = Z + N MJ'i 'A ^ - i IJ m i q f - b ?
b Nguyen to hoa hpc - D o n g v i : i
* Nguyen to'hoa hoc
+ Nguyen to' hoa hpc la tap hgip tat ca cac nguyen tu c6 ciing difn tich hat nhan V i dy: Nguyen to'O bao gom tat ca cac nguyen tu c6 di^n tich h^t
n5ng luQfng tu thap den cao
T h u ty sap xep cac phan lap theo chieu tSng cua nSng lugmg dvtqc xac djnh
bang thuc nghi^m va li thuye't: I s 2s 2p 3s 3p 4s 3d 4p 5s
Cau hinh electron nguyen tie ^^^^ ^
Budc 1: Xac djnh so electron cua nguyen tijr,
253
Trang 33dm nang 6n luy$n thi dji hgc 18 chuy§n dg H6a hgc - Nguygn Van HSi
Bu-oc 2: Dien cac electron vao cac phan miic nang lirgng theo t h i i t y :
Is 2s 2p 3s 3p 4s 3d 4p 5s , ••m%%mm*f>''W ''^'^-f
Biroc 3: Viet cau h i n h electron (sap xep theo trat t ^ cac lop tir trong
ngoai): Is 2s2p 3s3p3d 4s4p4d4f
V I D V M A U ' ^<M^\-: W\" ,
V i d\ 1: Trong hat nhan cac nguyen to'co so hi§u nguyen t u tvr 2 den 82, giug
tong so hat notron (N) va tong so hat proton (Z) c6 m o i quan h f la
V i d\ 2: Cation X^* c6 tong so h^t proton, notron va electron la 80, trong do ti
1? giiia so hat electron v o i so'h^t notron bang 4/5 V i t r i (chu k i , nhom) cua
X trong bang tuan hoan la
A chu k i 4, n h o m I I A B chu k i 4, n h o m IIB
C chu k i 4, n h o m VIIIB D chu k i 4, n h o m VIB
Lbi gidi:
Gpi so hat proton, notron va electron trong nguyen t u X Ian lug-t la P, N va
E T a c 6 : P = E
Nhan xet: Trong ion X^*, so'hat electron la E - 2
Theobai, trong ion X2*:P + N + E - 2 = 80 v a E - 2 = 0,8N |
^ 2P + N = 8 2 v a P = 0,8N + 2 '
P = 26 va N = 30
• ^ Cau h i n h electron cua X: Is^ 2s22p^ Ss^Sp^Sd* 4s2
^ X thuoc chu k i 4, n h o m VIIIB ,
- > D a p a n C ''issiof
V i dv 3: Cacbon c6 2 dong v i la ^^C chiem 98,9% v ^ ^^C chiem 1,1% ve so
l u g n g nguyen t u N g u y e n t u khoi trung binh ciia nguyen to' cacbon la:
Cty TNHH MTV UWH Khang Vigt
v i dv 4: X va Y la hai nguyen to' thupc ciing mpt chu ky, hai n h o m A lien tiep So'
proton cua nguyen t u Y nhieu hon so' proton ciia nguyen t u X Tong so' hat proton trong nguyen t u X va Y la 33 Nhan xet nao sau day ve X, Y la dung?
