các dạng điển hình và phương pháp giải nhanh bài tập trắc nghiệm hóa học

NGUYEN TUYENH A

CAC DANG fillN HiNH VA
PHJdNG PHAP GIAI NHANH BAI TAP TRAC NGH||M

I

Ji
1
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^

NHA XUAT BAN DAI HQC QUOC GIA TP. H OCHI MINH


CAC

DANG

O I E N HINH VA
BAI

PHl/dNG

P H A P GIAI

NHANH

TAP TRAC NGHIEM

L C l I N(&I D A U

H6AHOCI2
Nguyen T u y e n H a
N H A

X U A T

QUdc

DAI H O C

De h6 trg viec hoc tap va on t h i T U T A I - D A I 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 t h i dat ket qua tot nhat.

B A N

G I A TP 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
F a x : 38 239 172 - E m a i l :

Chiu track nhiem xudt bdn

TS H U Y N H BA L A N
T6 chiic bdn thao vd chiu track nkiem vi tac

quyin

DOAN V A N K H A N U
Bien tap

NGUYEN T H I NGOC H A N
S^a bdn in

T H A N T H I HONG

+ Phan I I : Cac phiicfng phap giai bai tap trac nghiem
+ Phan I I I : Gidi thieu 05 de t h i thuf Tu t a i va 06 de t h i thuf Dai hoc
de cac em thijf siJc minh.
Cac de t h i 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 h i 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
T U T A I - D A I HOC - CAO DANG s^p t d i . 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.
Tac gid

DIEM KHANH

Nguyen Tuyen Hd

TK.02.H(V)
494.2013/CXB/07-25
DHQG.HCM-13

TK.H.453-13(T)

In 1.000 cuon, khd 16 x 24cm.
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 c h i l u quy I V nSm

+ Phan I : Cac dang bai tap dp dung cong thiJc tinh nhanh

Xin tran trong cam cfn.

Trink bay bia

.

Ve noi dung cuon sach cd 3 phan:

2013.


T
CAC D4NG BAI

A P D^JNG
CACH TiNH NHANH
C a c h tinh nhanh so dong phan ciia:
- Ancol no, dctn chiiCc
- Andehit

dcfn chvCc, no

- Axit cacboxylic

(CnH2„0):

dcfn chiic, no

- Este no, dcfn chtic
- Ete darn chiic, no

(C„H2n02):

(1<

n <

2n-3

(2<

n < 7)

2n-3

(2<

n < 7)

2n-2

(C„H2„02):
(C„H2n+'20):

- Xeton dam chiic, no
- Amin dctn chiic, no

2n-2

(C„H2n+20):

(C„H2„0):

(1

n <

5)

(n-l)(n-2)
2

(2<

n < 6)

(n-2)(n-3)
2

(2<

n < 7)

(n<

5)

2n-l

(CnHzn+sN):

<

6)

Dang 1: KIM LOAI TAG DUNG VOl AXIT
A. H C I , H2SO4 ( l o a n g )
Doi vdi hat loai axit tren thi chi phdn

ling voi nhUng kim loai

diing

trade H trong day hoat d6ng hoa hoc.

K

Ca

Na

Hg

Ag

Au

Mg

Al

Zn

Cr

Fe

Ni

Sn

Pb

Phucfng trinh tong quat: (Vdi M la kim loai)
2M + nH2S04 (loang) -> M 2 ( S 0 4 ) n + nHzl

ta luon c6

2M + 2nHCl ^ 2MC1„+ nHg

n„. = 2. n^^

J

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,^

H

Cu


BAI

TAP

AP

DUNG

B a i 1: Cho l,04g hon hop hai k i m loai tan hoan toan trong dung dich
H2SO4 loang du thodt ra 0,672 Ht k h i 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
2 kim loai +

H2SO4I

dan

gidi

hh muoi sunfat +

B a i 4: Bok tan 1,19 gam hSn hop A gom A l , Zn b&ng dung dich HCl v i T a
du thu difoc dung dich X vk V l i t k h i 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.

Ap dung cong thiJc: m^ua-i ciorua

H2

4,03-1,19

+ 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, A l va Fe
trong dung dich HCl, c6 2,352 l i t k h i 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:
B. 10,985 gam

C. 11,195 gam

Hitdng dan gidi
nn^so, = 0,25.0,5 = 0,125 mol =^ n^,, = 0,125 . 2 = 0,25 mol
n„. = 0,25 mol

=> tong so mol H* = 0,25 + 0,25 = 0,5 mol

thuTc: m m u e i doma = nihSn iwp kim loai

q

+ 7 1 n„^

Chon ddp 6n B.
B a i 3: Hoa t a n hoan to^n 2,17 g a m h o n hap 3 k i m loai A, B, C trong
dung dich HCl da thu diTOc 2,24 l i t k h i H 2 (dktc) va m g a m muoi. Gia
t r i cua m 1^.
B. 5,72 gam.

C. 6,85 gam.

D. 6,48 gam.

Hiidng dan gidi

m„us-iciorua = nihSn h;>p kim lo^i + 71 n,,^ = 2,17

n„,

no

=1^

= 0,175 (mol)

Ta c6:

2H^ ^

Ban dau:

0,5 mol

PU:

0,35 <- 0,175 mol

Sau pif:

0,15 mol

H2

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

n„ =?^
= 0,l (mol)
22,4
Chon dap an A

= 0,25.1 = 0,25 mol

HHCI

= 3,53 + 0,105 . 71= 10,985 gam

A. 9,27 gam.

r^r^A ^ ^^

V„^ = 0,04 . 22,4 = 0,896 l i t

D. 7,2575 gam

= 0,105(mol)

Ap dung cong

+ 71 n^^

Chon dap an C.
B a i 5: Cho 5,35 gam h6n hop X gom Mg, Fe, A l vao 250ml dung dich Y
gom H 2 S O 4 0,5M va HCl I M thu duoc 3,921it k h i H 2 (dktc) va dung
dich A. Co can dung dich A trong dieu kien khong c6 khong k h i , 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.

HU&ng d&n gidi
n H . - | ^

= nihSn hap kim loai

nji^ = ^ ^ ^ p — = 0,04 ( m o l )

C O n g thufC: m m u t f i sunfat = mhSn hap kim lo?i

A. 12,405 gam

D. 1,792 l i t .

Hii&ng dan gidi

n „ . - - | | f = 0,03(mol)
Ap dung

C. 0,896 l i t .

= 5,35 + (0,125.96) + (0,25.35,5) = 26,225 gam

Chon dap an B
+ 0,1

. 71

= 9,27

gam

B a i 6: (DH khoi A.2010): Cho m gam hon hap bot X gom ba k i m loai
Zn, Cr, Sn c6 so mol b^ng nhau tac dung het vdi lirotng diT dung dich


H C l loang, nong t h u dugc dung dich Y
k h i Ha- Co can dung dich
Y t h u dtfdc 8,98 gam muoi k h a n . Neu cho m gam h o n hop X tac
dung hoan toan v d i O2 (du) de tao hon hop 3 oxit t h i the t i c h k h i

(dktc) p h a n ijrng la
A . 2,016 l i t .

B. 0,672 l i t .

C. 1,344 Ht.

Hii&ng dan

D. 1,008 l i t

= 0,4(mol)

" H , =|g

27x + 56y =11
Ta lap va g i a i he phuong t r i n h

- X

gidi
Fe

Chon dap a n B

- Con k h i tac dung O2, Zn tao +2, Cr tao +3, Sn tao +4. (X + O2 t h u

B a i 8: (De KA.2007):

V i cac k i m loai c6 so mol bSng nhau nen cAc muoi c6 so m o l b^ng

A . 1.

B. 6.

C. 2.
Hiidng

136x + 123x + 190x = 8,98 =>x = 0,02;

=> t d n g so m o l H^ = 0,25 + 0,25 = 0,5 m o l
5,32

22,4

Chon dap an D

^

nor,,-/ —

2H*

t h u dirge 8,96 l i t H2 (dktc). P h a n t r a m k h o i lugng cua Fe t r o n g h o n

Ban dau:

0,5 m o l

hop l a .

A. 49,09%.

Pu:

0,475

Sau pir:

0,025mol

B. 50,91%.
Hiiofng dan

C. 40,91%.

D. 59,09%.

gidi

Liiu y: K i m loai tac dung v d i m o t chat nao do ma sinh ra
le mol giufa k i m loai va H2 luon b^ng hod tri kim log,i chia 2

y mol

->

- H2
2

—>

— X

-4

H2

mol

y mol

thi ti

. V

„

= 0,2375(mol)

T a c6:

B a i 7: Cho 11 gam h o n hop A l va Fe tac dung h e t v6i dung dich H C l

Fe

gidi

nHci = 0,25 . 1 = 0,25 m o l => n „ . = 0,25 m o l

=> Vo^ = 0,045.22,4 = 1,008 l i t

2

dan

D . 7.

n„^o, = 0.25 . 0,5 = 0,125 m o l => n,j. = 0,125 . 2 = 0,25 m o l

3

"oj = 2 ' ' ^ 4 ^ ^ ' ' ' ^ 2,25x = 0,045 m o l

mol

0,1

Cho m gam hon hop M g , A l vao 250ml dung dich

nhau va bang x

X

y =

dung dich Y. Coi the tich dung dich khong doi. Dung dich Y c6 p H 1^

- Goi so m o l m o i k i m loai la x (mol) t h i :

Al

0,2

chufa hon hop H C l I M va H2SO4 0,5M t h u dugc 5,32 l i t k h i H2 (dktc) v^

daoc cac oxit: ZnO, CraOa, Sn02)

So do

+ y = 0,4

X =

0,1.56 X 100% = 5 0 , 9 1 %
11

=

3 k i m loai t r e n k h i p h a n i l n g 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 t h u di/Oc muoi: ZnClz, CrClz, SnClg.)

