Phần 1 tài liệu Các dạng điển hình và phương pháp giải nhanh bài tập trắc nghiệm Hóa học 12 giới thiệu tới người đọc các dạng bài tập áp dụng cách tính nhanh, một số phương pháp giải nhanh bài tập trắc nghiệm áp dụng các định luật. Mời các bạn cùng tham khảo.
Trang 1N G U Y E N T U Y E N H A
CAC DANG fillN HiNH VA
BIEN SOAN THEO CHUDNG TRINH MCJI DANH CHO HQC SINH BAN CO BAN VA NANG CAO
ON LUYEN THI TU TAI, DAI HQC VA CAO DANG
rH(i V!EN TINHBiNH THU-V^
N H A X U A T B A N D A I H Q C Q U O C G I A T P H O C H I M I N H
Trang 2C A C DANG O I E N HINH VA PHl/dNG P H A P GIAI NHANH
BAI T A P T R A C N G H I E M H6AHOCI2 Nguyen Tuyen Ha
N H A X U A T B A N
D A I H O C QUdc G I A T P H O C H I M I N H
Khu pho 6, phirdng Linh Trung, quan Thu Dutc, TP.HCM
So 3 Cong tradng Quoc te, quan 3, TP ROM
D T : 38 239 172, 38 239 170
Fax: 38 239 172 - Email: vnuhp@vnuhcm.edu.vn
Chiu track nhiem xudt bdn
S6' dang ky ke hogch xuat bin: 494-2013/CXB/07-25/DHQGTPHCM
Quyet dinh xuSt bhn so: 333/QD-DHQGTPHCM ngay 21 thang 6 nam 2013 cua Nha
xuat bdn DHQGTPHCM
In t?i Cong ty In Song Nguy6n, n^p lau chilu quy I V nSm 2013
L C l I N(&I D A U
De h6 trg viec hoc tap va on thi TU TAI - DAI HOC - CAO D A N G
hang nam, toi viet cuon sach n^y theo tinh than giiip cac em trang bi
du kien thiJc de tham diT k i thi dat ket qua tot nhat
Ve noi dung cuon sach cd 3 phan:
+ Phan I : Cac dang bai tap dp dung cong thiJc tinh nhanh + Phan I I : Cac phiicfng phap giai bai tap trac nghiem + Phan I I I : Gidi thieu 05 de thi thuf Tu tai va 06 de thi thuf Dai hoc
de cac em thijf siJc minh
Cac de thi thuf mang tinh he thong day du cau hoi l i thuyet va hki
tap Sau do cac em tham khao ddp an de rut kinh nghiem trong k i t h i sSp tdi nham dat ket qua nhuf mong doi Chiing toi hi vong cuon sdch nay la tai lieu tham khao hCJu ich giup cac em ren luyen k i nSng, nang cao kien thiJc cua minh va dat ket qua tot nhat trong k i t h i
TU T A I - DAI HOC - CAO DANG s^p tdi Mac du rat co g^ng de bign
soan nhuhg vdi thcfi gian c6 han chac ch^n van cbn nhufng khiem khuyet Tac gia rat mong nhan dacfc sir gop y xay diTng cua dong nghi?p, doc gid
gan xa de nhufng Ian sau tai ban se duac hokn thien hdn
Xin tran trong cam cfn
Tac gid Nguyen Tuyen Hd
Trang 3T
CACH TiNH NHANH
C a c h tinh nhanh so dong phan ciia:
- Ancol no, dctn chiiCc (C„H2n+20): 2n-2 ( 1 < n < 6)
- Andehit dcfn chvCc, no (CnH2„0): 2n-3 (2< n < 7)
- Axit cacboxylic dcfn chiic, no (C„H2n02): 2n-3 (2< n < 7)
- Este no, dcfn chtic (C„H2„02): 2n-2 (1 < n < 5)
- Ete darn chiic, no (C„H2n+'20): (n-l)(n-2)
Doi vdi hat loai axit tren thi chi phdn ling voi nhUng kim loai diing
trade H trong day hoat d6ng hoa hoc
De tinh khoi luong muoi thu diTcfc thi
•k Dung dich H2SO4: m^^^i sunfat = nihSn h(7p kim loai + 96 n,i^
* Dung dich HCI: nimudi dorua = nihSn hgp kim loai + 71 nj,^
Trang 4B A I T A P A P D U N G
B a i 1: Cho l,04g hon hop hai kim loai tan hoan toan trong dung dich
H2SO4 loang du thodt ra 0,672 Ht khi H 2 (dktc) Khoi luong hon hop
muoi sunfat khan thu di/oc la:
A 3,92g B l,96g C.3,52g D.5,88g
HUdng dan gidi
2 kim loai + H2SO4I hh muoi sunfat + H 2
n „ - - | | f = 0,03(mol)
Ap dung C O n g thufC: mmutfi sunfat = mhSn hap kim lo?i + 96 n„^
= 1,04 + 0,03.96 = 3,92 (gam) Chon dap an A
B a i 2: Hoa tan het 3,53 gam hon hop A gom ba kim loai Mg, Al va Fe
trong dung dich HCl, c6 2,352 l i t khi hidro thoat ra (dktc) va thu daoc
dung dich D Co c a n dung dich D, thu difdc m gam h6n hop muoi
k h a n T r i so ciia m la:
A 12,405 gam B 10,985 gam C 11,195 gam D 7,2575 gam
HU&ng d&n gidi
n H - | ^ = 0,105(mol)
Ap dung cong thuTc: m m u e i doma = nihSn iwp kim loai + 7 1 n„^
= 3,53 + 0,105 71= 10,985 gam Chon ddp 6n B
B a i 3: Hoa t a n hoan to^n 2,17 gam h o n hap 3 kim loai A, B, C trong
dung dich HCl da thu diTOc 2,24 l i t khi H 2 (dktc) va m gam muoi Gia
tri cua m 1^
A 9,27 gam B 5,72 gam C 6,85 gam D 6,48 gam
Hiidng dan gidi
n„ =?^ = 0,l (mol)
22,4
m„us-iciorua = nihSn h;>p kim lo^i + 71 n,,^ = 2,17 + 0,1 71 = 9,27 gam
Chon dap an A
B a i 4: Bok tan 1,19 gam hSn hop A gom Al, Zn b&ng dung dich HCl viTa
du thu difoc dung dich X vk V lit khi Y (dktc) Co can dung dich X
di/oc 4,03 gam muoi khan Gia t r i cua V 1^ ,
A 0,224 lit B 0,448 lit C 0,896 lit D 1,792 lit
Hii&ng dan gidi
Ap dung cong thiJc: m^ua-i ciorua = nihSn hap kim loai + 71 n^^
4,03-1,19 r^r^A ^ ^^
nji^ = ^ ^ ^ p — = 0,04 (mol)
^ V„^ = 0,04 22,4 = 0,896 l i t Chon dap an C
B a i 5: Cho 5,35 gam h6n hop X gom Mg, Fe, Al vao 250ml dung dich Y gom H 2 S O 4 0,5M va HCl I M thu duoc 3,921it khi H 2 (dktc) va dung dich A Co can dung dich A trong dieu kien khong c6 khong khi, thu diTOc m gam chat rSn khan Gia t r i cua m la
A 20,900 gam B 26,225 gam C 26,375 gam D 28,600 gam
Hitdng dan gidi
nn^so, = 0,25.0,5 = 0,125 mol =^ n^,, = 0,125 2 = 0,25 mol
Vi axit dir nen kho'i \\iang rin b^ng tong kho'i lifcfng kim loai va cdc
ion C O trong dung dich
mr^nkhan = 5,35 + (0,125.96) + (0,25.35,5) = 26,225 gam Chon dap an B
B a i 6: (DH khoi A.2010): Cho m gam hon hap bot X gom ba kim loai
Zn, Cr, Sn c6 so mol b^ng nhau tac dung het vdi lirotng diT dung dich
Trang 5H C l loang, nong t h u dugc dung dich Y k h i Ha- Co can dung dich
Y thu dtfdc 8,98 gam muoi khan Neu cho m gam hon hop X tac
dung hoan toan vdi O2 (du) de tao hon hop 3 oxit t h i the tich k h i
(dktc) phan ijrng la
A 2,016 l i t B 0,672 l i t C 1,344 Ht D 1,008 l i t
Hii&ng dan gidi
3 k i m loai t r e n k h i phan ilng v d i H C l loang nong deu h i oxi hoa
t h ^ n h so oxi hod +2 (X + H C l thu di/Oc muoi: ZnClz, CrClz, SnClg.)