A D o n chat X la chat k h i 6 dieu ki§n thudng
B D p a m dien cua X Ion hon dp am d i f n cua Y. ' 8''^'^i k tu > >; ; < = ^
C hop ngoai cung cua nguyen ttr Y (d trang thai co ban) c6 5 electron
D Phan lop ngoai cung cua nguyen t u X (6 trang thai co ban) co 4 electron
Lot gidi:
I4hqn xet: X va Y cung thupc mpt chu k i va a hai n h o m A lien tiep -> chung
hon kem nhau 1 d o n v i d i ^ n tich h^t nhan ' ' n V|, i -.^
Theo bai, so'proton cua nguyen t u Y nhieu h o n so proton cua nguyen t u X
Mat khac: ZY + Zx = 33 -> Zx = 16 ( L u u huynh) va ZY = 17 (Clo) !) is=
Loai A v i 6 dieu k i f n thuong, l u u h u y n h ton tai a the ran Q if
Lo^i B v i dp am d i f n cua S nho h o n ciia CI s M = '
Loai C v i 6 lop ngoai cung cua nguyen t u CI CO 7 electron ' A
D d u n g v i phan lop ngoai cung ciia S la 3p^ co 4 electron
—>Dap a n D
V i dv 5 (CD-10): D o n g co 2 dong v i ben la ^^Cu va ^^Cu N g u y e n t u khoi
trung b i n h ciia dong la 63,54 Thanh phan phan tram cua dong v i *^Cu la
A 30% B 27% C 73% D 37%
Lbi gidi: * ' v / '
Cach 1: Gpi phan tram ciia dong v j 29 Cu la a%
N g u y e n t u kho'i t r u n g binh ciia dong la: A = + 65 x (100—a) _ ^3 5 4
V^y a = 0,73 = 73% Dap an C
Cach 2:
Nhan xet: N e u t i 1? 2 dong v i la 50:50 thi nguyen t u khoi t r u n g b i n h ciia
dong la 64,00
Theo bai, nguyen t u khoi trung binh ciia dong la 63,54 < 64,00 -> dong v i
nh? hon, tuc 29 Cu chiem tren 50% -> chi dap an C thoa man
V i dy 6: Phan tram khoi l u p n g ciia nguyen to'R trong h p p chat k h i v o i hidro (R CO so oxi hoa thap nhat) va trong oxit cao nhat t u o n g l i n g la a% va b % ,
v 6 i a : b = l l :4 P h a t b i e u n a o s a u d a y l a d i i n g ? , , ,
A Phan t i i oxit cao nhat ciia R khong co c\rc , , ,
B Oxit cao nhat ciia R 6 dieu ki#n thuong la chat ran
C Trong bang tuan hoan cac nguyen to hoa hpc, R thupc chu k i 3
D N g u y e n t i i R (6 tr^ng thai co ban) co 6 electron s
255
Trang 34dm nang On luygn thi dji hgc 18 chuy6n dg H6a hQC - NguySn Van Hii
Lai gidi:
j, Nhan xet: Tong hoa trj cua nguyen to' R trong hqp cha't v o i k h i h i d r o vg
trong oxit bang 8
Gpi hoa tri cua R trong hgip chat v o i hidro I a n
-> H o a t r i cua R t r o n g oxit la 8 - n r , „ ,
Cong thuc h g p chat k h i a i a R v o i hidro: R H n , ^y., rj,./,, ,j
Cong thiic hgjp chat ciia R v o i oxi: R208.n ^ I ,•«,• '
Theo bai: -g- =H ^ 7R = 256 - 43n
R+n 2R+16(8-n) 4
Thay n = 1 den 4, nhan thay n == 4 va R = 12 (Cacbon) thoa m a n
A d i i n g V I phan t u CO2 k h o n g phan cue. v^>r>£U'j iioloiqaa 4^*1 r
Lo?i B v i a dieu k i ^ n t h u o n g CO2 la chat khi A- xS - yS
Loai C v i n g u y e n to cacbon 6 chu k i 2 n - X^K •< - SB ^ >
Loai D v i nguyen t u cacbon a trang thai co ban c6 4 electron s , 1,
—> D a p an A
V i dv 7: N g u y e n t u cua m g t nguyen to X c6 tong so hat proton, notron,
electron 46, t r o n g d o so hat m a n g d i ^ n nhieu h a n so' h?t k h o n g m a n g di^n
la 14 Cau h i n h electron nguyen t u X la U m, qsiJ •
A [Ne]3s23p3 B {Ne]3s' C [Ne]3s23p* D [Ne]3s23pi
A d u n g v i X va Z deu c6 ciing so khoi bang 13
B sai v i X v a Z c6 s6'hi?u nguyen t u khac nha u —> k h o n g la d o n g v i