1

3

.

H2
<-

Sau p h a n iJng a x i t diT H *

0,2375 m o l

I

0,025
0,25

= 0,1M

p H = - l o g [H^^ =1
Chon dap a n A
B a i 9: (De KB.2007): Cho 1,67 gam h o n hop 2 k i m l o a i d 2 chu k y ke
tiep nhau thuoc n h o m I I A tac dung het v d i dung dich H C l d i i , t h o a t
r a 0,672 l i t k h i H2 (dktc). H a i k i m loai do l a (Be = 9,Ca = 4 0 , M g = 24,
^ Sr = 87,Ba = 137)
A . M g va Ca.

B. Ca va Sr.

C. Sr va Ba.

D. Be va M g .


Hiictng dan

gidi

B. KIM LOAI T A G D y N G VC3l HNO3, H2SO4 (d|c)

D a t 2 k i m loai 1^ A

A x i t H2SO4 (dSc) + k i m loai = muoi sunfat + san p h a m khuf + H2O

T i le mol gifla k i m loai va h i d r o b^ng ho^ t r i k i m l o a i chia 2 n 6 n t i le

San phdm

m o l la 1: 1

*

= MZ2^ 0,03
22,4

(mol)

Ha

0,03mol

0,03mol

^

= 55,67 ^

Ca va Sr.

H C l I M va H2SO4 0,5M t h u dMc

dung dich B va 4,368 Ht

H2(dktc). P h a n t r S m k h o i laong M g va A l t r o n g X tucfng ufng la
A . 3 7 , 2 1 % M g va 62,79% A l .

B. 62,79% M g va 3 7 , 2 1 % A l .

C. 45,24% M g va 54,76% A l .

D . 54,76% M g va 45,24% A l .

Bai

khd c6 the la NO2 (nhan

muoi nitrat =

BAI TAP VAN

B ^ i nay l a m tifcfng t i f bai

Bai

lOe), NH4NO3 (nhan
H I k i m loai

3e), N2O

(nhan

8e).

+ molgpk.so e n h a n . 6 2

DUNG

1: Cho 1,86 gam h o n hdp A l va M g tac dung v d i dung dich HNO3

n h a t ) bay r a . K h o i liTOng muoi n i t r a t tao r a t r o n g dung dich l a :
A. 40,5 gam.

B. 14,62 gam.

C. 24,16 gam.

HtCclng dan

D. 14,26 gam.

gidi

n^o = ^ ^ - 0 , 0 2 5 m o l
N,o
22,4

3,65M (d = l , 1 9 g / m l ) t h u dugc m o t chat k h i va 1250g dung dich D.

A p dung cong thuTc: m musl nitrat = m kim loai + molspk-so e n h a n . 6 2
= 1,86 + 0,025.8.62 = 14,26 gam

V a y m c6 gia t r i :
B. 61,63 (g)
HUdng

C. 63,65 (g)
dan

D. 63,61 (g)

Chon dap a n D
Bai

gidi

m,,„c, = 1000.1,19 = 1190 ( g ) ; n „ c , = 3,65.1 = 3,65

2: Hoa t a n hoan toan 9,94 gam h o n hop X gom A l , Fe, Cu t r o n g

lugng du dung dich HNO3 t h u dugc 3,584 l i t k h i N O duy n h a t (dktc).
(mol)

T o n g k h o i lugng muoi k h a n tao t h a n h l a :
A. 39,7gam

Zn + 2HC1 ^ ZnCl2 + H2

B. 29,7gam

A p dung D L B T K L , t a c6: nihh.zn.Fe) + "^ddiici = "^udo + " ^ n ,
= 1250 + 2 ( ^ )

C. 39,3gam

Hiidng

Fe + 2HC1 ^ FeCl2 + H2

Chon dap a n C.

le), NO (nhan

loang dtf t h i t h u di/cJc 560 m l l i t k h i N2O (dktc, san p h a m khuf duy

5

1 1 : Hoa t a n m(g) h 6 n hop Zn va Fe can vCra du 11 dung dich H C l

A . 65,63 (g)

n h a n . i .96

Luu y: A l , Fe, Cr khong phan ting vcd HNO3 dSc nguoi va H2SO4 dac nguoi.

10: Cho 3,87gam h 6 n hop X g6m M g va A l v^o 250ml dung dich X

gom

phdm

8e), N2 (nhan

Chon dap a n B
Bai

6

6e)

a x i t HNO3 + k i m loai = muoi n i t r a t + san p h a m khijf + H2O

* m
MA =

"1 kin, loai + molgpk.so

m m u o i sunfat =

San

A

khii c6 thi Id SO2 (nhan 2e), H2S (nhan 8e),S (nhan

- 1190 ^ 63,65(g)

dan

D. 27,7gam

gidi

n„„=||M.o,16n.o.
A p dung cong thUc:

mmuoi

nitrat = m

k i m loai

+ molspk.so e n h a n . 6 2

= 9,94 + 0,16.3.62 = 39,7 gam
Chon dap a n A


B a i 3: Hoa tan 4,97 gam hon hap A l , 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 6: (BH khoi B.2008): Cho 2,16 gam M g tAc dung vdi dung dich
HNO3 (du). Sau khi cac phan iJng xay r a 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

B a i 4: Hoa tan hoan toan 13,68 gam hon hdp X gom A l , Cu, Fe bSng
dung dich HNO3 loang, diX thu diroc 1,568 lit khi N2O (dktc) va dung

n^g = 246 : 24 = 0,09 (moDin^o - 0,896 : 22,4 = 0,04 (mol)

dich chuTa m gam muoi. Gia tri cua m la.

Cdch 1:

A. 48,40 gam.

B. 31,04 gam.

C. 57,08 gam.

D. 62,70 gam.

3Mg + 8HNO3 ^ 3Mg(N03)2 + 2 N 0 + 4H2O

Bai nay gidi tuang tu bai 2.

0,06

B a i 5: (DH khoi A.2010): Nung 2,23 gam hon hop X gom cac kim loai

<<

0,06 <

0,04

4Mg + IOHNO3 > 4Mg(N03)2 + NH4NO3 + 3H2O
0,03
> 0,0075
0,03
>

Fe, A l , Zn, M g 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

Khoi liTcfng muoi khan thu diioc khi lam bay hoi dung dich X \k

Ht khi NO (san pham khuf duy nhat, d dktc). So mol HNO3 da phan

m = 0,09. 148 + 0,0075.80 = 13,92 (gam) ^ chon B.
Cdch 2: Diia vao dinh luat bao toan electron

ufng la
A. 0,12.

B. 0,14.

C. 0,16.

D. 0,18.

HiCdng ddn gidi

Mg

2e

0,09

2.0,09

-

Mg^2

N

+5

N

+5

0,09

3e

Tom t^t:
Fe, A l , Zn, Mg + O2 -> hh ; hh + HNO3
'

'

-> muoi +

NO

8e

N - ' NH4NO3)

^ U N O j (pu O trong oxit)

2.0,09 = 3.0,04 + 8x -> X = 0,0075 (mol)

= 0,03 (mol)

B a i 7: Hoa tan 23,4 gam G gom A l , Fe, Cu bang mot lirong vCra du dung
dich H2SO4 d5c, nong, thu di/dc 15,12 lit khi SO2 (dktc) va dung dich

^ H N O j (pu oxi ho4 - khut)

^NO

A. 153,0 gam.

B. 95,8 gam.

C. 88,2 gam.

Hii&ng ddn gidi

Mat khac:
HHNO^ (pu o t ^ n g oxiu =

( cho

n H N o , ,pu ox, ho. - uha, = 3.n^o

( Do

H^O);

N^^ —

+ 3nf^o+ n^o =

i^so,
^

=

15,12
22,4

= 0,675mol

NO)
nimuol simfat =

^T^o(<«iM+

= 2.0,03 + 4.0,03 = 0,18 mol
Chon dap a n D

X

chuTa m gam muoi. Gia tri cua m la.

Ta CO bao toan nguyen to N:

^ ^ ^ i i N o , = 2no,oxit)

-

m = 0,09. 148 + 0,0075. 80 = 13,92 (gam) -> chon B

mo = m , , - m„ = 2,71 - 2,23 = 0,48(g) -> n^ - ^

'^HNOj ~

0,04

Khoi lirong muoi khan thu dtfoc khi lam bay hoi dung dich X la

Bao toan khoi liTOng:

^

3.0,04
8.x

So mol electron dLfOc bao to^n

0.672 (l)«5|^=0,03(mol)

^ ' l a ^

N^2(N0)

-

HI kjn, loai

'^^m

+ molgpk.SO

6

nhan. — .96
2

= 23,4 + 0,675.2. \6 = 88,2 gam
Chon dap an C.

|

D. 75,8 gam.


3Cu + 8H* + 2NO3"

B a i 8: Kok t a n h o ^ n t o ^ n m gam h o n X gom A l , Fe, Cu vko dung dich
H N O 3 d S c nong du, t h u duoc dung dich Y chufa 39,99 gam m u o i 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.

