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 d
Trang 1Ca'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
-2K2Cr04 + H2SO4 > K2Cr207 + K2SO4 i / i i i t i x i > i t n ' >
Y la K2Cr04 va Z laK2Cr207 Dap an A
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
+ Khi cho X tac dung voi axit H2SO4 loang, nong: / >
2A1 + 3H2SO4 — ^ Al2(S04)3 + 3H2t ' ia«t>W,:!!I)a<iv'-^'
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 2elm 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:
Cu(OH)2 + 4NH3 > [Cu(NH3)4](OH)2 • Vnl '
Bao toan nguyen to Fe: n F e ( x ) = n F e ( O H ) j = ^ =a05 m o l - ^ ^ f t * '
Bao toan khoi lupng: m x = n i ^ y + mpg + mg
mcu = 12,4 - 0,05.56 - 0,1.32 = 6,4 gam TIQU ( X ) = — = 0,1 m o l
64 Qui doi X ve hon hpp cac don chat: Fe = 0,05 mol; Cu = 0,1 mol; S = 0,1 mol
(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 3ca'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
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)
—>DapanD OX^^-^M: l
iw:-man xet: Bai toan da "giau d i " san pham NH4NO3!
Vi d\ 7: Cho m gam bot sSt vao dung dich hon hgp gom 0,15 mol CuS04 va 0,2 mol HCl Sau khi cac phan ling xay ra hoan toan, thu dugc 0,725m gam
hon hgp kim loai Gia tri cua m la
A 16,0 B 18,0 C 16,8 D 11,2 I
Loigidi: ' « Nhan xet: Sau khi cac phan ling hoan toan thu dugc hon hgp kim logi -> Fe
Cac phan ung hoa hgc:
Fe + CuS04 > FeS04 + Cu Mol: 0,15 0,15 0,15
Trang 4dm 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
A 0,64 B.1,92 C 1,28 D 2,56
Cty TIM HH MTV DWH Khang Vigt
Lai gidi:
p h a n l i n g cua C u v a d u n g d i c h HNO3: nw.:,, ,1 i :K«u:> i.mi >nii*js
3Cu + 8HNO3 > 3Cu(N03)2 + 2 N O + 4H2O >I '-'f';!'
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 5dm 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^*)
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 6Ca'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 *
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 ~ > 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
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: ''^
Trong 1 mol FeO ho§c Fe304 deu chiia 1 mol Fe'^^ nen deu c6 kha nang
nhuong 1 mol electron, do do: ni^fir:^ >
= npe^Oy = 2nso2 "FexOy = 2-0/005 = 0,01 mol , , ,^ ^ , i ,
Bao toan nguyen to Fe: 2FexOy > xFe2(S04)3 d f^i, ^ihCi ^
Trang 7Ccim 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:
+ 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 8dm 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 9dm 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 10Ca'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
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 11dm 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 ;
2,08M 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 N 2 O 5 la ,
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 12C^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 13Ca'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^
loai do la
A N a v a K B L i v a N a C K v a R b D R b v a C s
Si.,
^264 •
Cty TIMHH MTV DVVH Khang Vi$t
P^i 4 (A-10):.Cac nguyen to t u L i den F, theo chieu tang cua d i ^ n tich hat nhan
thi •
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 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 14Cam 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(8-n) X X+8(8-n) 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
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
Bai 8:
Nhan xet: K h i gidm nhiet dp ciia he, can bang se chuyen dich theo chieu lam tang nhiet dp ciia he -> chieu thu nhiet < mi
Mat khac, theo bai ra, khi giam nhi|t dp, mau nau do ciia khi NO2 nhat
dan, chiing to can bang chuyen djch theo chieu thuan
Vay chieu thu nhif t la chieu thuan (A H > 0) ^ Dap an C ^ Bai 9:
(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 15dm 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:
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