C sai v i X v a Y c6 so' hi^u nguyen t u khac nhau —> thupc hai nguyen to khac nhau
Cac nguyen to' hoa hoc c6 ciing so l o p electron duoc sap xep thanh ciing mpt hang (chu k i )
Cac nguyen to hoa hpc c6 ciing so electron hoa t r i t r o n g nguyen tvr dupe sap xep thanh m o t cpt (nhom)
2 Cau tao cua b a n g h | t h o n g tuan hoan Dang bang tuan hoan dupe sir d u n g trong sach giao khoa hoa hpc p h o
thong hien nay la bang dang dai Cac khai n i e m co ban cua bang tuan hoan
bao gom:
6 nguyen to: L a v i t r i cua m p t nguyen to, so t h i i t u cua 6 bang so h i ^ u
nguyen tu' va bang so d o n v j di?n tich hat nhan nguyen tir
Chu ki: Co 7 chu ky, so' t h i i hf cua chu k i bang so lap electron cua nguyen
tvr gom:
+ Chu k i nho la cac chu k i 1, 2, 3 chi gom cac nguyen to s va cac nguyen to p
M o i chu k y nho g o m 8 nguyen t6'(tru' chu k y 1 chi co hai nguyen to)
+ Chu k i Ion la cac chu k i 4, 5, 6 ,7 gom cac nguyen t o s, p, d va f CK day, quan t r p n g la chu k y 4 va chu ky 5, m o i chu k y co 18 n g u y e n to
Nhom: Co 8 n h o m , so t h u t y ciia n h o m bang so electron hoa trj gom:
N h o m A : So' t h i i tir cua n h o m bang so electron l o p ngoai c i m g ( g o m cac nguyen t o s va p) N h o m A con dupe gpi la cac nguyen to n h o m chinh
N h o m B: So t h u t y ciia n h o m B bang so electron hoa t r i (gom can nguyen to
d va f) N h o m B con dupe gpi la cac nguyen to n h o m p h y
^- N h i m g t i n h c h i t b i e n d o i t u a n hoan
^ Cac dai l u p n g bien doi tang dan:
- Trong mot chu ki: d p am d i f n, nang l u p n g i o n hoa, t i n h p h i k i m , t i n h axit
ciia cac oxit va hidro).it
- Trong mot nhom: ban k i n h nguyen tvr, tin h k i m lo^i, t i n h baza ciia cac
oxit v a hidroxit
257
Trang 35Ca'm nang On luySn thi dji hpc 18 chuyfin 66 H6a hQC - NguySn Van H5i
+ Cac dai lugng bie'n doi giam dan:
- Trong mot chu kh ban kinh nguyen ttr, tinh kim loai, tinh baza ciia cac
oxit va hidroxit , • '
Si- Trongmot nhom: do am dien, nang luong ion hoa, tinh phi kim, tinh axit
cua cac oxit va hidroxit
V I D V M A U ' ' " " ^ ^ ' i " '
V i di^ 1: Mot ion M^* c6 tong so' hat proton, notron va electron la 79, trong (JQ
so' hat mang di#n nhieu hon so hat khong mang dien la 19 Ca'u hinh
electron cua nguyen tu M la
A [ Ar]3d34s2 B [Ar]3d54si C [Ar]3d64si D [Ar]3d^4s2
Nguyen tu M co: ,
Tong so' hgt proton, notron va electron la 79 + 3 = 82; so' hat mang dien
nhieu han so'hat khong mang dien la 19 + 3 = 22 Ins ffl c'»
- > So hat khong mang dien = (82-22)72 = 30 ^ N = 30 ^*
Vay s o h a i m a n g d i ^ n = 8 2 - 3 0 = 52 - > Z = E = 26
—> Dap an D
V i dv 2: Nguyen to'X c6 hoa trj cao nhat trong oxit la a; hoa tri trong hop chat
khi voi hidro la b Moi quan h$ giiia a va b la
A a - b = 8 B a + b = 8 C a = b D a < b
TO* oe
Lot giai:
Dap an B Day la ke't qua cac em hpc sinh can nho de van dyng trong cac bai
toan ve hoa tri ~, , i ,
Luu y: Khong ap dung cong thiic nay cho cac nguyen to chu ki 1 va 2
V i di;i 3: Oxit cao nhat cua nguyen to' R c6 cong thiic tong quat la RaOs Trong
hop cha't voi hidro, R chie'm 82,35% ve khoi lu^ng Nguyen to'R la
A Photpho B Nita C Oxi D L u u huynh
Lot giai:
Oxit cao nha't cua nguyen to R c6 cong thuc tong quat la R2O5 R c6 hoa tr!