Hiicfng ddn

Dau b a i :

P h a n iJng: 0,06 ^

D . 19,84 gam.

m
=^ m

0,16

0,08

0,16 ^

gidi

B a i 10: (DH

khoi

B-2009):

raM

nitrat

k i m loai

= m

= m

+ molspk.so e nhan.62

k i m loai

" molspk-SO

muoi nitrat

G

nhan.62
gam

A. 0,03 va 0,01

B. 0,06 va 0,02

C. 0,03 va 0,02

D . 0,06 va 0,01
Hiicfng ddn

Au

1. Cho 3,84 gam Cu p h a n iJng v d i 80 m l dung dich H N O 3 I M t h o d t r a

->

0,02

M hoa t r i 2 vifa du vao dung dich c h d a H N O 3 va H2SO4 va dun n o n g ,

thu dtroc 2,94 g a m h o n hap k h i B gom NO2 va SO2. The t i c h cua h 5 n

Quan he giOfa V i va V 2 la
B. V 2 = 2 V i .

C. V 2 = 2 , 5 V i .

Hiictng dan

0,08

0,08 ->

l i t (dktc). K h o i l u a n g m u o i k h a n t h u diroc l a .

A. 6,36g.

B. 7,06g.
HUdng

n„, = 0 , 0 8 m o l
,

3Cu + 8 H " + 2NO3-

D. V2 = l,5Vi.

hop k h i B l a 1,344

gidi

= 0,06 m o l

=0.08 mol

0,06

> A U C I 3 + N O + 2H2O

B a i 11: Hoa t a n 3 gam h o n hgfp A gom k i m loai R hoa t r i 1 va k i m l o a i

0,5 M t h o a t r a V 2 Ht N O . B i e t N O la san p h a m khijf duy

n h a t , cac the t i c h k h i do d cCing dieu k i $ n .

Dau b a i :

3HC1 + H N O 3

Chon dap a n B

2. Cho 3,84 gam Cu p h a n ling v d i 80 m l dung dich chufa H N O 3 I M va

nHN03

+

0,02 -> 0,06

Vi lit NO.

n c „ = ^
b4

gidi

• Ni/dc CLfdng toan l a t i le 3 : 1 giufa H C l va H N O 3

ThUc h i e n h a i t h i n g h i e m :

TNI:

K h i h o a t a n hoan toan 0,02 m o l A u b a n g

ntfdc cudng toan t h i so m o l H C l p h a n i l n g va so m o l N O (san p h a m

2007)

A. V 2 = V i .

0,04 m o l

khuf duy n h a t ) tao t h a n h I a n liToft l a

Chon dap a n A

H2SO4

->

Nhir vay V 2 = 2 V i . Chon dap a n B.

= 39,99 - 0,32.1.62 = 20,15

B a i 9: (DHKB.

0,04

-> Cu va H"^ p h a n iJng h e t

=> V 2 ti/ong urng v d i 0,04 m o l N O .

H M O = ^ ^ ^ = 0,32mol
"""^
22,4
A p dung cong thiJc:

0,06

>3Cu^* + 2NO^ + 4 H 2 O

^ 1^344 ^
22,4

n^^_ = 0,08 m o l

x + y = 0,06

p h a n iJng h e t

P h a n iJng: 0,03 <- 0,08 -> 0,02 ^

0,02 m o l

=> V i tirang ufng v d i 0,02 m o l N O .
TN2:

ncu = 0,06 m o l ; nn^o, = 0,08 m o l ; nn^so. = 0,04 m o l .

=> T o n g

n^^, = 0,16 m o l ; n ^ ^ . = 0,08 m o l .

•

dan

D . 12,26g.

gidi

oGmol dSt NO2 x m o l , SO2 y m o l

r46x + 64y = 2,94

> 3Cu^^ + 2 N 0 T + 4 H 2 O

C. 10,56g.

f x = 0,05

[ y = 0,01

nimuai = 3 + (0,05 .62 + 0 , 0 1 . - . 2 .96) = 7,06 gam
2
Dap a n B.

B a i 12: Cho 8,3 gam h 5 n hcfp A l va Fe tac dung v d i dung dich H N O 3
loang dtr t h i t h u dircfc 45,5 gam m u o i n i t r a t k h a n . The t i c h k h i N O
(dktc, san p h a m khuf duy n h a t ) t h o a t r a l a :
! A. 4,48 l i t .

B . 6,72 l i t .

C. 2,24 l i t .

D . 3,36 l i t .


Qua t r i n h khii:

Hiicmg dan giai
Ap dung cong thiJc:
^ DNO =
=>

m musi

nitrat

= ni

"-^-"^ ~
so e nhan.62

,0,1

=

0,3

+ molspk-so e nhan.62

N*^ + l e

= 0,2 mol

3.62

0,1

Chon dap an A
B a i 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 k h i NO2 (san pham khuf duy
nhat, d dktc). Phan trSm ve khoi liiong cua Cu trong hon hop X la
C. 68,05%.

D. 29,15%.

Hiidng ddn giai
n NO,

. Cu^^ + 2e
2x

X

Al •
y
Ta c6:

N""

0,2

S"^ + 2e

S*^

0,2

0,1

0,1
24x + 27y = 15

fx = 0,4 mol

2x + 3y = l , 4

[y = 0,2 mol

97 n 9

.%A1 = ^ ^ ^ . 1 0 0 % = 36%.
15
%Mg = 100% - 36% = 64%.
Bai 15: Cho 2,8 gam hon hop bot kim loai bac va dong tac dung v6i dung
Thanh phan phan tram cua bac va dong trong h6n hop Ian liTOt la:
A. 73% ; 27%.

B. 77,14% ; 22,86%

N"^ + 3e ->

C. 50%; 50%.

D. 44% ; 56%

0,18

Tuang tu bai 14.

0,06

Al'^ + 3e

3y
64x + 27y = l,23
|2x + 3 y - 0 , 0 6

Dang 2: OXIT KIM LOAI TAG DUNG VOI AXIT HQ, H2SO4 [bang)

fx = 0,015

Phi/dng trinh tdng quat

[ y = 0,01

MzOn + 2nHCl

%Cu=M15:6i.78,05%
1,23

M20„ + nH2S04

Chon dap an B

2MCln+ nHzO
(loang) ^

dich Y gom H N O 3 va H2SO4 dSc thu diioc 0,1 mol moi k h i SO2, NO,
NO2, N2O. Phan tr5m khoi Itfong cua A l va Mg trong X Ian lacft la

* Doi vdi axit H2SO4 (loang)
Khoi li/ong muoi thu dugc la:

A. 63% va 37%.

B. 36% va 64%.

Doi vdi a x i t H C l

C. 50% va 50%.

D. 46% va 54%.

Khoi luong muoi thu duoc la:

HU&ng dan gidi
HMg = X

mol;

M2(S04)„ + nHzO

Cach tmh nhanh cho trie nghi?m

B a i 14: Hoa tan 15 gam hon hop X gom hai k i m loai Mg va A l vao dung

Dat

0,8

2N*^

dich HNO3 dSc, dtr t h i thu duoc 0,896 l i t khi NO2 duy nhat (d dktc).

= M M = 0,06mol (Dat Cu X mol, A l y mol)
22,4
Cu

0,1

Theo dinh luat bao toan electron:

B. 78,05%.

2N*^ + 2 x 4e

=> Tdng so mol e nhan bkng 1,4 mol.

= 0,2.22,4 = 4,48 l i t

A. 21,95%.

+ 3e -> N*=^

DAI

x

2x

=> Tong so mol e nhudng b i n g (2x + 3y).

+

n,[^so,

-80

•
m,„„a-i ciorua = n i h o n hop oxit kim loai

+ 27,5 n,,^;,

BAl TAP VAN DgNG

= y mol. Ta c6:

Qud t r i n h oxi ho^: Mg -> Mg^* + 2e

m^^ai ^unfat = m h S n hap o x u k i m loai

Al
y

A l ' " + 3e
3y

B a i 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

HUcfng ddn
Ap dung

gidi

C O n g t h u f C : nimudi cloma = nihon h<7p oxit kim loai

+ 27,5

n^jp,

= 50 + (0,2.4.27,5) = 72 gam
Chpn dap an C.
B a i 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 t h i 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
Ap dung

gidi

C O n g thufC: mmum sunfat = nihSn hop oxit kim loai +

J^HaSOi

-^^

Chon dap an C.
B a i 3: De tac dung vCra du vdi 7,68g hon hop gom FeO, Fe304, FegOs can

dung 260 m l 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

oxit k i m loai + HCl
nnc. =
2H^

0,26

'

Taco:

= 0,03mol .nHci = 0,3.1 = 0,3 mol
2H^

> Hg

0,06

<-0,03

2H^+
0,24

O'-

m = mx-

mo(oxit)

>

A. 87,5 m l

B.125

So do hop thiJc:

2Fe ^

C.62,5 m l

D.175 m l

gidi

- 4 chat rin la Fe203
= Y | ^ = 0,01875 mol

So do hop thufc:
=>

0,13mol

ml

Hii&ng dan

nFe,03

>H20

5 6
=> mpe = 7,68 - 0,13.16 = 5,6 gam => ^fe =
=

= 12 - 0,12.16 = 10,08 gam

B a i 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?

gidi

> muoi clorua + H 2 O

O'-

H2O

->0,12mol

= 0,26.1 =0,26 mol
+

>

Chon dap an A.