- Con k h i tac dung O2, Zn tao +2, Cr tao +3, Sn tao +4 (X + O2 t h u
daoc cac oxit: ZnO, CraOa, Sn02)
- Goi so mol moi k i m loai la x (mol) t h i :
V i cac k i m loai c6 so mol bSng nhau nen cAc muoi c6 so mol b^ng
B a i 7: Cho 11 gam hon hop A l va Fe tac dung het v6i dung dich H C l
thu dirge 8,96 l i t H2 (dktc) Phan t r a m khoi lugng cua Fe trong hon
B a i 8: (De KA.2007): Cho m gam hon hop Mg, A l vao 250ml dung dich
chufa hon hop H C l I M va H2SO4 0,5M thu dugc 5,32 l i t k h i H2 (dktc) v^ dung dich Y Coi the tich dung dich khong doi Dung dich Y c6 p H 1^
A 1 B 6 C 2 D 7
Hiidng dan gidi
n„^o, = 0.25 0,5 = 0,125 mol => n,j = 0,125 2 = 0,25 mol
nHci = 0,25 1 = 0,25 mol => n„ = 0,25 mol
=> tdng so mol H^ = 0,25 + 0,25 = 0,5 mol
5,32 ^ n o r , , - / — V „
= 0,2375(mol) 22,4
Ta c6: 2 H * Ban dau: 0,5 mol Pu: 0,475 <-Sau pir: 0,025mol Sau phan iJng axit diT H *
=1
H2 0,2375 mol
I 0,025 0,25 = 0,1M
p H = - l o g [ H ^ ^
Chon dap a n A
B a i 9: (De KB.2007): Cho 1,67 gam hon hop 2 k i m loai d 2 chu k y ke
tiep nhau thuoc nhom I I A tac dung het vdi dung dich H C l dii, thoat
ra 0,672 l i t k h i H2 (dktc) H a i k i m loai do la (Be = 9,Ca = 40,Mg = 24,
^ Sr = 87,Ba = 137)
A M g va Ca B Ca va Sr C Sr va Ba D Be va M g
Trang 6Hiictng dan gidi
Dat 2 k i m loai 1^ A
T i le mol gifla k i m loai va hidro b^ng ho^ t r i k i m loai chia 2 n6n t i le
mol la 1: 1
= MZ2^ 0,03 (mol) 22,4
A Ha
0,03mol 0,03mol
MA = ^ = 55,67 ^ Ca va Sr
Chon dap an B
B a i 10: Cho 3,87gam h6n hop X g6m M g va A l v^o 250ml dung dich X
gom H C l I M va H2SO4 0,5M thu dMc dung dich B va 4,368 Ht
H2(dktc) Phan t r S m k h o i laong M g va A l trong X tucfng ufng la
A 37,21% M g va 62,79% A l B 62,79% M g va 37,21% A l
C 45,24% M g va 54,76% A l D 54,76% M g va 45,24% A l
B^i nay l a m tifcfng tif bai 5
B a i 1 1 : Hoa tan m(g) h6n hop Zn va Fe can vCra du 11 dung dich H C l
3,65M (d = l,19g/ml) t h u dugc mot chat k h i va 1250g dung dich D
B KIM LOAI TAG D y N G VC3l HNO3, H2SO4 (d|c)
A x i t H2SO4 (dSc) + k i m loai = muoi sunfat + san pham khuf + H2O
San phdm khii c6 thi Id SO2 (nhan 2e), H2S (nhan 8e),S (nhan 6e)
* m m u o i sunfat = "1 kin, loai + molgpk.so 6 n h a n i 96
axit HNO3 + k i m loai = muoi n i t r a t + san pham khijf + H2O
San phdm khd c6 the la NO2 (nhan le), NO (nhan 3e), N2O (nhan 8e), N2 (nhan lOe), NH4NO3 (nhan 8e)
* m muoi nitrat = H I kim loai + molgpk.so e nhan.62
Luu y: A l , Fe, Cr khong phan ting vcd HNO3 dSc nguoi va H2SO4 dac nguoi
BAI TAP VAN DUNG
B a i 1: Cho 1,86 gam hon hdp A l va M g tac dung vdi dung dich HNO3 loang dtf t h i thu di/cJc 560 m l l i t k h i N2O (dktc, san pham khuf duy
nhat) bay ra K h o i liTOng muoi n i t r a t tao ra trong dung dich la:
A 40,5 gam B 14,62 gam C 24,16 gam D 14,26 gam
HtCclng dan gidi
n ^ o = ^ ^ - 0 , 0 2 5 m o l N,o 22,4
Ap dung cong thuTc: m musl nitrat = m kim loai + molspk -so e nhan.62
= 1,86 + 0,025.8.62 = 14,26 gam Chon dap an D
B a i 2: Hoa tan hoan toan 9,94 gam hon hop X gom A l , Fe, Cu trong lugng du dung dich HNO3 t h u dugc 3,584 l i t k h i N O duy nhat (dktc)
Tong k h o i lugng muoi k h a n tao t h a n h la:
A 39,7gam B 29,7gam C 39,3gam D 27,7gam
Hiidng dan gidi
n„„=||M.o,16n.o
A p dung cong thUc: m m u o i nitrat = m k i m loai + molspk so e nhan.62
= 9,94 + 0,16.3.62 = 39,7 gam
Chon dap an A
Trang 7Bai 3: Hoa tan 4,97 gam hon hap Al, Cu, Fe trong dung dich HNOr,
loang dii thu dUcfc 1,792 Ut khi NO(dktc) Tong khoi luong muoi khan
tao thanh:
A 19,85 gam B 26,5 gam C 39,7 gam D 40,2 gam
Bai nay gidi tuang tii bai 2
B a i 4: Hoa tan hoan toan 13,68 gam hon hdp X gom Al, Cu, Fe bSng
dung dich HNO3 loang, diX thu diroc 1,568 lit khi N2O (dktc) va dung
dich chuTa m gam muoi Gia tri cua m la
A 48,40 gam B 31,04 gam C 57,08 gam D 62,70 gam
Bai nay gidi tuang tu bai 2
Bai 5: (DH khoi A.2010): Nung 2,23 gam hon hop X gom cac kim loai
Fe, Al, Zn, Mg trong oxi, sau mot thdi gian thu difdc 2,71 gam hon
hop Y Hoa tan hoan toan Y vao dung dich HNO3 (da), thu dtfOc 0,672
Ht khi NO (san pham khuf duy nhat, d dktc) So mol HNO3 da phan
Ta CO bao toan nguyen to N:
^ ' ^ H N O j ~ ^ U N O j (pu O trong oxit) ^ H N O j (pu oxi ho4 - khut) ^ N O
Mat khac:
HHNO^ (pu o t^ng oxiu = ( cho H^O);
n H N o , ,pu ox, ho - uha, = 3.n^o ( Do N^^ — ^ NO)
^ ^ ^ i i N o , = 2no,oxit) + 3nf^o+ n^o = ^T^o (<«iM+ '^^m
= 2.0,03 + 4.0,03 = 0,18 mol Chon dap an D
B a i 6: (BH khoi B.2008): Cho 2,16 gam Mg tAc dung vdi dung dich
HNO3 (du) Sau khi cac phan iJng xay ra hoan toan, thu duoc 0,896 lit
NO (d dktc) va dung dich X Khoi luong muoi khan thu diTOc khi lam
bay hoi dung dich X la
A 8,88 gam B.13,92 gam C 6,52 gam D.13,32 gam
Hii&ng dan gidi
n^g = 246 : 24 = 0,09 (moDin^o - 0,896 : 22,4 = 0,04 (mol)
Cdch 1:
3Mg + 8HNO3 ^ 3Mg(N03)2 + 2N0 + 4H2O
< 0,06 < 0,04 0,06 <
2e 2.0,09
m = 0,09 148 + 0,0075 80 = 13,92 (gam) -> chon B
B a i 7: Hoa tan 23,4 gam G gom Al, Fe, Cu bang mot lirong vCra du dung
dich H2SO4 d5c, nong, thu di/dc 15,12 lit khi SO2 (dktc) va dung dich
chuTa m gam muoi Gia tri cua m la
A 153,0 gam B 95,8 gam C 88,2 gam D 75,8 gam
Trang 8B a i 8: Kok tan ho^n to^n m gam hon X gom A l , Fe, Cu vko dung dich
H N O 3 dSc nong du, thu duoc dung dich Y chufa 39,99 gam muoi va
7,168 Ht k h i N O 2 (dktc) Gia t r i cua m la
A 20,15 gam B 30,07 gam C 32,28 gam D 19,84 gam
Hiicfng ddn gidi
HMO = ^ ^ ^ = 0,32mol
"""^ 22,4
Ap dung cong thiJc: m raM nitrat = m kim loai + molspk.so e nhan.62
=^ m kim loai = m muoi nitrat " molspk-SO G nhan.62
= 39,99 - 0,32.1.62 = 20,15 gam Chon dap an A
B a i 9: (DHKB 2007)
ThUc hien hai t h i nghiem:
1 Cho 3,84 gam Cu phan iJng v d i 80 m l dung dich HNO3 I M thodt ra
V i l i t NO
2 Cho 3,84 gam Cu phan ling vdi 80 m l dung dich chufa H N O 3 I M va
H 2 S O 4 0,5 M thoat ra V 2 Ht NO.Biet NO la san pham khijf duy
nhat, cac the tich k h i do d cCing dieu ki$n
Quan he giOfa V i va V2 la
A V2 = V i B V2 = 2Vi C V2 = 2,5Vi D V 2 = l , 5 V i
Hiictng dan gidi
T N I : n c „ = ^ = 0,06 mol b4 n„, =0,08 mol n^^_ = 0,08 mol
3Cu + 8 H " + 2NO3- > 3Cu^^ + 2 N 0 T + 4H2O
Dau bai: 0,06 0,08 0,08 -> phan iJng het
Phan iJng: 0,03 <- 0,08 -> 0,02 ^ 0,02 mol
=> V i tirang ufng vdi 0,02 mol NO
TN2: ncu = 0,06 mol; nn^o, = 0,08 m o l ; nn^so = 0,04 mol
=> Tong n^^, = 0,16 mol; n ^ ^ = 0,08 mol