cao nhat bang 5 -> Hoa tri trong hop chat voi hidro bang 3 - » cong thtH
hgip chat cua R voi hidro c6 dang R H 3 Ta c6:
%R 82,35 R 82,35 „ , X T- x r ^ - - '
= — ^ — = — > R = 14(Nito) Dap a n B 'i^^m-''''
% H 17,65 3 17,65 \ V
V i dv 4 (B-08): Cong thuc phan tu cua hgp chat khi tao boi nguyen to R v^'
hidro la RHs Trong oxit ma R c6 hoa trjcao nha't thi oxi chie'm 74,07%
khoi luQ-ng Nguyen to R la ,, ; ,
A L u u huynh B Asen C Nita D Photpho
A Tinh phi kim giam dan, ban kinh nguyen tu tang dan
B Tinh kim loai tang dan, do am dien tang dan ^
C D 9 am dien giam dan, tinh phi kim tang dan ' " '^''^" ^ ' " • , ^
D Tinh kim loai tang dan, ban kinh nguyen tu giam dan
Latgtat:
A dung: Tinh phi kim giam dan, ban kinh nguyen tu tang dan '
B sai vi tinh kim loai tang dan, do am dien giam dan '
C sai vi do am dien giam dan, tinh phi kim giam dan * ^ ' ' '
D sai vi tinh kim loai tang dan, ban kinh nguyen tu tang dan
A K i m logi va kim loai B Phi kim va kim loai wa^;,
C K i m loai va khi hie'm D Phi kim va phi kim '
X CO muc nang lugng cao nha't la 3p ^ X c6 it electron han Y Theo bai ->
ca'u hinh lap ngoai ciing cua X la 3s23p* (it hon 2 electron so voi Y) j , -> X la phi kim Dap an B
Vi dy 7 (A-09): Nguyen tu ciia nguyen to X c6 ca'u hinh electron lop ngoai cung la ns^np* Trong hgp chat khi ciia nguyen to X voi hidro, X chie'm 94,12% khoi lugng Phan tram khoi lugng ciia X trong oxit cao nha't la
A 50% >fu-,sy B.27% C.60% D 40%
Lot giai:
X CO ca'u hinh electron lap ngoai cimg la ns^np"* -> X thugc nhom V I A - » Hoa tri cao nha't trong oxit ciia X bang 6 -> Hoa tri trong hgp chat voi hidro bang 8 - 6 = 2 - » Cong thuc hgp chat ciia R voi hidro c6 d^ng R H 2 ^,,,
• , ^ • • „ • 259
Trang 36dm nang On luy^n thi Jgi hpc 18 chuySn 66 H6a hpc - Nguyin Van Hil
Toe d o p h a n u n g la d p b i e h t h i e n n o n g d p cua m p t t r o n g cac cha't p h a n ling
hoac san p h a m t r o n g m p t d a n v i t h a i gian
b Toe dp trung b i n h ciia p h a n ling , , ,
c C a c y e u to a n h Huong den toe dp phan u n g
+ Nong do: K h i t a n g n o n g dp chat p h a n u n g , toe d p p h a n u n g t a n g
+ Ap suat: V a i p h a n u n g eo chat k h i , k h i tang ap sua't, toe d p p h a n u n g tang
+ Nhiet do: K h i t a n g n h i e t dp, toe d p p h a n u n g tang
+ Dien tich be mat: K h i t a n g d i ^ n tich be m a t cua chat p h a n l i n g (tiic l a m kich
t h u d e hat n h o d i ) , toe d p p h a n u n g tang
+ Chat xuc tac: Cha't xiie tae la chat l a m tang toe d p p h a n l i n g
v i D V MAU
V i dv 1: C h o cac y e u to: (1) n o n g dp; (2) n h i # t d p ; (3) ap suat; (4) d i f n t i c h tiep
xiie; (5) xiie tac So' y e u to' c6 a n h h u a n g d e n toe d p p h a n u n g la
A 2 B 3 C 4 D 5
Loi giai:
Nhqn xet: Cac y e u to a n h h u o n g de'n toe d p p h a n u n g g o m : n o n g d p chat
p h a n u n g , n h i ^ t dp, ap suat, d i # n tich tiep xiie va chat xiie tae
- > D a p an D
V i d v 2: C h o chat xiic tac M n 0 2 vao 100ml d u n g dieh H2O2, sau 60 giay thu