= 2,81 + (0,3.0,1.80) = 5,21 gam

Hii&ng dan

n„^ =

mo,ox,t)

2Fe

Fe203

0,0375 <-

0,01875

= 2,8 - (0,0375.56) = 0,7 gam

no,oxit)=^-0,04375mol

0>l"iol

FeaOg

0,1 -> 0,05 mol
=> mp^^o^ = 160.0,05 = 8 gam
Chon dap an C.
B a i 4: Oxi hoa cham m gam Fe ngoai khong k h i 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
Ht khi (dktc). Tinh m?
A.10,08
B.8,96
C.9,84
D.10,64

2H^

+

0,0875 <-

O'-

>

H2O

0,04375

=> V = 87,5 m l
Chon dap an A.
B a i 6: Hoa tan hoan toan 5,4 gam mot oxit sSt vao dung dich H N O 3 dii
thu dagc 1,456 l i t hon hop NO va N O 2 (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


Hiidng
mkhi = 5,4

Lap

-

2,49

= 2,91

x + y =

he:

1,456

gam

ddn

(NO

x mol va NO2 y

= 0,065

22,4

a

y = 0,06

+ 3e

O

3a

b

^

N O 3 ' + 3e

^

0,015

NO

<-

J
no

^ MZ5
^ 1
0,075
1

T h u c c h a t p h a n ufng khuf cac o x i t t r e n l a
CO

^

0,075

0,075

+

H2

2e

B. BAI T A P A P

2b
+

le

^

NO2

0,06

<-

0,06

0 ( t r o n g oxit)

>

0 ( t r o n g oxit)

N e n t a l u o n c6:

O 2-

^

>

-

CO2
H2O

= n„^o >

no,t„,„g„,it) = " H ,

nctrongoxit) =

^co

=

^co,

DUNG

B a i 1 : Khijf h o a n t o a n

17,6g

h o n hap

gom

Fe, FeO,

FegOs, c a n 4,48

lit

H 2 ( d k t c ) . K h o i l u o n g s a t t h u di/oc l a :
A . 14,5 g

B . 15,5 g

,
,
^
,
=> {a = b = 0 , 0 7 5 m o l
^

C. 14,4 g

Hitdng
4
no(trongoxit) = "^11,

FeO

mpe = 17,6 -

ddn

D . 16,5

g

gidi

48

= »H,0 =

=

0.2

HlOl

g

3,2 = 14,4

g

C h o n d a p d n C.
B a i 2: H 6 n h o p A g o m sSt v a o x i t sSt c6 k h o i luofng 2,6 g . Cho k h i C O d i
3nN(Fe(N03)3

+

^N(NO)

+

^NCNO^)

= 3.0,075 + 0,005 -t- 0,06 = 0,29
n„N03

+

mo = 16 . 0,2 = 3,2

Fe(N03)3

njjfHNo,)

+

NO3-

oxit sat la

FeO

B T N T NitO:

D o i v d i t r u d n g h o p c h a t khuf l a C O , H 2 t h i

mol

0,005

56a + 16b = 5,4
Tac6:<^
[ 3 a - 2 b = 0 , 0 6 + 0,015

*

mol).

X = 0,005 m o l

30x + 46y = 2,91
Fe - > Fe

giai

= nj,,HN03)

^

^imo,

=

0.29

qua A d u n n o n g , k h i d i r a sau p h a n ufng di/gfc d a n vao b i n h d i i n g nirdfc v o i
mol

t r o n g diT, t h u diTcfc l O g k e t t u a t r S n g . K h o i Itfcfng sSt t r o n g A l a :
A. 1 g

mol.

B . 1,1 g

C. 1,2 g

Hi£&ng ddn

C h o n d a p a n C.

D . 2,1 g

gidi

Ca(0H)2 + C O 2 -> CaCOsi + H2O

Dang 3: KHLf CAC GXIT KIM LOAI BANG (CO. C. Ha. Al)

10
=

^ 0 0 ,

•

ncaco3

•

J^odrongoxit) =

=

^co

=

—

=

0,1

mol

A. LfTHUYET
t

I . Phxftfng t r i n h p h a n i ^ n g t o n g q u a t :

K h o i li/gfng sit

= n^o^

=0,1

mol

t r o n g h 5 n h a p A l a : mpe = 2,6 -

16 . 0 , 1 = 1 g.

C h o n dap a n A.
B a i 3: Cho V l i t ( d k t c ) k h i Hg d i qua h o t C u O d u n n o n g , t h u duoc 32 g C u .
N e u cho V l i t H 2 d i qua h o t F e O d u n n o n g t h i ItfOng Fe t h u duoc l a :
A. 24g

B. 2 6 g
HUitng

D 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 C O ( H 2 ) t h i C O ( H 2 ) l a y o x i
ctja o x i t k i m l o a i r a k h o i o x i t .

32
= ncu = npe = —
D4

mpe = 56 . 0,5 = 2 8

C h o n d a n a n C.

g

= 0,5

C. 2 8 g
dan
mol

gidi

D.30g


B a i 7: N u n g n o n g h o n hop X gom PbO v^ FeO v d i m o t l u o n g C vijfa du.

B a i 4: Khuf ho^n toan 32g h o n hop CuO va FezOg b^ng k h i H2, t h a y tao
r a 9 g nude. K h o i lirong h 6 n hop k i m loai t h u duoc l a :
A . 12 g

B. 16g

C. 24 g

D. 26 g.

Hitdng d&n gidi
9
= n„^o

n o d r o n g oxit) =
niO (trong oxit) =

m

kim i o , i

16

. 0,5

=
=

=

"^'^

Sau k h i p h a n ufng xay r a hoan t o a n . t h u d U o c h o n hop chat r ^ n Y va
k h i k h o n g mau Z. Dem can h o n hop rSn Y t h a y k h o i liTOng g i a m 4,8g

so v d i h o n hop X. Cho h o n hop Y tac dung v d i 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 v o i t r o n g d i i di/oc k e t
tua trSng. The t i c h k h i A (dktc) va k h o i liTOng k e t tua t h u d i r o c 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

8g

= 32 - 8 = 24 g
2PbO + C — ^ 2 P b + C 0 2

Chon dap a n C.
Bai

5:

AI2O3

Cho

khi

CO

qua

ong

d i T n g a (g)

hon

hop

gom

CuO,

Fe.304,

FeO,

nung nong. K h i t h o a t r a duoc cho vao niidc v o l t r o n g du t h a y c6

30g k e t tua t r a n g . Sau p h a n

ufng,

chat rSn t r o n g ong

suf

c6 k h o i luong

202g. K h o i lirong a (g) cua h o n hop cac oxit ban dau l a :
A . 200,8g

B. 216,8g

C. 206,8g

D. 103,4g.

Ca(0H)2 + CO2 ^ CaCOgi + H2O

•

= "c;o

nourongoxit) = ^ c o

=

—

= 0,3

= ^ 0 0 , = 0.3

= 4,8g =>

'

ncaco3 = n c o ,

viia du V l i t k h i CO (d dktc), sau p h a n

ufng

A. FeO va 0,224

B. FeaOg va 0,448

C. Fe304 va 0,448

D. Fe304 va 0,224.

HiCdng dan gidi
= " c o , = 0,02

mol)

npe =

X la Fe304

can

t h u dugc 0,84 gam Fe va

0,02 mol k h i CO2. Cong thufc cua X va gia t r i V I a n luot l a .

V = 0,02.22,4 = 0,448 ( l i t )

= 0,3 (mol)

lb

^ n^j,^ =

n^hv

=

(mol)

1

^ V,,^ = 0,3.22,4 = 6,72 ( l i t )

CO2 + C a ( 0 H ) 2 ^ CaC03 + H2O

Khuf hoan toan m o t oxit s a t X d n h i e t do cao

x
0,015
3
Fe.Oy: - = ^ ^ ^ ^ = ' y
0,02
4

4 8

no - - V

• n„^ = n^hY = 0 , 3 (mol)

"^ol;

Chon dap a n C.

no(irongoxit) = "^co

- 0,3

oxit da b i C lay d i tao CO2.

Fe + 2 H C l ^ Fe + H2

mol

ma = 202 + 0,3.16 = 206,8 g
B a i 6: (CD -2009):

O trong

Pb + 2 H C l - > PbClz + H2

30
"co,

li/Ong

" c o , - - ^ n o =0,15(mol)

• a = m c h a t r^n + ^ l o d r o n g oxit)

=

K h o i li/ong chat r ^ n Y g i a m so v d i h 6 n hop X 1^ k h o i

=> m o

HU&ng dan gidi

nc«co3

2FeO + C — ^ 2 F e + C 0 2

0,84

-b^

56

= 0,15 (mol) =>

mc^co,

= 0,15.100 = 15(g)

Vay dap a n diing la A.
B a i 8: Cho 0,3 m o l Fe^Oy t h a m gia p h a n ufng n h i e t n h o m t h a y tao r a 0,4

m o l A I 2 O 3 . Cong thufc oxit ski l a :
A. FeO
B. Fe203
D. K h o n g xac d i n h

diTOc

C. Fe304

v i k h o n g cho biet so m o l Fe tao r a .

Hiidng dan gidi
= 0,015

(mol)

A l lay di oxi cua Fe^Oy de tao ra
trong

AI2O3

AI2O3.

V i vay so mol nguyen tuf O

va t r o n g FoxOy p h a i bang nhau.

Do do: 0,3 y = 0,4 . 3 = 1,2

Chon dap an C.