3Cu + 8H* + 2NO3" >3Cu^* + 2NO^ + 4H2O
Dau bai: 0,06 0,16 0,08 -> Cu va H"^ phan iJng het Phan iJng: 0,06 ^ 0,16 ^ 0,04 -> 0,04 mol
=> V2 ti/ong urng v d i 0,04 mol NO
Nhir vay V2 = 2 V i Chon dap an B
B a i 10: (DH khoi B-2009): K h i hoa tan hoan toan 0,02 mol Au bang
ntfdc cudng toan thi so mol H C l phan ilng va so mol NO (san pham
khuf duy nhat) tao t h a n h Ian liToft l a
B a i 11: Hoa tan 3 gam hon hgfp A gom k i m loai R hoa t r i 1 va k i m loai
M hoa t r i 2 vifa du vao dung dich chda HNO3 va H2SO4 va dun nong,
thu dtroc 2,94 gam hon hap k h i B gom NO2 va SO2 The tich cua h5n
hop k h i B la 1,344 l i t (dktc) K h o i luang muoi k h a n thu diroc la
A 6,36g B 7,06g C 10,56g D 12,26g
HUdng dan gidi
^ 1^344 ^ oGmol dSt NO2 x mol, SO2 ymol
22,4 r46x + 64y = 2,94 fx = 0,05
x + y = 0,06 [ y = 0,01
nimuai = 3 + (0,05 62 + 0,01.-.2 96) = 7,06 gam
2
• Dap an B
B a i 12: Cho 8,3 gam h5n hcfp A l va Fe tac dung v d i dung dich HNO3
loang dtr t h i thu dircfc 45,5 gam muoi n i t r a t khan The tich k h i N O (dktc, san pham khuf duy nhat) thoat ra la:
! A 4,48 l i t B 6,72 l i t C 2,24 l i t D 3,36 l i t
Trang 9Hiicmg dan giai
Ap dung cong thiJc: m musi nitrat = ni ,0,1 + molspk-so e nhan.62
^ D N O = "-^-"^ ~ = = 0,2 mol
so e nhan.62 3.62
=> = 0,2.22,4 = 4,48 l i t
Chon dap an A
Bai 13: Hoa tan hoan toan 1,23 gam hon hop X gom Cu va A l vao dung
dich HNO3 dSc, nong thu dixoc 1,344 l i t khi NO2 (san pham khuf duy
nhat, d dktc) Phan trSm ve khoi liiong cua Cu trong hon hop X la
Bai 14: Hoa tan 15 gam hon hop X gom hai kim loai Mg va A l vao dung
dich Y gom H N O 3 va H2SO4 dSc thu diioc 0,1 mol moi khi SO2, NO,
NO2, N2O Phan tr5m khoi Itfong cua A l va Mg trong X Ian lacft la
A 63% va 37% B 36% va 64%
C 50% va 50% D 46% va 54%
HU&ng dan gidi
Dat H M g = X mol; D A I = y mol Ta c6:
Qud trinh oxi ho^: Mg -> Mg^* + 2e A l
0,1 0,1 0,2 0,1
=> Tdng so mol e nhan bkng 1,4 mol
24x + 27y = 15 fx = 0,4 mol Theo dinh luat bao toan electron:
Phi/dng trinh tdng quat
MzOn + 2nHCl 2MCln+ nHzO
M20„ + nH2S04 (loang) ^ M2(S04)„ + nHzO
Cach tmh nhanh cho trie nghi?m
* Doi vdi axit H2SO4 (loang)
Khoi li/ong muoi thu dugc la: m^^ai ^unfat = m h S n hap o x u k i m loai + n,[^so, -80
Doi vdi a x i t H C l • Khoi luong muoi thu duoc la: m,„„a-i ciorua = n i h o n hop oxit kim loai + 27,5 n,,^;, BAl TAP VAN DgNG
Bai 1: Cho 50g hon hop hot oxit kim loai gom ZnO, FeO, Fe203, Fe304, MgO tac dung het vdi 200ml dung dich H C l 4 M (vCra du) thu duoc
dung dich X Luong muoi c6 trong dung dich X
bSng.-A 79,2g B 78,4g C 72g D 72,9g
Trang 10HUcfng ddn gidi
Ap dung C O n g thufC: nimudi cloma = nihon h<7p oxit kim loai + 27,5 n^jp,
= 50 + (0,2.4.27,5) = 72 gam Chpn dap an C
Bai 2: Hoa tan hoan toan 2,81 gam hon hop A gom FezOg, MgO, ZnO
bkng 300ml dung dich H 2 S O 4 0,1M (vCra du) Co can can than dung
dich thu diioc sau phan ufng thi thu difoc lirong muoi sunfat khan la:
A 3,81 gam B 4,81 gam C 5,21 gam D 4,8 gam
HU&ng dan gidi
Ap dung COng thufC: mmum sunfat = nihSn hop oxit kim loai + J^HaSOi - ^ ^
= 2,81 + (0,3.0,1.80) = 5,21 gam Chon dap an C
Bai 3: De tac dung vCra du vdi 7,68g hon hop gom FeO, Fe304, FegOs can
dung 260 ml dung dich HCl I M Dung dich thu diTOc cho tac dyng vdi
NaOH du, ket tua thu Auac mang nung trong khong khi den khoi
iLfOng khong doi duoc m gam chat r^n Gia t r i cua m la:
A.6g B 7g C.8g D.9g
Hii&ng dan gidi
oxit kim loai + HCl > muoi clorua + H 2 O
nnc = = 0,26.1 =0,26 mol
2H^ + O'- > H 2 0
0,26 > 0,13mol
5 6
=> mpe = 7,68 - 0,13.16 = 5,6 gam => ^fe = = 0>l"iol
So do hop thiJc: 2Fe ^ FeaOg
0,1 -> 0,05 mol
=> mp^^o^ = 160.0,05 = 8 gam
Chon dap an C
Bai 4: Oxi hoa cham m gam Fe ngoai khong khi sau mot thcJi gian thu
duoc 12 gam hon hop X g6m(Fe, FeO, FezOg, Fe304) De hoa tan het
X, can vCra du 300 ml dung dich HCl I M , dong thdi giai phdng 0,672
0,24 ->0,12mol
m = m x - mo(oxit) = 12 - 0,12.16 = 10,08 gam Chon dap an A
Bai 5: Hoa tan hoan toan 2,8 gam hon hop FeO, Fe203 va Fe304 can vCra
du V ml dung dich HCl I M , thu di/oc dung dich X Cho tii tii dung dich
NaOH dir vao dung dich X thu dugfc ket tua Y Nung Y trong khong
khi den khoi liiOng khong doi thu dirge 3 gam chat r^n Tinh V?
A 87,5 m l B.125 m l C.62,5 m l D.175 m l
Hii&ng dan gidi
- 4 chat rin la Fe203
Bai 6: Hoa tan hoan toan 5,4 gam mot oxit sSt vao dung dich H N O 3 dii
thu dagc 1,456 lit hon hop NO va NO2 (dktc va khong con san pham khuf nao khac) Sau phan ufng khoi lUOng dung dich tang len 2,49 gam
so vdi ban dau Cong thufc cua oxit s^t va so mol H N O 3 phan ufng la:
A FeO va 0,74 mol B Fe304 va 0,29 mol
C FeO va 0,29 mol D Fe304 va 0,75 mol
Trang 11D i e u k i e n : (M Id kim lo<^i diKng sau Al trong day di?n hod)
- K h i khuf o x i t k i m l o a i b S n g cac c h a t khuf CO ( H 2 ) t h i C O ( H 2 ) l a y o x i
B a i 3: Cho V l i t (dktc) k h i Hg d i qua h o t CuO dun n o n g , t h u duoc 32 g Cu
N e u cho V l i t H 2 d i qua hot FeO d u n n o n g t h i ItfOng Fe t h u duoc l a :
Trang 12B a i 4: Khuf ho^n toan 32g hon hop CuO va FezOg b^ng k h i H2, tha y tao
r a 9 g nude K h o i lirong h6n hop k i m loai t h u duoc la:
A 12 g B 16g C 24 g D 26 g
Hitdng d&n gidi
9
nodrong oxit) = = n„^o = = "^'^
niO (trong oxit) = 16 0,5 = 8g
m kim i o , i = 32 - 8 = 24 g
Chon dap an C
B a i 5: Cho k h i CO qua ong diTng a (g) hon hop gom CuO, Fe.304, FeO,
A I 2 O 3 nung nong K h i thoat ra duoc cho vao niidc vol trong du thay c6
30g ket tua trang Sau phan ufng, chat rSn trong ong suf c6 k h o i luong
202g K h o i lirong a (g) cua hon hop cac oxit ban dau la:
B a i 6: (CD -2009): Khuf hoan toan mot oxit s a t X d nhie t do cao can
viia du V l i t k h i CO (d dktc), sau phan ufng thu dugc 0,84 gam Fe va
0,02 mol k h i CO2 Cong thufc cua X va gia t r i V Ian luot la
B a i 7: Nung nong hon hop X gom PbO v^ FeO v d i mot l u o n g C vijfa du
Sau k h i phan ufng xay ra hoan toan thu dUoc hon hop chat r ^ n Y va
k h i khong mau Z Dem can hon hop rSn Y thay k h o i liTOng giam 4,8g
so vdi hon hop X Cho hon hop Y tac dung vdi dung dich H C l duf, t h u
di/oc chat k h i A Sue k h i Z vao dung dieh nude voi trong dii di/oc ket
tua trSng The tich k h i A (dktc) va k h o i liTOng ket tua t h u diroc l a :
A 6,72 l i t va 15g B 3,36 l i t va 30g