d u p e 33,6 m l k h i O2 (dkte).Toc dp t r u n g b i n h ciia phan u n g (tinh theo H2O2)
Toe d p t r u n g b i n h ciia p h a n u n g ( t i n h theo H2O2) t r o n g 60 giay tren la
V = (3.102 mol/l)/60s = 5,0.10^ mol/(I.s) j ;i<
-—> D a p a n A V/J.'B
"i/i^-V i dv 3: C h o p h a n u n g : Br2 + H C O O H > 2HBr + C O 2
N o n g d p b a n d a u ciia Br2 la a mol/1, sau 50 giay n o n g dp Br2 con lai la 0,0^
mol/1 Toe d p t r u n g b i n h ciia p h a n u n g t r e n t i n h theo Br2 la 4.10-* mol/(l.s)
Gia t r i ciia a la ,„ , fiOff Jinfed n'i'y d > i ;
Xet p h a n l i n g p h a n h i i y Na2S203 t r o n g d u n g dieh axit H2SO4 loang a 25°C:
Na2S203+ H2SO4 > Na2S04 + S + SO2 + H2O ,
Ban d a u n o n g d p ciia Na2S203 la 0,2M, sau 100 giay n o n g d p ciia Na2S203
2 6
Trang 37C^m nang On luy^n thi dgi hpc 18 chuySn 6i H6a hgc - Nguygn Van HSi
la 0,02M Toe do trung binh cua phan ling tinh theo N2O5 la
A 2,0.10-3 mol/(l.s) B 1,0.10-3 mol/(l.s)
C 1,8.10-3 mol/(l.s) ' D 1,8.10-4 mol/(l.s)
Laigidi *
Nong do N2O5 da phan ung: AC = 0,2 - 0,02 = 0,18 mol/1
Toe do trung binh ciia phan ung ti'nh theo Na2S203: ';;
G Can bang hoa hoe la trang thai eiia phan ung thugn nghieh khi toe dp phan
' ' ung thuan bang toe do phan ung nghieh
b Stf chuyen dich can bing hoa hgc
Nguyen li La Sa-ta-li-e: Mot phan ung thuan nghieh dang 6 trang thai can
bang khi ehiu mot tac dong tu ben ngoai (nhu thay doi nong do, ap suat,
nhiet do) thi can bang se chuyen dich theo chieu cho'ng Iqi tac dong ben
ngoai do
+ Khi tang nong do ciia mpt chat, can bang se chuyen dich theo chieu lam
giam nong do chat do —> tue chieu chat do tham gia phan ung
+ Khi tang ap suat chung cua h§, can bang se chuyen dich theo chieu lam giam
ap suat ciia he —> tiie chieu lam giam so'mol khi
+ Khi tang nhi^t dp cua h§, can bang se chuyen dich theo chieu lam giam nhiet
dp cua he tue chieu thu nhiet
VI D U M A U
Vi du 1: Cho can bang hoa hoc: N2 (k) + 3H2 (k) < = ± 2NH3 (k); phan ung
thuan la phan ung toa nhiet Can bang hoa hpc khong bi chuyen dich khi:
i; A Thay doi ap suat cua he B Thay doi nong dp N2
C Thay doi nhiet dp D Them chat xiic tac Fe
Lbi giai:
Nhan xet: Them chat xiie tac chi lam thay doi toe dp phan ung, khong lam
chuyen dich can bang hoa hpc
—> Dap an D
Vi du 2: Cho can bang hoa hpc: 2SO2 + O2 < > 2SO3; phan ung thuan la
phan ung toa nhiet Can bang chuyen dich theo chieu thuan khi
A Tang nhi^t dp B Giam nong dp O2
C Giam ap suat he phan ung D Giam nong dp SO3
Cty TIMh, TV DVVH Khang Vigt
Lai giai:
Loai A vi khi tang nhiet dp cua he, can bang se chuyen dich thao chieu lam giam nhiet dp ciia he -» tue chieu thu nhiet -> chieu nghieh
Loai B vi khi giam nong dp O2, can bang se chuyen dich theo chieu lam tang
nong dp O2 -> tue chieu tao ra 0 2 ^ chieu nghieh ' ( ' ' S'^'
Loai C vi khi giam ap suat chung ciia h$, can bang se chuyen dich theo chieu lam tang ap suat -> hie chieu lam tang so'mol khi -> chieu nghieh
Phuong an D: Khi giam nong dp SO3, can bang se chuyen dich theo chieu lam tang nong dp SO3 -> tiic chieu tao ra SO3 -> chieu thuan