=> y = 4 =^ Fe304


B a i 9: Dot chay k h o n g ho^n toan 1 luong sSt da dung h e t 2,24 l i t O2 d
dktc, t h u 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 d i r a sau p h a n l i n g duoc dSn vao b i n h difng
n i / d c voi t r o n g d U . K h o i l i i O n g k e t tua t h u dUOc l a :
A. 10 g

B. 20g

C. 30g

D. 40 g

Hiidng ddn giai

B a i 12: Khuf 39,2g m o t h o n hop A gom FegOa va FeO b&ng k h i CO t h u
duoc hon hop B gom FeO va Fe. B t a n vUa du t r o n g 2,5 l i t dung dich
H2SO4 0,2M cho r a 4,48 l i t k h i (dktc). T i n h k h o i luong FezOg va FeO
t r o n g 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

Ca(0H)2 + CO2 ^ CaCOai + H2O
2 24
n a t r o n g oxit) =

n c o = n^o^ = n^^co^ =

^ . 1 - ,
. [FeoOg : x(mol)
Goi h o n hap A<^ ^ '
^
^
F e O : y(mol)

-2 = 0,2 m o l

mc,co3 = 100-0.2 = 20g

H o n hgp B + H2SO4: FeO + H2SO4

Chon dap An B.

A p dung D L B T N T Fe t r o n g h a i h o n hgp A va B t a c6:

vao a x i t H C l t h i the t i c h k h i H2 (dktc) t h u dUcJc l a :
B. 1,12 l i t

C.3,36 l i t

160x + 72y = 39,2

D.2,24 l i t

J ^ H j = n(hh

kim loai) =

2.24

.

„ ,

= u,i

TCr (1), (2)

,

moi

x = 0,2
,y =

o,i

= 0,lmol

Chon dap a n D.

B a i 11: T h o i m o t luong k h i CO dtf d i qua ong dung hon hap h a i oxit
Fe304 v a CuO nung nong den k h i p h a n ufng xay ra hoan toan t h u dugfc

oxit k i m loai ban dau l a :
C. 4g

D . 4,2g

Hii&ng dan giai
nO(trong oxit) =

H c o = HCQ^

moxitkimio^i = n i k i m io,i +

Chon d a p d n A .

^

m

"caCOa
oxi

(g).

B a i 13: De khuf hoan toan 45 gam h o n hgp gom CuO, FeO, Fe304, Fe va
M g O 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 p h a n u"ng l a :
A. 39g

2,32 g hon hop k i m loai. K h i thoat r a duac dua vao b i n h dung dung
d i c h Ca(0H)2 du t h a y c6 5g k e t tua t r i n g . K h o i lugfng h 5 n hap

B. 3,21g

(g);

Vay dap a n dung la A.

VH^= 22,4.0,1 = 2,24 l i t

A. 3,12g

mp,^o^ = 0 , 2 . 1 6 0 = 32

mp^o = 0,1.72 = 7,2

K h i hoa t a n h 6 n h g p k i m loai vao a x i t t h i :
h h k i m loai

(2)

" 2 x + y = 0,5

Hii&ng dan giai

n„ = n

FeS04 + H2O

n„^so^ = 0 , 2 . 2 , 5 = 0,5 (mol);

h t H2 (ot dktc).Neu dem h o n hop k i m loai t h u d i / a c hoa t a n hoan toan

nhhoxit =

(1)

Fe + H2SO4 ^ FeS04 + H2

B a i 10: De khijT hoan toan h 6 n hop FeO va ZnO t h a n h k i r n loai can 2,24

A. 4,48 l i t

160x + 72y = 39,2

= 0,05(mol)

= 2,32 + 0,05.16 = 3,12 ( g )

B. 38g

C. 24g

D . 42g

Hii&ng ddn giai

hai

K h o i lugng chat r ^ n sau p h a n iJng l a : 45 - 1 6 . - ^ ^ = 39g
Chon dap a n 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 t h u dugc la
A. 5,6 gam.

B. 6,72 gam.

C. 16,0 gam.

D . 8,0 gam.


Hii&ng dan

gidi

Hii&ng dan

2 24
K h o i luong chat r a n s a u p h a n ufng l a : 17,6 - 1 6 . - ~ -

K h o i luong nguyen tuf oxi = do g i a m cua chat r ^ n

= 16,0 gam

m = 31,9 - 28,7 = 3,2 g

Chon dap a n C.

B a i 15: De k h i l hoan t o ^ n 30 gam h o n hop CuO, FeO, FezOg, Fe304,

3 2
So m o l nguyen tuf O = ^
= 0,2 mol= so m o l C O

M g O can dung 5,6 l i t k h i CO (6 dktc). K h o i lifong chat r a n s a u p h a n
ufng la
A. 28 gam.

B. 26 gam.
HUdng

C. 22 gam.
dan gidi

16
CO

D. 24 gam.

CO2 + 2e

(ZnO + 2e ^ Zn ^ Zn^* + 2e

5 6
K h o i l i i a n g chat r a n sau p h a n ufng l a : 30 - 1 6 - ^ ^

gidi

F e O + 2e -> F e ^ Fe^* + 2e)

= 26 gam

2W

Chon dap a n B.
B a i 16: D a n tCf t\i V l i t k h i CO (6 dktc) d i qua m o t ong s i l difng lUdng d i i
h o n hop r a n gom CuO, FezOs (d n h i e t do cao). Sau k h i cac p h a n ufng

+ 2e -> H2

Vay so m o l nguyen tuf O = so m o l C O = so m o l H2 = 0,2 m o l
Chon dap a n B.

xay r a hoan t o a n , t h u duoc k h i X. D i n toan bo k h i X d t r e n vao lUOng

C . BAI T A P V A N

diT dung dich Ca(0H)2 t h i tao t h a n h 4 gam k e t tua. Gia t r i cua V la

B a i 1: Cho luong k h i C O d i qua m gam Fe203 dun ndng, t h u di/oc 39,2
gam h 6 n hop gom bon chat r a n la sat k i m loai va ba oxit cua n6,
dong thcfi C O h o n hop k h i t h o a t r a . Cho h 6 n hop k h i nay hap t h u vao
dung dich nudc v o i t r o n g c6 dii, t h i t h u di/Oc 55 gam k e t t u a . T r i so
cua m l a :

A. 1,120 l i t .

B. 0,896 l i t .

C. 0,448 l i t .
Hii&ng dan gidi
nco = nco, = nc«co, = ^
= ^'^^

D. 0,224 l i t .

=> Vco = 0,04. 22,4 = 0,896 l i t
Chon dap a n B.
B a i 17: T h o i m o t luong k h i CO d i qua ong suf ditog m gam h o n hcfp
Fe304 va CuO nung nong t h u diTOc 2,32 gam h o n hap r a n . Toan bo
k h i t h o a t r a cho hap t h u het vao b i n h dung dung dich Ca(0H)2 du t h u
dtfoc 5 gam k e t tua. Gia t r i cua m l a :
A. 3,22 gam.

B. 3,12 gam.

C. 4,0 gam.

Hii&ng dan
m

hSnhop =

2,32

+

16.

=

3,12

C. 64 gam

D. T a t ca deu sai, v i se k h o n g xac d i n h duoc.
B a i 2: Cho luong k h i H2 c6 dif d i qua ong suf c6 chufa 20 gam hon hop A

t r o n g hon hop A l a :

gam

gom

AI2O3, ZnO, FeO va CaO t h i t h u difcfc 28,7 gam hon hgp chat r a n (Y).
Cho toan bo h 6 n hop chat r a n (Y) tac dung v d i dung dich H C l dif t h u
duoc V l i t H2 (dkc). Gia t r i V la

A. 2gam; 18gam

B. 4gam; 16gam

C. 6gam; 14gam

D. 8gam; 12gam.

B a i 3: Cho luong k h i CO (du) d i qua 9,1 gam hon hop gom CuO va AI2O3
nung nong den k h i p h a n ufng hoan t o a n , t h u dUdc 8,3 gam chat r a n .
K h o i luong CuO c6 t r o n g h o n hop ban dau la

A. 0,8 gam.

C. 6,72 l i t .

B. 40 gam

l a i , t h a y k h o i liXOng chat r a n g i a m 3,2 gam. K h o i liicfng m o i cha't

gidi

B a i 18 Cho dong k h i CO du d i qua h o n hap (X) chufa 31,9 gam

B. 4,48 l i t .

A. 48 gam

gom MgO va CuO nung nong. Sau k h i p h a n iJng hoan toan, dem can

D . 4,2 gam.

Chon dap a n B.

A. 5.60 l i t .

DUNG

D. 2,24 l i t .

B. 8,3 gam.

C. 2,0 gam.

D. 4,0 gam.


Fe(N03)2 + AgNOa
0,1

Dang 4: KIM LOAI TAG DUNG VOI DUNG DICH MUOl
Ca'^ N a "
F e ' * Ag*

Mg'^ A l ' " Zn'*

Au'*

__

•
Tinh

K

Ca

Na

F e ' * Ag

Au

Mg

Al

Zn

Cr

oxi hod ion kim log,i
Ni

Fe

Sn

Pb

Fe

——

.

Tinh
- Chieu p h a n ufng: Chat oxi hod m^nh

H

0,1

=

nAgNo,

C r ' * Fe^* N i ' * Sn'* Pb'* Fe^* H * Cu'*

tang
Cu

khvC kim loi^i

gidm

+ Chat khii mg,nh ->

Chat

-

0,3

=

0,2

chufa h o n hop gom AgNOs 0,1M va Cu(N03)2 0,5M. Sau k h i cac p h a n
ufng xay ra hoan t o a n , t h u diTOc dung dich X va m gam chat r a n Y.
Gia t r i ciia m l a .
A. 2,80.