C 6,72 l i t va 30g D 3,36 l i t va 15g
Hitdng dan gidi
2PbO + C — ^ 2 P b + C 0 2 2FeO + C — ^ 2 F e + C 0 2
K h o i li/ong chat r ^ n Y giam so vdi h6n hop X 1^ k h o i l i / O n g O trong
oxit da b i C lay di tao CO2
• n„^ = n^hY = 0 , 3 (mol) ^ V,,^ = 0,3.22,4 = 6,72 (lit)
CO2 + Ca(0H)2 ^ CaC03 + H2O
ncaco3 = n c o , = 0,15 (mol) => mc^co, = 0,15.100 = 15(g) Vay dap an diing la A
B a i 8: Cho 0,3 mol Fe^Oy t h a m gia phan ufng nhiet nhom thay tao ra 0,4
mol A I 2 O 3 Cong thufc oxit ski la:
A FeO B Fe203 C Fe304
D Khong xac d i n h diTOc v i khong cho biet so mol Fe tao ra
Hiidng dan gidi
A l lay di oxi cua Fe^Oy de tao ra A I 2 O 3 V i vay so mol nguyen tuf O trong A I 2 O 3 va trong FoxOy phai bang nhau
Do do: 0,3 y = 0,4 3 = 1,2 => y = 4 =^ Fe304
Chon dap an C
Trang 13B a i 9: Dot chay khong ho^n toan 1 luong sSt da dung het 2,24 l i t O2 d
dktc, thu dtfoc h6n hop A gom cac oxit sat va s^t du Khuf hoan toan
A bang k h i CO diS, k h i di ra sau phan ling duoc dSn vao b i n h difng
n i / d c voi trong d U K h o i l i i O n g ket tua thu dUOc la:
B a i 10: De khijT hoan toan h6n hop FeO va ZnO t h a n h kirn loai can 2,24
h t H2 (ot dktc).Neu dem hon hop k i m loai thu d i / a c hoa tan hoan toan
vao axit H C l t h i the t i c h k h i H2 (dktc) thu dUcJc l a :
A 4,48 l i t B 1,12 l i t C.3,36 l i t D.2,24 l i t
Hii&ng dan giai
2.24 „ , ,
n h h o x i t = J ^ H j = n(hh kim loai) = = u , i moi
K h i hoa tan h6n h g p k i m loai vao axit t h i :
n„ = n hh kim loai = 0 , l m o l
VH^= 22,4.0,1 = 2,24 l i t
Chon dap a n D
B a i 11: Thoi mot luong k h i CO dtf d i qua ong dung hon hap hai oxit
Fe304 v a CuO nung nong den k h i phan ufng xay ra hoan toan t h u dugfc
2,32 g hon hop k i m loai K h i thoat ra duac dua vao b i n h dung dung
d i c h Ca(0H)2 du thay c6 5g ket tua t r i n g K h o i lugfng h5n hap hai
oxit k i m loai ban dau l a :
A 3,12g B 3,21g C 4g D 4,2g
Hii&ng dan giai
nO(trong oxit) = Hco = HCQ^ ^ " c a C O a = 0,05(mol)
m o x i t k i m i o ^ i = n i k i m i o , i + m o x i = 2,32 + 0,05.16 = 3,12 (g)
Chon d a p dn A
B a i 12: Khuf 39,2g mot hon hop A gom FegOa va FeO b&ng k h i CO t h u
duoc hon hop B gom FeO va Fe B tan vUa du trong 2,5 l i t dung dich
H2SO4 0,2M cho ra 4,48 l i t k h i (dktc) T i n h k h o i luong FezOg va FeO
trong hon hop A
A 32g FezOs; 7,2g FeO B 16g Fe203; 23,2g FeO
C 18g FeaOg; 21,2g FeO D 20g Fe203; 19,2g FeO
HUdng dan giai
mp^o = 0,1.72 = 7,2 (g)
Vay dap an dung la A
B a i 13: De khuf hoan toan 45 gam hon hgp gom CuO, FeO, Fe304, Fe va MgO can dung vCra du 8,4 l i t CO d (dktc) K h o i lugng chat r i n t h u
dugc sau phan u"ng la:
A 39g B 38g C 24g D 42g
Hii&ng ddn giai
K h o i lugng chat r ^ n sau phan iJng la: 45 - 1 6 - ^ ^ = 39g Chon dap an A
B a i 14: Khuf hoan to^n 17,6 gam hon hgp X gom Fe, FeO, Fe203 can
2,24 l i t CO (or dktc) Klio" lugng sSt thu dugc la
A 5,6 gam B 6,72 gam C 16,0 gam D 8,0 gam
Trang 14Hii&ng dan gidi
2 24
K h o i luong chat r a n sau phan ufng la: 17,6 - 1 6 - ~ - = 16,0 gam
Chon dap a n C
B a i 15: De k h i l hoan to^n 30 gam hon hop CuO, FeO, FezOg, Fe304,
MgO can dung 5,6 l i t k h i CO (6 dktc) K h o i lifong chat r a n sau phan
ufng la
A 28 gam B 26 gam C 22 gam D 24 gam
HUdng dan gidi
5 6
K h o i liiang chat r a n sau phan ufng la: 30 - 1 6 - ^ ^ = 26 gam
Chon dap an B
B a i 16: Dan tCf t\i V l i t k h i CO (6 dktc) di qua mot ong sil difng lUdng d i i
hon hop r a n gom CuO, FezOs (d n h i e t do cao) Sau k h i cac phan ufng
xay ra hoan toan, thu duoc k h i X D i n toan bo k h i X d t r e n vao lUOng
diT dung dich Ca(0H)2 t h i tao t h a n h 4 gam ket tua Gia t r i cua V la
A 1,120 l i t B 0,896 l i t C 0,448 l i t D 0,224 l i t
Hii&ng dan gidi
nco = nco, = nc«co, = ^ = ^'^^
=> Vco = 0,04 22,4 = 0,896 l i t
Chon dap an B
B a i 17: T h o i mot luong k h i CO di qua ong suf ditog m gam hon hcfp
Fe304 va CuO nung nong thu diTOc 2,32 gam hon hap r a n Toan bo
k h i thoat ra cho hap t h u het vao b i n h dung dung dich Ca(0H)2 du t h u
dtfoc 5 gam ket tua Gia t r i cua m la:
A 3,22 gam B 3,12 gam C 4,0 gam D 4,2 gam
Hii&ng dan gidi
m h S n h o p = 2,32 + 16 = 3,12 gam
Chon dap an B
B a i 18 Cho dong k h i CO du di qua hon hap (X) chufa 31,9 gam gom
AI2O3, ZnO, FeO va CaO t h i th u difcfc 28,7 gam hon hgp chat r a n (Y)
Cho toan bo h6n hop chat r a n (Y) tac dung vdi dung dich H C l dif t h u
duoc V l i t H2 (dkc) Gia t r i V la
A 5.60 l i t B 4,48 l i t C 6,72 l i t D 2,24 l i t
Hii&ng dan gidi
K h o i luong nguyen tuf oxi = do giam cua chat r ^ n
FeO + 2e -> Fe ^ Fe^* + 2e)
2W + 2e -> H2
Vay so mol nguyen tuf O = so mol C O = so mol H2 = 0,2 mol
Chon dap an B
C BAI T A P V A N D U N G
B a i 1: Cho luong k h i C O di qua m gam Fe203 dun ndng, t h u di/oc 39,2
gam h6n hop gom bon chat r a n la sat k i m loai va ba oxit cua n6,
dong thcfi CO hon hop k h i thoat ra Cho h6n hop k h i nay hap t h u vao
dung dich nudc voi trong c6 dii, t h i t h u di/Oc 55 gam ket tua T r i so cua m la:
A 48 gam B 40 gam C 64 gam
D Tat ca deu sai, v i se khong xac d i n h duoc
B a i 2: Cho luong k h i H2 c6 dif di qua ong suf c6 chufa 20 gam hon hop A gom MgO va CuO nung nong Sau k h i phan iJng hoan toan, dem can
l a i , thay k h o i liXOng chat r a n giam 3,2 gam K h o i liicfng moi cha't trong hon hop A la:
A 2gam; 18gam B 4gam; 16gam
C 6gam; 14gam D 8gam; 12gam
B a i 3: Cho luong k h i CO (du) di qua 9,1 gam hon hop gom CuO va AI2O3 nung nong den k h i phan ufng hoan toan, thu dUdc 8,3 gam chat r a n
K h o i luong CuO c6 trong hon hop ban dau la
A 0,8 gam B 8,3 gam C 2,0 gam D 4,0 gam
Trang 15Dang 4: KIM LOAI TAG DUNG VOI DUNG DICH MUOl
Ca'^ Na" Mg'^ A l ' " Zn'* Cr'* Fe^* N i ' * Sn'* Pb'* Fe^* H * Cu'*
Tinh khvC kim loi^i gidm
- Chieu phan ufng: Chat oxi hod m^nh + Chat khii mg,nh -> Chat
oxi hod yeu + Chat khvi yeu
P T : C u ' * + F e F e ' * + C u
Dicing bai tap nay can lUu y den quy tdc a
Ddu a cang Ian khd ndng phdn ling xdy ra cdng manh
B A I T A P A P D U N G
B a i 1: Cho O,lmol Fe vao 500 ml dung dich AgNOs I M thi dung dich thu
difOc chufa:
A AgNOg B AgNOs va Fe(N03)2
C Fe(N03)3 D AgNOa va Fe(N03)3
Hitcfng dan gidi
Vay muoi gom c6 Fe(N03)3 va AgNOs dii
B a i 2: (DH khoi B.2009): Cho 2,24 gam hot sSt vao 200 m l dung dich