-> Dap an D '
Vi du 3: Cho cac can bang hoa hpc: ^
(1) N2 (k) + 3H2 (k) : i = > 2NH3 (k) »i d ^ '.V J , , (2) H2 (k) + l2(k) i = ± 2HI (k) " ' (3) 2SO2 (k) + O2 (k) ^=i± 2S03 (k)
(4) 2NO2 (k) <==± N2O4 (k) Khi thay doi ap suat, nhiSng can bang hoa hpc bi chuyen dich la
A (2), (3), (4) B (1), (3), (4) C (1), (2), (4) D (1), (2), (3)
Lot giai
Nhan xet: Khi thay doi ap suat, nhung can bang c6 su khac nhau ve tong so
mol khi cac chat 6 2 ve se bi chuyen dich ,
-> Cac can bang bi chuyen dich khi thay doi ap suat: (1), (3), (4) -> Dap an B
Vi du 4: San xuat amoniac trong cong nghiep dua tren phuang trinh hoa hpc:
2N2(k) + 3H2(k) 4=z± 2NH3(k) AH < 0
Can bang hoa hpc se chuyen dich ve phia tao ra amoniac nhieu hon neu
A Giam ap suat chung ciia he B Giam nong dp khi nito
C Tang nhi^t dp ciia h§ D Tang ap suat chung ciia h$
Lai giai:
Loai A v i khi giam ap suat chung ciia he, can bSng se chuyen dich theo chieu lam tang ap suat -> tiic chieu lam tang so mol khi -> chieu nghieh
Loai B vi khi giam nong dp N2, can bang se chuyen dich theo chieu lam tang
nong dp N2 tiic chieu tao ra N2 -> chieu nghieh
Loai C vi khi tang nhiet dp ciia he, can bang se chuyen dich theo chieu lam giam nhi?t dp ciia he -» tue chieu thu nhi^t chieu nghieh
Phuong in D: Khi tang ap suat chung ciia h$, can bang se chuyen dich theo
ifjl^: chieu lam giam ap suat chieu lam giam so mol khi -> chieu thuan
-)• Dap an D
Trang 38Ca'm nanQ On luyjn t h i d ? i hpc 18 chuy6n d g H6a tiQC - Nguygn van HSi
V i d^ 5: Phan u n g : N 2 + 3 H 2 2NH3 dugc xiic tac bc« bpt Fe Tac dpn
ciia bot ¥2 den can bang la
:, A Chuyen dich can bang sang trai B Chuyen dich can bang sang phj^j
C Tang toe d p phan l i n g D Tang hang so can bang
Loi gidi:
Nhan xet: Bot Fe d o n g vai tro la chat xiic tac - > them b p t Fe khong 1^^^
chuyen dich can bang hoa hoc, khong l a m tang hang so can bang, chi l a ^
tang toe d p phan l i n g
- > D a p a n C ' ' ' ^ • rrr,hr<:i
V i dy 6: Cho can bang hoa hpc: H 2 (k) + h (k) < > 2 H I (fc); A H > 0
Can bang k h o n g b i chuyen dich k h i i , ' ! ' •
A G i a m ap suat chung cua h^ B Tang nong dp H2 ) , }
C Tang nhiet d p cua he D Giam nong d p H I
Lai gidi Nhan xet: Can bang hoa hpc da cho c6 su bang nhau ve tong so m o l k h i cac
chat 6 2 v e K h i thay d o i ap sua't, can bang khong b i chuyen dich
-> Dap an A
5 B A I T A P O N L U Y E N
Bai 1: X la nguyen to thuoc n h o m chinh trong bang tuan hoan Phan t r a m khoi
l u p n g cua nguyen to X trong oxit cao nhat va trong h a p chat k h i v a i h i d r o
t u a n g u n g la a% va b%, v a i a : b = 0,425 Phat bieu nao sau day la sai?
A X thuoc chu k i 3 trong bang tuan hoan
B N g u y e n to'X a dieu k i f n t h u a n g la chat ran
C Phan t u oxit cao nhat cua X c6 cue
D N g u y e n t u X (6 trang thai ca ban) c6 6 electron s
Bai 2: X va Y la h a i nguyen to thuoc cung m o t chu k y , hai n h o m A lien tiep
Tong so hat proton trong nguyen t u X va Y la 3 9 va Zx < ZY N h a n xet nao
sau day ve X, Y la dung?