B. 4,08.

nFe

dan

D. 0,64.

gidi

= 0,04 m o l , n^^. = 0,02 m o l , n^^^. = 0,1 mol

Chat khuf m a n h n h a t se tac dung v d i chat oxi hoa m a n h n h a t triTdc:
+ 2Ag*^Fe'*

+

0,02

2Ag
0,02mol

+ Cu'* ^

0,03
manh

C. 2,16.

Hiidng

Fe

Dicing bai tap nay can lUu y den quy tdc a

dif

Cho 2,24 gam hot sSt vao 200 m l dung dich

B a i 2: (DH khoi B.2009):

0,01

Fe'* + Cu

mol

Vay muoi gom c6 Fe(N03)3 va AgNOs d i i

Fe

Ddu a cang Ian khd ndng phdn ling xdy ra cdng

0,5

•

oxi hod yeu + Chat khvi yeu
PT: Cu'* + Fe

Fe(N03)3 + A g

Fe'* + Cu

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 m o l FeCls.

Sau k h i p h a n ufng xay r a hoan toan t h u dtfoc 3,36 gam chat rkn.

Gia

t r i cua m la
A. 5,04

B. 4,32

C. 2,88

HiCdng dan
npe = ^
BAI T A P A P

56

B a i 1 : Cho O,lmol Fe vao 500 ml dung dich AgNOs I M thi dung dich thu
difOc chufa:

gidi

= 0,06 mol

M g + 2FeCl3

DUNG

MgCl2 + 2 F e C l 2

0,06 -> 0,12 ^

0,12

M g + FeClz ^ MgCl2 + Fe

A. AgNOg

B. AgNOs va Fe(N03)2

0,06

C. Fe(N03)3

D. AgNOa va Fe(N03)3

m = (0,06 + 0,06).24 = 2,88 (g). Chon dap a n C.

Hitcfng dan
"AgNo^

=

0.5-1

=

0,5

0.2

0.1

gidi

0,06

<-

B a i 4: (DH khoi A.2010):

0,06 (mol)

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, t h u duoc m gam k i m loai. Gia t r i cua m la

mol

Fe + 2AgN03 ^ Fe(N03)2 + 2Ag
0.1

D. 2,16.

0.2

A . 6,40g

B. 16,53g

C. 12,00g

D. 12,80g.


Hii&ng dan
Goi

X

^

la so mol cua Zn
65x + 64.2x
X = 0,1 (nzn = 0,1
n^,^,. = 0,2.2 = 0,4

Ta c6:

<-

> 2Fe^^ + Cu^^

B a i 5: (CD - 2009;.- Cho mi gam A l vao 100 ml dung dich gom Cu(N03)2
0,3M va AgNOs 0,3M. Sau k h i cac phan ufng xay ra hoan toan t h i thu
dtfgc m2 gam chat rAn X. Neu cho m2 gam X tac dung vdi lugng du
dung dich HCl t h i thu duac 0,336 l i t k h i (d dktc). Gia t r i cua mi va mg
Ian lUOt la

A. 8,10 va 5,43

B. 1,08 va 5,16

C. 0,54 va 5,16

D. 1,08 va 5,43.

2Ag^

0,1

^

0,2

+

0,03

+

2Ag

0,1

^

0,2

^

+

Ag

0,05

->

Ag
0,03

Cu^^ + 2e ^ Cu
0,03

0,06

0,03

z:> Tong so mol A l = 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
B a i 6: (DH KA.2008): Cho hon hdp bot gom 2,7 gam A l va 5,6 gam Fe
vao 550 ml dung dich AgNOa I M . Sau k h i cac phan ufng xay ra hoan
toan, thu dufcfc m gam chat r ^ n . Gia t r i cua m la (biet thuf t u trong
day the dien hoa: Fe^VFe^' diJng trUdc AgVAg).
C. 59,4

Fe^*

0,05

Khoi liigfng Ag = 0,55.108 = 59,4 gam
Chon dap an C.
B a i 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 k i m 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:
B. 64g; 25,6g

C. 32g; 12,8g

CM(ZnS04) = 2,5CM(FeS04) ^

Ap dung dinh luat bao toan e: 3x = 0,03 + 0,06 => x = 0,03 mol.

B. 54,0

^

Fe^*

HiC&ng dan
0,015

0,03

0,3 mol

Ag"

gidi

0,01

A. 64,8

+

A. 12,8g; 32g

Theo bai ra A l con diX: 2A1 + 6HC1 -> 2AICI3 + 3H2

3x

3Ag

Sau cac phan ilng chat ran la Ag c6 so mol 0,3 + 0,2 + 0,05 = 0,55 mol.

Chon dap an A.

X

Fe

mol

Ag^ + le

+

0,3

Fe'"

Al^^ + 3e

Al'^

Dir Ag^= 0,25 - 0,2 = 0,05 mol

0,1 mol

Hi^&ng dan

^

vdi Fe

Khoi luong k i m loai con lai la khoi ItfOng cua Cu: 0,1.64 = 6,4g

Al

3Ag*

gidi

Sau phan ufng vdi A l , Ag* c6n 0,55 - 0,3 = 0,25 mol dung phan ufng

0,1 mol

ncucondu = 0,1

+

0,1

> 2Fe^" + Zn^^

2Fe^^ + Cu
0,2

Al

t h i so mol cua Cu la 2x
= 19,3
mol ; n c u = 0,2 mol)
mol

2Fe'^ + Zn
0,2

Hii&ng dan

gidi

D. 32,4

Zn

+ CUSO4

2,5x

2,5x

gidi
=

'^MF.SO,

ZnS04 + C u i

2,5x

2,5x
+ Cui

Fe

+ CUSO4

-^FeS04

X

<—

<—

X

n^^^o,

D. 25,6g; 64g.

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 Z n = 64g; m c u bAm len t h a n h F e = 25,6g. Chon dap an B.
B a i 8: Nhiing mot thanh graphit dUdc phu mot Idp k i m 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


AgNOa t h i k h i p h a n uTng xong k h o i luang t h a n h g r a p h i t t a n g
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

Hiidng

dan

len

HiCctng dan

H A g N O , (ban dSu, =

gidi

M ' " + Cui

1 mol

Cu + 2AgN03

1 m o l —> g i a m M - 64 (g)

-^^(mol)
M - 64

4-Amgian,

M + 2 Ag^ ->

2 Agi

1 mol

2 mol

0,015 <- 0,03

= 0,24 (g)

Amtang = 0,52

= _ M 2 _ ^ M
216-M

= 112

B a i 9: N g a m m o t t h a n h Cu t r o n g dung dich c6 chiJa 0,04 m o l AgNOa,
sau m o t t h d i gian lay t h a n h k i m loai ra t h a y k h o i li/dng tSng hdn so
vdfi luc dau la 2,28 gam. Coi toan bo k i m loai sinh ra deu b a m het vao
t h a n h Cu. So m o l AgNOs con l a i t r o n g dung dich la
HUdng

C. 0,02.
dan

D . 0,015.

2x
64x

=

M g + CuCl2

(tan)

gam.

C. 2,43 gam.

D . 4,13

gam.

gidi

MgCla + Cu
X

y

y

Ta c6:

2,28

(64 - 24).(x + y) = 0,8
=> 40(x + y) =0,8
=>

da =0,04 - (0,015.2) =0,01 mol

X

+ y = 0,02

K h o i lirong muoi tao t h a n h la 3,23 + (0,02.24) - (0,02.64) = 2,43 gam
Chon dap an C.

Chon dap an A.
B a i 10: N g a m m o t v a t bSng dong c6 k h o i li/cJng 15 gam t r o n g 340 gam
dung dich AgNOs 6%. Sau m o t t h d i gian lay v a t ra t h a y k h o i luong

AgNOa t r o n g dung dich g i a m 25%. K h o i lUcfng cua v a t sau p h a n iJng la
A. 3,24 gam.

B. 1,43 gam.
Hii&ng dan

X = 0,015mol
n^^No^

(b^m) ~

M g + Cu(N03)2 ^ C u ( N 0 3 ) 2 + Cu

2x

108.2X -

A. 1,15 gam.

X

gidi

Cu + 2AgN03 - 4 Cu(N03)2 + 2Ag
X

ban d i u +

B a i 11: Hoa t a n 3,23 gam h o n hgtp gom CuCl2 va Cu(N03)2 vao niTdc
difcfc dung dich X. N h u n g t h a n h k i m loai M g vao dung dich X den k h i

dung dich m a t mau xanh r o i lay t h a n h M g ra, can l a i t h a y tSng t h e m
0,8 gam. K h o i lU(?ng muoi tao ra t r o n g dung dich la

Vay dap a n diing la B: Cd.

B. 0,005.

^v&t

0,03 m o l

Chon dap an C.

(g)

V i cung m o t t h a n h g r a p h i t t h a m gia p h a n iJng n e n :

A . 0,01.

^

= 15 + (108.0,03) - (64.0,015) = 17,28

tSng 2.108 - M = 216 - M (g)
^

_a2^
M-64

> Cu(N03)2 + 2 A g i

n i v a t sau phan ilng =

(mol)

,
mol;

25
n A . N O 3 < « = 0 , 1 2 . — =0,03 mol.