chufa hon hop gom AgNOs 0,1M va Cu(N03)2 0,5M Sau k h i cac phan ufng xay ra hoan toan, t h u diTOc dung dich X va m gam chat r a n Y
Gia t r i ciia m la
A 2,80 B 4,08 C 2,16 D 0,64
Hiidng dan gidi
nFe = 0,04 mol, n^^ = 0,02 mol, n^^^ = 0,1 mol
Chat khuf m a n h n h a t se tac dung vdi chat oxi hoa m a n h n h a t triTdc:
Fe + 2 A g * ^ F e ' * + 2Ag 0,01 0,02 0,02mol
Fe + Cu'* ^ Fe'* + Cu 0,03 0,03 0,03 mol
m = 0,02.108 + 0,03.64 = 4,08g Chon dap an B
B a i 3: (CD -2009): Cho m gam M g vao dung dich chufa 0,12 mol FeCls
Sau k h i phan ufng xay ra hoan toan thu dtfoc 3,36 gam chat rkn Gia
B a i 4: (DH khoi A.2010): Cho 19,3 gam hon hop bot Zn va Cu c6 t i le mol
tuong ijfng la 1: 2 vao dung dich chufa 0,2 mol Fe2(S04)3 Sau k h i cac phan
ufng xay ra hoan toan, thu duoc m gam k i m loai Gia t r i cua m la
A 6,40g B 16,53g C 12,00g D 12,80g
Trang 16Hii&ng dan gidi
Goi X la so mol cua Zn thi so mol cua Cu la 2x
Bai 5: (CD - 2009;.- Cho mi gam A l vao 100 ml dung dich gom Cu(N03)2
0,3M va AgNOs 0,3M Sau khi cac phan ufng xay ra hoan toan thi thu
dtfgc m2 gam chat rAn X Neu cho m2 gam X tac dung vdi lugng du
dung dich HCl thi thu duac 0,336 lit khi (d dktc) Gia t r i cua mi va mg
Ian lUOt la
Ạ 8,10 va 5,43 B 1,08 va 5,16
C 0,54 va 5,16 D 1,08 va 5,43
Hi^&ng dan gidi
Theo bai ra Al con diX: 2A1 + 6HC1 -> 2AICI3 + 3H2
0,01 0,015
Al Al^^ + 3e Ag^ + le Ag Cu^^ + 2e ^ Cu
Ap dung dinh luat bao toan e: 3x = 0,03 + 0,06 => x = 0,03 mol
z:> Tong so mol Al = 0,04(mol) mi = 27.0,04 = 1,08 (g)
=> m2 = mAi + m c u + niAg = 0,01.27 + 0,03.108 + 0,03.64 = 5,43 (g)
Chon dap an D
Bai 6: (DH KẠ2008): Cho hon hdp bot gom 2,7 gam Al va 5,6 gam Fe
vao 550 ml dung dich AgNOa I M Sau khi cac phan ufng xay ra hoan
toan, thu dufcfc m gam chat r^n Gia t r i cua m la (biet thuf tu trong
day the dien hoa: FêVFế diJng trUdc AgVAg)
Ạ 64,8 B 54,0 C 59,4 D 32,4
Hii&ng dan gidi
Al + 3Ag* ^ A l ' ^ + 3Ag 0,1 0,3 0,3 mol Sau phan ufng vdi Al, Ag* c6n 0,55 - 0,3 = 0,25 mol dung phan ufng vdi Fe
Fe + 2Ag^ Fê* + 2Ag 0,1 ^ 0,2 ^ 0,1 ^ 0,2 Dir Ag^= 0,25 - 0,2 = 0,05 mol
Fé" + Ag" ^ Fê* + Ag
0,05 -> 0,05 Sau cac phan ilng chat ran la Ag c6 so mol 0,3 + 0,2 + 0,05 = 0,55 mol Khoi liigfng Ag = 0,55.108 = 59,4 gam
Chon dap an C
Bai 7: Nhiing mot thanh kem va mot thanh sat vao cung mot dung dich
C U S O 4 Sau mot thdi gian lay hai thanh kim loai ra thay trong dung dich con lai c6 nong do mol ZnS04 hang 2,5 Ian nong do mol FeS04
Mat khac, khoi li/ofng dung dich giam 2,2g I ^ o i Itrofng dong bam len
thanh kem va thanh sSt Ian lucft la:
Ạ 12,8g; 32g B 64g; 25,6g C 32g; 12,8g D 25,6g; 64g
HiC&ng dan gidi
CM(ZnS04) = 2,5CM(FeS04) ^ n^^^o, = '^MF.SO,
Zn + C U S O 4 ZnS04 + Cui 2,5x 2,5x 2,5x 2,5x
Fe + C U S O 4 - ^ F e S 0 4 + Cui
X <— X <— x —> X
Do giam khoi lagfng cua dung dich la:
Am = mcu (Mm) - nizn (tan) " " I P e (tan)
o 2,2 = 64(2,5x + x) -65.2,5x- 56x
Ax = 0,4 (mol) Vay mcu b^m len tiianh Zn = 64g; m c u bAm len thanh Fe = 25,6g Chon dap an B
Bai 8: Nhiing mot thanh graphit dUdc phu mot Idp kim loai hoa t r i I I
vao dung dich C U S O 4 dtf Sau phan ufng khoi iMng cua thanh graphit giam di 0,24g Cung thanh graphit nay neu diTdc nhiing vao dung dich
Trang 17AgNOa t h i k h i phan uTng xong khoi luang thanh graphit tang len
0,52g K i m loai hoa t r i I I la k i m loai nao sau day:
A Pb B Cd C Fe D Sn
Hiidng dan gidi
Goi k i m loai c6 hoa t r i I I do la M c6 khoi lifcfng m(g)
B a i 9: Ngam mot thanh Cu trong dung dich c6 chiJa 0,04 mol AgNOa,
sau mot thdi gian lay thanh k i m loai ra thay khoi li/dng tSng hdn so
vdfi luc dau la 2,28 gam Coi toan bo k i m loai sinh ra deu bam het vao
thanh Cu So mol AgNOs con lai trong dung dich la
A 0,01 B 0,005 C 0,02 D 0,015
HUdng dan gidi
Cu + 2AgN03 - 4 Cu(N03)2 + 2Ag
1 0 8 2 X - 64x = 2,28
X = 0,015mol n^^No^ da =0,04 - (0,015.2) =0,01 mol
Chon dap an A
B a i 10: Ngam mot vat bSng dong c6 khoi li/cJng 15 gam trong 340 gam
dung dich AgNOs 6% Sau mot thdi gian lay vat ra thay khoi luong
AgNOa trong dung dich giam 25% Khoi lUcfng cua vat sau phan iJng la
A 3,24 gam B 2,28 gam C 17,28 gam D 24,12 gam
HiCctng dan giai
B a i 11: Hoa tan 3,23 gam hon hgtp gom CuCl2 va Cu(N03)2 vao niTdc
difcfc dung dich X Nhung thanh k i m loai M g vao dung dich X den k h i dung dich mat mau xanh roi lay thanh M g ra, can lai thay tSng them 0,8 gam Khoi lU(?ng muoi tao ra trong dung dich la
A 1,15 gam B 1,43 gam C 2,43 gam D 4,13 gam
Hii&ng dan gidi
B a i 12: Ngii6i ta phu mot Idp bac tren mot vat b^ng dong c6 khoi lifOng
8,48 gam bSng each ngam vat do trong dung dich AgNOa Sau mot
th6i gian \zy vat do ra khoi dung dich, rufa nhe, lam kho can dufeJc 10
gam Khoi luong A g da phu tren be mSt cua vat la
A 1,52 gam B 2,16 gam C 1,08 gam D 3,2 gam
Trang 18Hii&ng dan gidi
Cu + 2AgN03 -> Cu(N03)2 + 2Ag
B a i 13: Cho m gam hon hcjp hot Z n va Fe vao li/cfng du dung dich
C U S O 4 Sau k h i ket thiic cac phan ufng, loc bo phan dung dich t h u
duac m gam bot r ^ n T h a n h phan % theo khoi li/ong cua Zn trong h6n
hop ban dau la
B a i 14: Cho 1,12 gam bot Fe vh 0,24 gam bot M g tac dung v d i 250 m l
dung dich C U S O 4 x M , khuay nhe cho den k h i dung dich m a t mau
xanh nhan thay k h o i liicfng k i m loai sau phan ufng la 1,88 gam Gia
K (Z=19) Is22s22p^3s^3p^4s' hay [Ar]4s'
Deu CO 1 electron d Idp ngoai cung
2 T a c d u n g v d i a x i t ( H C l , H2SO4 loang): tao muoi H2
T h i du: 2Na + 2HC1 ^ 2NaCl + H 2 t
3 T a c d u n g v d i n\i6c: tao dung dich k i e m va H2
T h i du: 2 N a + 2H2O 2 N a O H + Hgt
Trang 19III Dieu che:
1 Nguyen tSc: khuf ion kim loai kiem thanh nguyen tur
2 Phifcfng phap: dien phan nong chay muoi halogen hoSc hidroxit cua
chung
Thi du: dieu che Na b^ng each dien phan n6ng chay NaCl NaOH
PTDP: 2NaCl —^^^^^ 2Na + CI2
4NaOH ) 4Na + 2H2O + O2
B M Q T SO H O P C H A T Q U A N T R Q N G CUA K I M LOAI K I E M :
I Natri hidroxit - NaOH
+ Tac dung vdi axit: tao muoi va niTdc
Thi du: NaOH + HCl ^ NaCl + H2O
+ Tac dung vdfi oxit axit: tao mtioi va niidc
Thi du: 2NaOH + CO2 -> Na2C03 + H2O
+ Tac dung vdi dung dich muoi:
Thi du: 2NaOH + CUSO4 ^ Na2S04 + Cu(0H)2i
n Natri hidrocacbonat - NaHCOg
1 Phan v(ng phan hiiy:
T h i du: 2NaHC03 — ^ Na2C03 + COot + H2O
2 Tinh litofng tinh:
+ Tac dung vdi axit:
NaHCOs + HCl ^ NaCl + C02t + H2O
+ Tac dung vdi dung dich bazd: ?