A O x i t ciia X c6 tinh baza m a n h h a n oxit cua Y
B D p a m di^n cua X Ion h o n dp am d i f n cua Y
C T i n h k h u cua Y m a n h h o n cua X
D Ban k i n h n g u y e n t u cua X nho h o n cua Y
Bai 3: H o a tan hoan toan 0,3 g a m h o n hpp hai k i m lo?i (thuoc n h o m l A va 0
hai chu k i lien tiep) vao nude, thu dupe 0,224 l i t k h i h i d r o (6 dktc) H a i Vv^
A Ban k i n h nguyen t u tang, d p am di^n giam
B Ban k i n h nguyen t u va dp am d i ^ n deu tang S
C Ban k i n h nguyen t u giam, d p am di^n tang uff"!* - / I : : S; -,
D Ban k i n h nguyen t u va dp a m dien deu giam wvh' ; r =
Bai 5: Ban k i n h nguyen t u cua cac nguyen to': sLi, sO, 9F, i i N a dupe xep theo
t h i i t y tang dan t u trai sang phai la
A F , 0 , L i , N a B F, Na, O, L i C F , L i , 0 , N a D L i , N a , 0 , F
Bai 6: H o a tan 3,68 g a m h o n h o p hai m u o i cacbonat ciia h a i k i m loai M , M '
(thuoc n h o m I I A va 6 hai chu k i ke tiep) bang d u n g dich H C l , t h u dupe
0,896 l i t C O 2 (a dktc) va d u n g dich X H a i k i m loai M va M ' la
A Be va M g B M g va Ca C Ca va Sr D Sr va Ba
Bai 7: Phan h u y m o t l u p n g ozon trong b i n h k i n eo d u n g tich 2 l i t (khong doi),
sau 30s t h u dupe 0,045 m o l oxi Toe dp trung b i n h cua phan l i u g (tinh theo ozon) t r o n g 30s tren la
A 2,5.10^ mol/(l.s) B 7,5.10"^ mol/(l.s) 3MA i v
C 1,0.10-3 mol/(l.s) D 5,0.10^ mol/(l.s) N-,nuh
Bai 8: Cho can bang sau trong b i n h k i n : 2NO2 (fc) ^ = ± N2O4 (k) A H
Trang 39Cam naiig 611 iuy(.n llii ciai hpc 18 chuySn dl Hba hgc Nijiiydn Van Hii
Cong thiic hop chat khi cua X v6i hidro: XHn
Cong thuc hop chat ciia X v6i oxi: XjOg.,^
Theo bai: 0,425 -> 0,425 ''^'^ '
2X+16(8n) X X+8(8n) w i i y i ,
-^ 5 , 7 5 R = 2 7 2 - 4 4 n ' !
Thay n = 1 deh 4, nhan thay n = 2 va X = 32 (Luu huynh) thoa man
A diing vi nguyen to luu huynh 6 chu ki 3 sa;
B dung vi 6 dieu kien thuong, luu huynh la chat ran 4
C sai vi phan tu SOa khong phan cue 0' <
f D diing vi nguyen tu cacbon 6 trang thai co ban c6 6 electron s <>!; i
Dap an C o
Bai 2: lufesv •(X:) tif
Nhan xet: X va Y ciing thuoc mot chu ki va 6 hai nhom A lien tiep Chung
hon kem nhau 1 don vi dien tich hat nhan
Theo bai, so proton cua nguyen tu Y nhieu hon so' proton cua nguyen tir X
nen: Z Y - Zx = 1
Mat khac: Z Y + Zx = 39 Zx = 19 (Kali) va Z Y = 20 (Canxi)
A diing vi K hoat dong manh hon nen oxit se c6 tinh bazo Ion hon
Loai B vi do am dien ciia K nho hon cua Ca
Loai C vi tinh khii ciia C a yeu hon K
Loai D vi ban kinh nguyen tu ciia K Ion hon C a (trong mot chu ki, ban kinh
nguyen tu giam dan)
Nhan xet: Cac nguyen to'tir L i den F (la cac nguyen to trong cung mpt chi-'
ki 2), theo chieu tang ciia dien tich hat nhan thi ban kinh nguyen tii- giani
dp am di^n tSng ; , , , >
- » Dap an C
Cty TNHH MTV DWH Khang Vi$t
Li O va F nam trong cung mpt chu ki ma trong mpt chu ki tu trai sang
phai ban kinh nguyen tix giam dan -> ban kinh: F < O < L i
Li va Na nam trong ciing mpt phan nhom chinh ma trong mpt phan nhom thi ban kinh nguyen tii tang dan -> ban kinh: L i < N a
_ > D a p a n A •••^'^^^-'h'vf-'V^r-iHy-;' ;''', ' ' ' '
Gia su 2 kim loai trong 2 chu ki lien tie'p c6 ki hieu chung la M n 1