Goi k i m loai c6 hoa t r i I I do la M c6 k h o i lifcfng m(g)
M + Cu^^ ^

340.6
^^Q^QQ = 0.1^

D . Sn

giai

B. 2,28 gam.

C. 17,28 gam.

D . 24,12

gam.

B a i 12: Ngii6i ta phu m o t Idp bac t r e n m o t vat b ^ n g dong c6 k h o i lifOng

8,48 gam bSng each n g a m vat do t r o n g dung dich AgNOa. Sau m o t
t h 6 i gian \zy v a t do ra k h o i dung dich, rufa nhe, l a m kho can dufeJc 10
gam. K h o i luong A g da phu t r e n be mSt cua vat la
A .1,52 gam.

B .2,16 gam.

C. 1,08 gam.

D . 3,2 gam.


Hii&ng dan
Cu + 2AgN03 -> Cu(N03)2

Fe

2x

T a c6: 108.2x - 64x = 10 X

+ Cu
0,01

+ CuS04

T a c6:

^FeS04

+ Cui
X

64x - 56x = 0,12
X

= 0,015 m o l

Z

gam

Chon dap a n B.
Sau k h i k e t thiic cac p h a n ufng, loc bo p h a n dung dich t h u

duac m gam bot r ^ n . T h a n h p h a n % theo k h o i li/ong cua Zn t r o n g h6n

^cuso,

= 0 , 0 1 + 0,015 = 0,025 m o l

_ 0,025

B a i 13: Cho m gam h o n hcjp hot Z n va Fe vao li/cfng d u dung dich
CUSO4.

Mg2"

X

8,48

= 0,01 m o l

m A g = 0,01.2.108 = 2,16

+ Cu^^ ^

0,01

+ 2Ag

2x

X

Mg

gidi

Chon dap a n C.

hop ban dau la
A. 90,27%.

B. 85,30%.

C. 82,20%.

Hiidng

Zn +

CUSO4

dan

D . 12,67%.

Dang 5 . CDs. SOa T A G D U N G VOl D U N G D|CH K I E M

gidi

A. K I M L O A I K I E M : T O M T A T LI T H U Y E T

1. V i t r i t r o n g b a n g t u a n h o a n , c a u h i n h e l e c t r o n

-> ZnS04 + C u i

X

K i m loai k i e m gom: L i t i ( L i ) , N a t r i (Na), K a l i (K), Rubidi (Rb), Xesi
(Cs), F r a n x i ( F r ) .

X

Fe +

CUSO4

^ FeS04 + C u i

y

Thuoc n h o m l A

y

V i m hon hop r ^ n dau b^ng m h o n hdp r ^ n sau

Cau h i n h electron:

N e n : 64x + 64y = 65x + 56y <=> x = 8y

N a ( Z = l l ) l s ' 2 s ' 2 p ' 3 s * hay [Ne]3s'

% Zn = — ^ ^ - ^ y
.100% = 90,27%
65.8y + 56y

K (Z=19) Is22s22p^3s^3p^4s' hay [ A r ] 4 s '
Deu CO 1 electron d Idp ngoai cung

Chon dap An A .
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

CUSO4

x M , khuay nhe cho den k h i dung dich m a t

mau

x a n h n h a n t h a y k h o i liicfng k i m loai sau p h a n ufng la 1,88 gam. Gia
t r i cua x la
A . 0,04M.

B. 0,06M.

C. O . I M .

Hii&ng ddn
n i k i m io,i sau phan ling =

1,88

-

1,12

=> Am =0,01.(64-24) = 0,4 gam
n i c b n i a i = 0,52

L i (Z=3) ls'2s* hay [He]2si

-

0,4

=

0,12

-

0,24

D . 0,025M.

gidi
= 0,52

I I . T i n h c h a t ho a h o c
Co t i n h khut m a n h : M ->

+ e

1- T a c d u n g v6i p h i k i m :
T h i du:

4Na + 0 2 - ^ 2 N a 2 0

2Na + C ] 2 ^ 2 N a C l
2. T a c d u n g v d i a x i t ( H C l , H2SO4 l o a n g ) : tao muoi

gam

T h i du:

2Na + 2HC1 ^ 2 N a C l + 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

H2


C. DANG T O A N CO2 (HOAC SO2) VAO DUNG DICH KIEM NaOH (HOAC
KOH)

I I I . Dieu che:
1. Nguyen tSc: khuf ion kim loai kiem thanh nguyen tur.

C a c pht^ofng trinh xay ra:

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
PTDP:
B.

MQT

SO

HOP

CUA

CO2 + O H - ^

KIM

LOAI

CO2 + 2 OH" ^

KIEM:

+ Tac dung vdi axit: tao muoi va niTdc

<1

NaOH + HCl ^ NaCl + H2O

• CO2
BAI T A P V A N

2NaOH + CUSO4 ^ Na2S04 + Cu(0H)2i

Na2C03 + COot + H2O

^^^^

NaHCOs + HCl ^ NaCl + C 0 2 t + H2O
?

NaHCOa + NaOH ^ Na2C03 + H2O
I I I . Natri cacbonat - Na2C03

•*

+ Tac dung v d i dung dich axit manh:
Na2C03 + 2HC1 ^

I V . K a l i nitrat: KNO3
Tinh chat: c6 phan ling nhiet phan
2KNO3

= 1

du

2KNO2 + O2

• NaHCOa
• NaaCOa

• NajCOi

• NajCOa
• NaOH dir

dan

gidi

ehi tao muoi NaHCOg.

B a i 2: Hap thu hoan toan 4,48 l i t 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

2NaCl + CO2T + H2O

Muoi cacbonat cua kim loai kiem trong nadc cho moi triTdng kiem

>2

CO2
+
NaOH ^
NaHC03
0,1 moi
->
0,1 moi
^ii^ico, = 0.1-84 = 8,4 gam.

+ Tac dung v d i axit:
+ Tac dung vdi dung dich bazd:

=2

B a i 1: Sue 2,24 l i t k h i CO2 vao 100ml dung dich NaOH I M , t i n h khoi

lUdng muoi thu duac.

2. T i n h litofng tinh:

Thi du:

• NaHCOa

HUdng

2NaHC03 — ^

1 <2

DUNG

1. P h a n v(ng p h a n hiiy:

T h i du:

=1

• NaHCOa

2NaOH + CO2 -> Na2C03 + H2O

n. Natri hidrocacbonat - NaHCOg
T h i du:

CO3'- + H2O

San phani

+ Tac dung vdi dung dich muoi:
T h i du:

HCO3-

^NaOH

+ Tac dung vdfi oxit axit: tao mtioi va niidc
T h i du:

(2)

Ta can lap t le

I. Natri hidroxit - NaOH
T h i du:

(1)

Thiic chat ta dung hai pt sau

) 4Na + 2H2O + O2

CHAT QUAN TRQNG

NaHCOa

CO2 + 2NaOH ^ Na2C03 + H2O

NaOH

2NaCl — ^ ^ ^ ^ ^ 2Na + CI2
4NaOH

CO2 + N a O H

Hii&ng dan

2NaOH+ SO2 ->
0,4
m

0,2
muoi

NazSOg
0,2

= 0,2.126 = 25,2 gam

Chon dap an D.

C.23,0g
gidi

D.25,2g.

B a i 3: DSn 10 l i t h 5 n 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 t h u difgrc I g k e t t u a . T i n h p h a n t r a m theo
the t i c h CO2 t r o n g h 5 n hap k h i .
A.2,24% va 15,68%

B. 2,24%

C. 15,68%

D. 2,24% va 7,84%.
Hitcfng ddn

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 , t h u duac dd "
A. Cho 100 m l dung dich Ba(0H)2 I M vao dung dich A di/ac m gam

giai

k e t tua. Gia t r i m bang:
A. 19,7g

nco^ = n c , c o 3

=Y^ =

B. 15,76g

0,01mol

C. 5 9 , l g

Hiidng

= 0,02.2 = 0,04mol

nca.oH),

m^uo. = 0,1.23 + 0,1.39 + 0,1 .81 + 0,05.80 = 18,3 gam
Chon dap a n D.

Triforng hofp 1:
•

Theo d i n h luat bao toan nguyen tuT (tdng khoi lugng ion trong muoi
bang long khoi lugng muoi)

"N^OH

= M i : ^ . 1 0 0 % = 2,24%

= Hj^^, = n^,,^ = 0,38

dan

mol

gidi
"^n^iom, = n^^^, = 0 , 1

E n ^ j ^ . = 0,38 + 0,2 = 0,58 m o l

nco, =

-^Q

CO.