NaHCOa + NaOH ^ Na2C03 + H2O
III Natri cacbonat - Na2C03 •*
+ Tac dung vdi dung dich axit manh:
Thi du: Na2C03 + 2HC1 ^ 2NaCl + CO2T + H2O
Muoi cacbonat cua kim loai kiem trong nadc cho moi triTdng kiem
IV Kali nitrat: KNO3
Tinh chat: c6 phan ling nhiet phan
Thi du: 2KNO3 2KNO2 + O2
C DANG TOAN CO2 (HOAC SO2) VAO DUNG DICH KIEM NaOH (HOAC KOH)
Cac pht^ofng trinh xay ra:
Thiic chat ta dung hai pt sau
Bai 1: Sue 2,24 lit khi CO2 vao 100ml dung dich NaOH I M , tinh khoi
lUdng muoi thu duac
HUdng dan gidi
^^^^ = 1 ehi tao muoi NaHCOg
0,1 moi -> 0,1 moi
^ii^ico, = 0.1-84 = 8,4 gam
Bai 2: Hap thu hoan toan 4,48 lit khi SO2 (d dktc) vao dung dich chufa
16g NaOH thu dugc dung dich X Khoi liicfng muoi tan thu duac trong dung dich X la bao nhieu?
A.20,8g B.18,9g C.23,0g D.25,2g
Hii&ng dan gidi
2NaOH+ SO2 -> NazSOg
0,4 0,2 0,2
m muoi = 0,2.126 = 25,2 gam Chon dap an D
Trang 20B a i 3: DSn 10 l i t h5n hap k h i gom N2 va C d 2 do or dktc sue vao 2 l i t
dung dich Ca(OH)2 0,02M thu difgrc I g ket tua.Tinh phan t r a m theo
the tich CO2 trong h 5 n hap k h i
B a i 4: Hap thu 3,36 l i t SO2 (dktc) vac 0,5 l i t h 5 n hgp gom N a O H 0,2M
va K O H 0,2M Co can dung dich sau phan ufng t h u duc/c k h o i I'Jcfng
muoi khan la
Theo d i n h luat bao toan nguyen tuT (tdng khoi lugng ion trong muoi
bang long khoi lugng muoi)
m^uo = 0,1.23 + 0,1.39 + 0,1 81 + 0,05.80 = 18,3 gam
Chon dap an D
B a i 5: Cho 6,72 l i t k h i CO2 (dktc) vao 380 m l dd N a O H I M , thu duac dd "
A Cho 100 m l dung dich Ba(0H)2 I M vao dung dich A di/ac m gam
ket tua Gia t r i m bang:
A 19,7g B 15,76g C 5 9 , l g D.55,16g
Hiidng dan gidi
"N^OH = Hj^^, = n^,,^ = 0,38 mol "^n^iom, = n^^^, = 0 , 1 mol
E n ^ j ^ = 0,38 + 0,2 = 0,58 mol nco, = - 0,3 mol
K h o i lirang k e t tua la: mB^cOj = 0,1.197 = 19,7 gam Chon dap an A
B a i 6: Dot chay hoan toan 0,1 mol etan r o i hap t h u toan bo san pham chay vao b i n h chufa 300 m l dd N a O H I M K h o i liiong muoi t h u difgrc sau phan iJng?
Trang 21COa + 2 N a O H ^ NazCOg + H2O
D DANG TOAN CO2 (HOAC SO2) VAO DUNG DjCH Ca(0H)2 hogc
Ba(0H)2 KIM LOAI PHAN NHOM CHINH NHOM II
T6M TAT L I T H U Y E T
1 Vj tri trong bang thong tuan hoan, tinh chat v^t If
a) Vi tri
K i m loai phan n h o m I I gom:
Beri (Be); Magie (Mg); Canxi (Ca);
Stronti (Sr); Bari (Ba) va Radi (Ra)
Trong cac chu k i cac nguyen to nay diing lien sau k h i loai k i e m
b) Tinh chdt vat li
- N h i e t do nong chay nhiet do soi thap
- La k i m loai mem (mem hon nhom)
- K h o i li/Ong rieng tiidng doi nho
2 Tinh chat hoa hoc
Cac nguyen to phan nhom chinh nhom I I c6:
- 2 electron hoa t r i (s^)
- Co ban k i n h nguyen tuf \6n
- La nhOfng chat khuf m a n h M - 2e -> M^^
Trong cac hap chat c^c nguyen to nay c6 so oxi hoa +2
a) Tdc dung vai phi kim
- Vdi oxi k h i dot nong:
2 M + O2 2 M 0 ( M la nguyen tut k i m loai)
2Ca + 0 2 ^ 2CaO
- Vdi CI2: M + CI2 -> MCI2
M g + CI2 ^ MgCl2
b) Tdc dung vai axit
_ De dang khilf ion trong dung dich axit (HCl, H2SO4 loang) t h a n h
4 M + IOHNO3 4M(N03)2 + NH4NO3 + 3H2O
c) Tdc dung vai H2O
T r o n g H2O, Be k h o n g p h a n urng, M g khuf c h a m , cac k i m loai con l a i khuf m a n h
M + 2H2O ^ M(0H)2 + H2t
C a + 2H2O Ca(0H)2 + H2T
d) Tdc dung vai dung dich muoi
- M g day cac k i m loai hoat dong yeu hcfn r a k h o i dung dich muoi
MQT SO HOP CHAT QUAN TRONG CUA CANXI
1 Canxi oxit: CaO
Trang 222 C a n x i h i d r o x i t : Ca(OH)2
L a chat r S n it t a n trong H2O
Dung dich C a (0H)2 c6 t i n h bazO yeu hon N a O H
- T a c dung vdi axit v a oxit axit tao muoi tUofng ting
- T a c dung vdi dung dich muoi
C a (0H)2 + NaaCOs CaCOai + 2 N a O H
C a ' ^ + CO^- ^ CaCOai
3 C a n x i c a c b o n a t C a C O a
C a n x i cacbonat l a c h a t r S n m a u t r a n g k h o n g tan trong H2O
C a C O s la muoi cua axit yeu v a k h o n g ben
CaCOa + 2HC1 CaCl2 + H2O + C O z t
CaCOa + 2CH3COOH -> Ca(CH3COO)2 + H2O + C O a t
ot n h i e t do t h a p CaCOa t a n dan trong H2O c6 CO2
CaCOa + H2O + CO2 -> Ca(HC03)2
Nude curng chia t h a n h 3 loai
1 Nifde ciJng tarn t h d i : la nirdc curng c6 chura ion HCO3"
2 NiXdc curng v i n h cijfu: la niJdc curng c6 chufa ion CI" hoSe SO^
3 Nifdc curng toan phan: La niidc cufng c6 chiJa dong thcfi cac ion HCO3", C r hoSe SO^-
3 T a c h a i c i i a nitofc ciifng
- Xa phong khong tan
- V a i soi mau muc n a t
- Nau thurc an lau chin, giam m u i v i
- Tao chat c&n t r o n g noi h a i l a m lang p h i n h i e n lieu
4 C a c h l a m m e m ni^cfc
Nguyen tSc: L a m giam nong do cac ion Ca^^ va Mg^* t r o n g nufdc bang
each chuyen nhij'ng ion t i i do nay vao t h a n h phan chat khong t a n Phirong phap: Phuong phap hoa hoc va phuong phap trao ddi ion
a) Phuang phap hod hoc
* Doi v d i ntrdc cufng t a m t h d i Dun nong trUdc k h i dung
Ca(HC0a)2 — ^ CaCOai + H2O + C02t
Loc bo chat k h o n g t a n , dirge nifdc mem
- Dung Ca(0H)2 v t o du de trung hoa
Ca(HC03)2 + Ca(0H)2 2 C a C 0 a i + 2H2O
Loc bo chat k h o n g t a n dtrgc nude mem
* Doi vdi nude cufng v i n h cufu va ni/dc cufng toan phan Dung dung dich Na2C0a, Na3P04
C a S 0 4 + Na2C03 -> CaC03i + Na2S04
Ca(HC03)2 + Na2C03 ^ CaCOgi + 2NaHC03
Trang 23Cả* + CO3 ^ CaC03>l
Mg'* + CO3 ^" MgCOa
b) Phuang phdp trao doi ion
Cho nutdfc cufng di qua chat trao doi ion (ionit) chat nay se hap thu
cac ion Câ^ va Mg^* the vao do la ion Na\a duoc nude mem
D^ng bai tap CO2 vao dung dich CăOH)2 hoqc BăOH)2 thi:
Dang 1: Biet nc^o,,,,^ ncaco, "co,
^Că0H)2
T H I : n^^ô - nc^cô
T H 2 : n^ô - 2.nc^,o„)^ -''^c^cô
Dang 2 : Biet n^ô, nc^oH)^ tim nc^cô
^^CatOlllj
T H I : nco, = nc.co,
T H 2 : ncacos - •^^cMom, ~ "cô
Dang 3: Biet nc^cộ "00, tim
nCaC03 _ ,
^ "CăOIl)o ~ •
Neu n^o, ^ ric^co, =^ ^CMOMU
BAI T A P V A N D U N G
B a i 1 Hap thu toan bo x mol CO2 vho dung dich chiJa 0,03 mol Că0H)2
diroc 2 gam ket tuạ gia t r i x?