n n r , n = ^ ' ^ ^ ^ =0,04 (mol) Ta CO phuong trinh: , *'
M C O 3 + 2HC1 > MCI2 + CO2 + H2O 11 <
Mol: 0,04 <- vV : 4 0 4 ^ Vay: M = =32
0,04 -» Mpt kim loai can tim c6 M < 32 va mpt kim loai c6 M > 32
Hai kim loai can tim la Mg va C a - > Dap an B
Bai 7:
Phuong trinh hoa hpc phan hiiy O3: 2O3 > SOi Nong dp O2 tao thanh la 0,045/2 = 0,0225 mol/1 ^ Nong dp O 3 da phan ling: A C = (2/3).0,0225 = 0,015 mol/1
Toe dp trung binh ciia phan ling (tinh theo Os) trong 30 giay tren la
V = (0,015 mol/1) / 30s = 5,0.10-^ mol/(l.s)-> Dap an D
(1) K h i tang nhi^t dp ciia h^, can bang se chuyen dich theo chieu lam giam
nhiet dp ciia he tiic chieu thu nhi^t -> chieu thuan
(4) Khi gidm ap suat chung ciia he, can bang se chuyen dich theo chieu lam tang ap sua't ciia he tiic chieu lam tang so mol khi -> chieu thuan
Trang 40dm nanq On luyQn thi dgi hgc 18 chuyfln dg H6a hgc - NguySn van H5i
Chat dien li la nhiing chat khi tan trong niroc phan li ra ion
+ Chat dien li manh /i jito:S/.^:tau-sMi nt::;! ,
Chat di^n li manh la chat khi tan trong nuoc, cac phan tu hoa t?n deu phan
li ra ion, vi du:
H C l > + CI- ) Na2S04 > 2Na* + S O ^ ^
i ^ ' •
' '<f
Su di?n l i ciia cac ion k h i hoa tan N a C l trong nuac. t ((,0;
+ Chat dien li yeu
Cha't dien l i yeu la chat k h i tan trong nuoc chi c6 m o t phan so' phan t u hoa
tan phan l i ra ion, phan con lai van ton tai d u o i dang phan t u , v i d y :
CHaCOOH < = ± CH3COO- + ^
Sy phan l i a i a chat di?n l i yeu la qua t r i n h thuan nghich Can bang dien H
la m o t can bang d o n g va cung tuan theo nguyen l i chuyen djch can bang
La Sa-to-li-e
D Q di?n l i a (anpha) cua chat d i f n l i la t i 1§ giiia so phan t u phan l i ra iof
tren tong so' phan t u hoa tan
H a n g so'phan l i
Giong n h u m p i can bang hoa hpc, can bSng di§n l i d i n g c6 hang so caf
bang, gpi la hang so' phan l i
Xet can bang: CH3COOH ^=± CHaCOO" + K
T a c 6 : K = ' ^ ^ C O O ^ l l H * !
^ [ C H 3 C O O H I
Cty TNHH MTV DWH Khang Vi^t
Xrong t r u o n g hgip nay, hang so' can bang dug-c gpi la hang so' phan l i axit
Qia t r i ciia h l n g so'Ka chi p h u thuQC vao ban chat ciia axit va nhi?t dp
A X I T , B A Z O , M U O I Khai n i # m axit theo thuyet A-re-ni-ut , ? 1 , j 1 j ; x i >;
y^xit la chat k h i tan trong nuoc phan l i ra cation H"^: ; j i h n i , < <
CH3COOH < = ± CHsCOO- + KinsvidomfeM:-:H/fi I O C , i i , , ; a f : : / > :
Axit nhieu na'c: A x i t k h i tan trong nuoc m a phan t u phan l i nhieu Ian ra ion H* gpi la axit nhieu na'c • ' ;
Vi d u H3PO4 la axit 3 na'c:
H3PO4 ; ± H2PO4 +
H2PO; < = t H P O r + H
2-H P O ; 2- ± PO4" + )\ ] H i r.ilfji n\
b Khai ni^m axit theo thuyet A-re-ni-ut
+ Bazo la chat k h i tan trong nuoc phan h ra anion O H - : > i en, ^n'' mu
N a O H > Na^ + O H
-1 "
A
5, m
M i n h hoa khai niem axit-bazo ciia A-re-ni-ut
lidroxit luong tinh
^Hidroxit l u o n g t i n h la hidroxit k h i tan trong nuoc, vira c6 the phan l i n h u axit, vira c6 the phan l i n h u bazo
V i d u , Zn(OH)2 la h i d r o x i t l u o n g tinh: ' Phan l i theo kieu bazo: Zn(OH)2 < = ± Zn^- + 2 0 H -
Phan l i theo kieu axit: Zn(OH)2 Z n O ^ + 2H*
Cac h i d r o x i t l u o n g tinh t h u o n g gap la: Zn(OH)2, Sn(OH;2, Pb(OH)2, Al(OH)3, Cr(OH)3