Ạ 0,02 mol va 0,04 mol B 0,02 mol va 0,05 mol
C 0,01 mol va 0,03 mol D 0,03 mol va 0,04 mol
HiC&ng dan gidi
2
n r , r o = = 0,02 mol
TrUang hap 1: rico, = ntettua = 0,02 mol
TrUdng hop 2: H C O , = 2nb«,„ - ntettua = 2.0,03 - 0,02 = 0,04 mol
Chon dap an Ạ
Bai 2 : (DH A - 2008): Hap thu hoan toan 4,48 lit khi CO2 (d dktc) vao
500 ml dung dich hon hop gom NaOH 0,1M va Bă0H)2 0,2M, sinh ra
m gam ket tuạ Gia t r i cua m lạ
Ạ 19,70 g B 17,73 g C 9,85 g D 11,82 g
Hiidng dan gidi
nNaoii = " N a * = " o H " " ^ ' ^ ^ ^^^'^ "^u^iom, = ^ 3 , =0,1 mol
I n^„ = 0,05 + 0,2 =0,25 mol; n^o, M | = 0,2 mol
Bai 3: (BH A - 2009): Cho 0,448 l i t khi CO2 (d dktc) hap thu het vao
100 ml dung dich chuTa hon hop NaOH 0,06M va Bă0H)2 0,12M, thu
duoc m gam ket tuạ Gia t r i cua m lạ
Ạ 1,182 g B 3,940 g C 1,970 g D 2,364 g
Hit&ng dan gidi
nNaOH = n^^ = nQ„_ =0,006 mol; H B ^ I O H ) , = n^^,, = 0,012mol
In^jj =0,006 + 0,012.2 = 0,03 mol; nco, = = 0,02mol
Trang 24B a i 4: (DH A - 2007;.- Hap t h u hoan toan 2,688 l i t k h i CO2 (or dktc) vao
2,5 l i t dung dich Ba(0H)2 nong do a mol/1, t h u difdc 15,76 gam k e t
tua Gia t r i cua a la
A 0,032 M B 0,048 M C 0,06 M D 0,04 M
Hiiafng d&n gidi
2,688 ^ , , „ „ , _ _ 1 5 ^
n^f> = = 0,12mol; ng^co, = = 0,08mol
nco, > neaco, chi xay r a tri/dng hap 2
C„^,„„, = - ^ = 0,04M
Chon dap an D
B a i 5: Dan V h t CO2 (dkc) vao 300ml dung dich Ca(0H)2 0,5 M Sau
phan ling duac lOg k e t tua Gia t r i cua V bSng:
A 2,24 l i t B 3,36 l i t C 4,48 l i t D Ca A, C deu dung
Hildng d&n gidi
TriCang hap 1: n^o^ = nkettoa = 0,1 m o l
Vco = 0,1 22,4 = 2,24 l i t
TriCdng hap 2: n^^^ = 2nb,,o - n t e t t u a = 2.0,15 - 0,1 = 0,2 mol
Vco, = 0,2.22,4 = 4,48 l i t
Chon dap a n D
B a i 6: Thoi V l i t (dktc) CO2 vao 100 m l dung dich Ca(0H)2 I M , t h u diroc
6 gam ket tiia Loc bo ket tua lay dung dich thu ducfc dun nong l a i c6 4gam ket tiia nufa Gia t r i V la:
A.3,136 l i t B 1,344 l i t
C 1,344 l i t hoSc 3,136 l i t D 3,36 l i t hoSc 1,12 l i t
Hit6ng dan gidi
Ca(0H)2 + CO2 CaCOsi + H2O
0,06 <- 0,06 Ca(0H)2 + 2CO2 ^ Ca(HC03)2
0,08 <- 0,04 Ca(HC03)2 — ^ CaCOai + H2O + C02t 0,04 < - 0,04
I n c o , =0,08 + 0,06 = 0,14 mol => Vco,= 0,14 22,4 = 3,136 l i t Chon dap an A
B a i 7: Hap thu toan bo 0,3 mol CO2 vao dung dich chijfa 0,25 mol Ca(0H)2
Khoi lifofng dung dich sau phan uTng tang hay giam bao nhieu gam?
A Tang 13,2gam B Tang 20gam
C Giam 16,8gam D Giam 6,8gam
Hii&ng dan gidi
Be tinh khoi lixang chat sau phan i2ng tang hay giam a lay khoi liigng
vao trie khoi lilgng ra khdi dung dich niu > 0 la tang n^u < 0 la giam
Ta c6: n^o^ > n^^^Q^^^^ nen chi xay ra T H 2
J^COj - 2nbaza ~ ^kgt tua
nc,co, = 2nc„oH,, - ^co, = 2.0,25 - 0,3 = 0,2mol
Am = mco, - ixic^co, = 0,3.44 - 0,2.100 = - 6 , 8 gam
Chon dap an D
Trang 25B a i 8: Cho 5,6 h t hon hdp X gom Ng va CO2 (dktc) di cham qua 5 Kt
dung dich Ca(0H)2 0,02M de phan ilng xay ra hoan toan t h u dUdc 5
gam ket tua T i n h t i k h o i hoi cua h6n hcfp X so v d i H2
Khac dap an loai
Truang hap 2: n^^^ = 2nb^,„ - nkeuaa = 2.0,1 - 0,05 = 0,15 mol
j ^ > , 1 5 4 4 0 , 1 2 8 ^ = 3 ^ = 18,8
0,25 ""'"^ 2
Chon dap an A
B a i 9: Sue CO2 vao 200 m l h5n hgfp dung dich gom K O H I M va Ba(0H)2
0,75M Sau k h i k h i h i hap t h u hoan toan thay tao 23,6 gam k e t tua
T i n h Vco^ da dung d (dktc)
A 8,512 l i t B 2,688 l i t C 2,24 l i t D Ca A va B dung
HU&ng dan gidi
= 0,2 mol; ri^^mn, = ^Ba^* = ^' 1^"^°^
Chon dap an D
I n , , , , =0,2 + 0,15.2 = 0,5 mol;
197 = 0,12mol
a i 10: Sue 4,48 h t (dktc) CO2 vao 100ml hon hop dung dich gom K O H
I M va Ba(OH)2 0,75M Sau k h i k h i h i hap t h u hoan toan thay tao m
gam ket tua T i n h m
A 23,64g B 14,775g C 9,85g D 16,745g Giai tiicfng t i i bai t r e n
B a i 11: Thoi V m l (dktc) CO2 vao 300 m l dung dich Ca(0H)2 0,02M, t h u
duac 0,2g ket tua Gia t r i V m l la:
A 44,8 hoSc 89,6 B.44,8 hoSc 224 C 224 D 44,8
TMng tuf cac bai t r e n
Dang 6 MUQI CACBONAT TAG DUNG VO! DUNG D|CH
HCI, H 2 S 0 4 loang
Nguyen tdc cua phuang phdp nay Id dua vdo sU tang, gidm khoi lugng khi chuyen tii chat nay sang chdt khac Ung vdi 1 mol chdt
Pt: XCO3 + 2HC1 -> XCI2 + H2O + CO2T (1)
Y2(C03)3 + 6HC1 2YCI3 + 3H2O +3CO2T (2)
Cach t i n h nhanh dung cho tr&c nghiem
Theo 2 phuong t r i n h (1),(2) ta thay
n c f = 2n^^,_ (ma M^, = 35,5; M^^, = 60)
•^muo'i clorua ~ ^ h h rauoi cacbonat -'^cOo
( 7 1 - 6 0 ) Muoi cacbonat tac dung vdi dung dich H 2 S O 4 loang Pt: A C O 3 + H 2 S O 4 (loang) ^ A S O 4 + CO2T + H2O
B 2 C O 3 + H 2 S O 4 (loang) ^ A 2 S O 4 + CO2T + H2O
Cach t i n h nhanh cho muoi sunfatv
nirauo'i sunfat — rnmuo'i cacbonat ^CO^
BAl TAP VAN DUNG
B a i 1: Cho 115g hon hcrp gom A C O 3 , B 2 C O 3 , R 2 C O 3 tac dung het vdi dung
dich HCI thay thoat ra 0,4481 CO2 (dktc) Khoi liicfng muoi clorua tao ra
trong dung dich la:
Trang 26• " c o , = nH,o = = 0,02 (mol);
• rijici = 2ncô = 2.0,02 = 0,04 (mol)
Ap dung DLBTKL: m„,„6i cacbonat + rnnci - nimuó; clorua + ^Hfi ' ^ C O j
^ rriniuoi clorua = nimuoi cacbonat + n i H C l ~ ^^H^O ~ ^ C O ^
= 115 + 0,04.36,5 - 0,02 (18 + 44) = 115,22 (g) Chon dap an Ạ
Cach giai 2: Ap dung cong thiic
rtlmuoi clorua = hh muoi cacbonat + I^cOj -C^-^ ~
= 115 + 0,02.11 = 115,22 (g) Chon dap an Ạ
Bai 2: Hoa tan 5g h6n hop 2 muoi cacbonat kim loai hoa t r i I va hoa tri
I I bSng dung dich HCl thu diigfc dung dich M va 1,121 khi CO2 (dktc)
Khi CO can dung dich M thu dircKc khoi lufdng muoi khan bang:
Ạ l l l g B 5,55g C 16,5g D 22,2g
HiC&ng dan gidi
1 12
n^o = —— = 0,05mol CO, 22,4
Ap dung cong thufc: rtir^^ii clorua = nihh muoi cacbonat + n^ọ^ (71 - 60)
= 5 + 0,05.11 = 5,55 gam Chon dap an B
Bai 3: Cho 42 gam hon hap muoi MgCOg, CuCOs, ZnCOg tac dung vdi
dung dich H2SO4 loang, thu diTcrc 0,25 mol CO2, dung dich A va chat
ran B Co can dung dich A, thu diTdc 38,1 gam muoi khan Dem nung
liiong chat rSn B tren cho den khoi liiong khong doi thi thu duac 0,12
mol CO2 va con lai cac chat rSn B' Khoi lUcfng cua B va B' 1^:
Ạ 10,36 gam; 5,08 gam B 12,90 gam; 7,62 gam
C 15, 63 gam; 10,35 gam D 16,50 gam; 11,22 gam
Hitdng dan gidi
Dat h5n hop muoi la A CO3
A CO3+ H2SO4 ^ A SO4+ H2O + CO2T
Dang 7 KIM LOAI SAT B| 0X1 HOA THANH HON HPP
(Fe, FeO, FegOạ Fe304)
Dang nay c6 rat nhieu each giai khac nhau nhitog qua dai nen chi dUa
ra 2 each giai saụ
Cdch 1: Vi Fe bi oxi hoa thanh cac exit s^t c6 the con s^t dif (Fe,
FeO, FeaOs, Fe304) Tom lai ta c6 the coi hon hop chi c6 sat va oxi nen ta c6 CTPT chung la FêOy
FêOy - (3x - 2y)e xFê^ + yỐ
Ta luon c6 he phiiong trinh sau
56x + 16y - m,5„
hop oxit sat
3x - 2y = mol san pham khuf.so e nhan
Cdch 2: Van dung cong thitc tinh nhanh nipe = 0,7.mh6n hop oxit sit
+ (molsan phim kM-SO e nhan.5,6)
BAI T A P V A N DyNG
Bai 1: (De KA-2008): Cho 11,36 gam mot hon hap gom Fe, FeO, FegOs,
Fe304 phan ufng het vdi dung dich HNOg loang da, thu dtfac 1,344 lit
NO (la san pham khuf duy nhat d dktc) va dung dich X Co can dung dich X thu dufoc m gam muoi khan Gia tri cua m lạ
Ạ 49,09 gam B 34,36 gam C 35,50 gam D 38,72 gam
Hii&ng dan gidi Cdch 1: Ap dung cong thitc
mpe = 0,7.mh6n hap oxit s2it + (mol san phám khii-so 6 nhan.5,6)
= 0,7.11,36 + ( i ^ 3 5 , 6 ) = 8,96 gam
Trang 27- B a i 2: (DH KA 2007): Nung m gam hot s^t trong oxi t h u dtfoc 3 gam
hon hop chat r ^ n X Hoa tan het X trong dung dich HNO3 dU thay
thoat ra 0,56 l i t (dktc) k h i NO duy nhat Gia t r i cua m?
Hitdng dan gidi
V g " dung cong thiic
m p e = 0,7 vcihSn hgp oxit sdt + (moljdn pArfm * A i ? - s o 6 nhdn.5,6)
16,8-(0,25.2.5,6)
=> mx = — ^-^ — = 20 gam
0,7 Chon dap an C
B a i 4: De m gam bot sSt ngoai khong k h i , sau mot t h d i gian se chuyen
th&nh hon hop B gom 4 chat rAn c6 k h o i Itfong 12 gam Cho hon hap
B phan iJng het v d i dung dich HNO3 du thay thoat r a 2,24 l i t N O
(dktc) T i n h m va k h o i iuong HNO3 da phan iJng?
I A.10,08 g va 34,02 g B 10,8 g va 34,02 g C.10,8 g va 40,32 g D 10,08 g va 40,32 g
Hiidng ddn gidi Vdn dung cong thiic
mpe = 0,7.mA<5„ hifp out sdt + (molsdn phdm khu-sd e nhdn.5,6)
Trang 28B a i 5: Hoa tan het m gam hon hop X gom Cu va hai oxit s^t can vCfa du
500ml dung dich H C l 1,2M Co can dung dich sau phan urng t h u daoc
38,74 gam hon hop h a i muoi khan, m nhan gia trỉ
Ạ 22,24 B 20,72 C 23,36 D 27,04
Hiiofng dan giai
Hai muoi k h a n d day la CuCl2 va FeCl2
Ta c6: 2H^ + Ố > H2O ( H C l ^ H^ + C D
0,6 0,3mol
B T K L : mcu va Fe = m^.ai - m^, = 38,74 - 0,6.35,5 = 17,44 gam
^ mx = mcuvi Fe + mo(ox.t) = 17,44 +0,3.16 = 22,24 gam
Chu y: A1(0H)3 khong tan dUdc trong dung dich NH3, trong axit cacbonic
- CO2 day duoc goc aluminat ra k h o i muoi
NaA102 + CO2 + 2H2O ^ A l ( 0 H ) 3 i + NaHC03 CO2 khong hoa tan diioc A1(0H)3 nen phan ling dCrng l a i d ket
BAI TAP VAN DUNG
B a i 1: Cho 3,42gam A l 2 ( S 0 4 ) 3 tac dung vdi 25 m l dung dich N a O H tao r a di/oc 0,78 gam ket tuạ T i n h nong do mol cua N a O H da diing
B a i 2: Cho 150 m l dung dich K O H 1,2M tac dung vdi 100 m l dung dich
A I C I 3 nong do x mol/1, t h u difOc dung dich Y va 4,68 gam két tuạ Loai
bo k e t tua, t h e m tiep 175 m l dung dich K O H 1,2M vao Y, t h u difoc 2,34 gam ket tuạ Gia t r i cua x la
Ạ 1,2 M B 0,8 M C 0,9 M D 1,0 M
Hiicfng dan giai
T N I : H K O H = 0,18 mol; n^^oH), = 0.06 mol
04
Chon dap an Ạ
Trang 29B a i 3: Trong mot coc dUng 200ml dung dich AICI3 2M Rot vao coc
200ml dung dich NaOH c6 nong do a mol/lit, ta duac mot ket tua;
Dem say kho va nung den khoi Itfgrng khong ddi diioc 5,lg chat r^n
Tron dung dich chiia a mol AICI3 vdi dung dich chufa b mol NaOH De
thu daoc ket tua thi can cd t i le
A a: b = 1: 4 B a: b < 1: 4
C a: b = 1: 5 D a: b > 1: 4
Hiidng dan giai
Trpn a mol AICI3 vdi b mol NaOH de thu dtfac ket tua thi
Al=*^ + 30H- > A1(0H)3^
A1(0H)3 + OH- > AlO- + 2H2O
Al'" + 40H- > A10-+2H,0
a 4 mol
De ket tua tan hoan toan thi > 4 ^ - > 4
Vay de c6 ket tua thi - < 4 => a : b > 1 : 4
a Chon dap ^n D
B a i 5: Cho 200ml dung dich AICI3 1,5M tdc dung vdi V lit dung dich
NaOH 0,5M, Itfong ket tua thu diTcfc la 15,6 gam Tinh gia t r i Idn nhat
cua V
A 2 lit B 2,5 lit C 1,5 lit D 1 lit
Hii&ng dan giai
15 6
nAi(0H)3 = = 0,2mol n^ic, =0,2.1,5 = 0,3 mol
nAici, > nAuoH)3 chi xay ra trirdng hop 2 Tacd: _ = 4 n ^ , 3 - n A „ 0 H ) 3
nxaOH = no„- = 4.0,3 - 0,2 = 1 mol
VN«OH T T ^ = 2 lit Chon dap an A
U,5
B a i 6: Cho V lit dung dich NaOH 0,3M vao 200 ml dung dich A l 2 ( S 0 4 ) 3
0,2M thu dtroc mot ket tua tr^ng keo Nung ket tua nay den khoi
lugng lUOng khong ddi thi diigfc l,02g r^n Tinh the tich dung dich
NaOH da dung
A 0,4 lit hoSc 1,2 lit B 0,2 lit ho^c 1,2 lit
C 0,2 lit hoac 1 lit D 0,4 lit hoSc 1 lit
HUcfng dan giai
nA.,,so,,3= 0,2.0,2 = 0,04mol 1,02 ^ ,
n., n = — — = 0,01mol
n^,3 =2.0,04 = 0,08mol '"^"^ 102 pt: A P " + 30H- ^ A1(0H)3; 2A1(0H)3 ^ AI2O3
Trang 30B a i 7: M o t dung dich hon hop chufa a mol NaAlOg va a mol N a O H tac
dung vdi mot dung dich chdfa b mol H C l Dieu k i e n de t h u ducfc ket
tua sau phan uTng la
A a = b B a = 2b C b = 5a D a < b < 5a
HUdng dan gidi
Phaong t r i n h phan ufng:
N a O H + H C l > N a C l + HzO
a mol -> a mol
NaA102 + H C l + H 2 O > A l ( 0 H ) 3 i + NaCl
A1(0H)3 + 3HC1 > AICI3 + 3H2O
NaAlOa + 4HC1 > AICI3 + N a C l + 2H2O
a mol -> 4a mol
Dieu k i e n de khong c6 k e t tua k h i nnci ^ 4nN^,o^ + nNaOH = 5a
Vay suy ra dieu k i e n de c6 ket tua:
HNaOH < H H C I < 4 n N a A i o , + nNaOH => a < b < 5a
Chon dap an D
B a i 8: K h i cho 130 m l AICI3 0,1M tac dung vdi 20 m l dung dich N a O H ,
t h i thu dLfOc 0,936gam ket tua T i n h nong do mol/1 cua dung dich
N a O H
A 0,6M hoac 1,95M B 0,6M hoac 1,8M
C 1,95M hoSc 1,8M D 1,8M hoSc 2 M
Hiiofng dan gidi
Van dung cong thiic
B a i 9: Cho 3,42gam Al2(S04)3 tac dung v d i 25 m l dung dich N a O H tac ra
dtfOc 0,78 g a m k e t tua Nong do mol cua N a O H da dung la?
B a i 10: T r o n 10,8g hot A l v d i 34,8g hot Fe304 roi t i e n h a n h phan iJng nhiet nhom trong dieu k i e n khong c6 khong k h i Hoa tan hoan toan
h6n hop r a n sau phan ufng bSng dung dich H2SO4 loang (dtf) t h u di/gc 10,752 l i t k h i H2 (dktc) Hieu suat cua phan uTng nhiet nhom l a
A 80% B 90% C 70% D 60%
Hiicfng dan gidi
8A1 + 3Fe304 — ^ 4AI2O3 + 9Fe 0,4 0,15
8x 3x 4x 9x
(0,4-8x) (0,15-3x) 4x 9x
K h i phan iJng vdi H2SO4 loang:
A l + H^SO, -> A l ' * + H2 ; Fe + H^SO, Fe'* + H
^ n n = ^ ? 7 r ^ = 0.48mol
22,4 Theo dinh luat bao toan electron
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