TrvLUns hc/p 2:

•

- n^^co^ = 2.0,04 - 0,01 = 0,07mol

n^o^ =

^ 0,07.22,4

^

^

g

^

CO2 + O H X

^

D.55,16g

22,4

- 0,3

mol

mol

HCO3-

X

X

CO2 + 2 0 H - -> C O 3 ' - + H 2 O

y

Chon dap a n A
B a i 4: H a p t h u 3,36 l i t SO2 (dktc) vac 0,5 l i t h 5 n hgp gom N a O H 0,2M

Ta c6:

va K O H 0,2M. Co can dung dich sau p h a n ufng t h u duc/c k h o i I'Jcfng
muoi k h a n l a .
A. 9,5gam

B. 13,5g
HU&ng dan

nwaOH

nKOH

= n ^ , = n^j,, = 0,5.0,2 = 0,1 m o l

SO2 + 0 H - - >
X

HSO3

D. 18,3g

y

x + y = 0,3

fx = 0,02

x + 2y = 0,58

[ y = 0,28

B a ' " + CO^-

BaCOsi

0,1

0,1

0,28

K h o i lirang k e t t u a l a : m B ^ c O j = 0,1.197 = 19,7 gam
Chon dap a n A .
B a i 6: D o t chay hoan toan 0,1 m o l etan r o i hap t h u t o a n bo san p h a m
chay vao b i n h chufa 300 m l dd N a O H I M . K h o i liiong m u o i t h u difgrc
sau phan iJng?
A. 8,4g va 10,6g

B. 84g va 106g

C. 0,84g va l , 0 6 g

D. 4,2g va 5,3g.

X

Hiidng

SO2 + 2 0 H - - ^ S0^-+ H 2 O
y

gidi

= n^^. = n^„_ - 0,5.0,2 = 0,1 mol

Z n ^ ^ . = 0,1 + 0,1 = 0,2 m o l

X

C. 12,6g

2y

2y

y

Taco:

C2H6 ^ 2CO2
0,1 - > 0,2mol

x + y = 0,15

fx = 0,1

" x + 2y = 0,2

|y = 0,05

CO2 + N a O H -> NaHCOa
X

X

X

ddn

gidi


b) Tdc dung vai axit

COa + 2 N a O H ^ NazCOg + H2O
y

2y

y

Ta c6: <

fx = 0,1
=> <
x + 2y = 0,3
,.y = 0 , l

V4y

-mN,HC03

H2 t\i do.

rx + y - 0 , 2

= 0.1-84 = 8,4

M + 2H^

gam

M^* + H2t

- D o i v d i axit c6 t i n h oxi hod m a n h nhiT HNOs, H2SO4.

gam

-> N"'(N02), N*^(NO), N-^(NH4N03)...

Chon dap a n A.

S^*^ ^ S^'(S02), S-=^(H2S), S°(S).

D. DANG TOAN CO2

(HOAC SO2)

VAO

DUNG DjCH Ca(0H)2 h o g c

Ba(0H)2. KIM LOAI PHAN NHOM CHINH NHOM II
TAT

MSO4 + H z t

M + H2SO4

mNa,co, - 0,1.106 = 10,6

T6M

trong dung d i c h a x i t ( H C l , H2SO4 loang) t h a n h

_ D e d a n g khilf ion

LITHUYET

4 M + IOHNO3
c) Tdc dung vai
T r o n g H2O,

4M(N03)2 + NH4NO3 + 3H2O

H2O

B e k h o n g p h a n urng, M g khuf c h a m , c a c k i m loai con l a i

khuf m a n h .

1. Vj tri trong bang

thong t u a n hoan, tinh chat v^t If

M + 2H2O ^

a) Vi tri

M(0H)2 +
Ca(0H)2 +

C a + 2H2O

K i m loai p h a n n h o m I I gom:
B e r i (Be); Magie (Mg); Canxi

(Ca);

T r o n g cac chu k i cac nguyen to nay diing l i e n sau k h i loai k i e m .
b) Tinh chdt vat li

- M g day c a c k i m loai hoat dong y e u hcfn r a k h o i dung d i c h muoi
M g + CUSO4 ^

- C a c k i m loai con l a i t a c dung vdi H2O trong dung dich

D i e n p h a n n o n g c h a y muoi halogenua cua chiing

- La k i m loai m e m (mem hon n h o m )

MX2

- K h o i li/Ong r i e n g tiidng doi nho

X : halogen

Jig" phan n6ng chay

^ M

+

X2

hoc

Cac nguyen to p h a n n h o m c h i n h n h o m I I c6:
- 2 electron hoa t r i (s^)
- Co ban k i n h nguyen tuf \6n
- L a nhOfng chat khuf m a n h M - 2e -> M^^
T r o n g cac hap chat c^c nguyen to nay c6 so oxi hoa +2.
a) Tdc dung vai phi

kim

2 M 0 ( M la nguyen tut k i m loai)
2CaO

1. C a n x i oxit: CaO
C a x i oxit l a oxit b a z a
- T a c dung m a n h l i e t v d i H2O tao b a z a

C a O + H2O ^

Ca(0H)2

C a O + 2HC1 ^ CaCl2 + H2O
- T a c dung vdi oxit a x i t tao muoi tuang iJng
C a O + CO2

- V d i CI2: M + CI2 -> MCI2
M g + CI2 ^

MQT SO HOP CHAT QUAN TRONG CUA CANXI

- T a c d u n g vdi n h i e u a x i t tao muoi tuang ufng

- V d i oxi k h i dot nong:
2Ca + 0 2 ^

MgS04 + C u i

3. D i e u che

- N h i e t do n o n g chay n h i e t do soi thap

2 M + O2

H2T

d) Tdc dung vai dung dich muoi

S t r o n t i (Sr); B a r i (Ba) va Radi (Ra)

2. T i n h chat hoa

H2t

MgCl2

CaC03

- C a n x i oxit dirge dieu che bkng phijang phap nhiet p h a n muoi cacbonat.
C a C 0 3 — ^ C a O + CO2


5. Nifofc ciJtng

2. C a n x i h i d r o x i t : C a ( O H ) 2

1. Nx^oTc cufng

L a c h a t r S n it t a n t r o n g H2O

Nadc curng la nUdc c6 chura n h i e u ion Ca^*, Mg^*. Nifdc k h o n g chiJa
hoSc chura i t nhOfng ion t r e n , goi la niTdc m e m .

D u n g d i c h C a ( 0 H ) 2 c6 t i n h bazO y e u h o n N a O H
- T a c dung v d i a x i t v a oxit a x i t tao m u o i tUofng ting

Ca(0H)2 + 2HC1 ^
C a ( 0 H ) 2 + CO2 ^

2. P h a n l o a i nvCdc

C a C l a + 2H2O

Nude curng chia t h a n h 3 loai

C a C O a i + H2Q

1. Nifde ciJng tarn t h d i : la nirdc curng c6 chura ion HCO3"

C a ( 0 H ) 2 + 2CO2 ^ C a ( H C 0 3 ) 2
, CaCOH),

N e u t i le m o l

CO2

2. NiXdc curng v i n h cijfu: la niJdc curng c6 chufa ion CI" hoSe SO^

1 ^

< — tao muoi a x i t
2 •

3. Nifdc curng toan p h a n : L a niidc cufng c6 chiJa dong thcfi cac ion
HCO3", C r hoSe SO^-

N e u t i le m o l ^^^^^^2 < ^ ^.^Q jjjy5'j t r u n g t i n h
CO2
N e u t i le m o l 9^^^k
CO2

trong k h o a n g

i < ^^^^^
2

3. T a c h a i c i i a nitofc ciifng
< 1

tao d6ng

- Xa phong k h o n g t a n
- V a i soi mau muc n a t

t h d i 2 muoi

- Nau thurc a n lau c h i n , g i a m m u i v i

- T a c dung v d i dung d i c h muoi

- Tao chat c&n t r o n g n o i h a i l a m l a n g p h i n h i e n lieu...

Ca(0H)2 + NaaCOs

C a ' ^ + CO^- ^
3. C a n x i c a c b o n a t

cufng

CaCOai + 2 N a O H

4. C a c h l a m m e m ni^cfc

CaCOai

Nguyen tSc: L a m g i a m n o n g 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 p h a n chat k h o n g t a n .

CaCOa

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 t a n trong

H2O

C a C O s l a muoi c u a a x i t y e u v a k h o n g b e n

CaCOa + 2 H C 1

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 d a n trong H2O c6 CO2

CaCOa + H2O + CO2 -> C a ( H C 0 3 ) 2
4. C a n x i s u n f a t : C a S 0 4
C a S 0 4 con goi l a t h a c h cao, m ^ u t r ^ n g , i t t a n t r o n g H2O
C a S 0 4 . 2 H 2 0 : t h a c h cao song
2 C a S 0 4 . H 2 0 : t h a c h cao n u n g n h o lufa
C a S 0 4 : t h a c h cao k h a n
T h a c h cao n u n g duoc dung de due tiiOng.dung trong y hoc de bo bot...

Phirong phap: Phuong phap hoa hoc va phuong phap trao ddi ion.

a) Phuang

phap

hod

hoc

* D o i v d i ntrdc cufng t a m t h d i . Dun nong trUdc k h i dung
Ca(HC0a)2 — ^

C a C O a i + H2O + C 0 2 t

Loc bo chat k h o n g t a n , dirge nifdc m e m
- D u n g Ca(0H)2 v t o du de t r u n g 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 m e m
* D o i v d i nude cufng v i n h cufu va ni/dc cufng t o a n p h a n
Dung dung dich Na2C0a, Na3P04
C a S 0 4 + Na2C03 -> C a C 0 3 i + Na2S04
Ca(HC03)2 + Na2C03 ^ CaCOgi + 2NaHC03


Ca?* + 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 Ca^^ va Mg^* the vao do la ion N a \a duoc nude mem.
D^ng bai tap CO2 vao dung dich Ca(OH)2 hoqc Ba(OH)2 thi:
Dang 1: Biet nc^o,,,,^ .ncaco,

"co,

TrUang hap 1: rico, =

ntettua

TrUdng hop 2:

2nb«,„

HCO,=

= 0,02 mol

- ntettua = 2.0,03 - 0,02 = 0,04 mol

Chon dap an A.
B a i 2 : (DH A - 2008): Hap thu hoan toan 4,48 l i t k h i CO2 (d dktc) vao
500 m l dung dich hon hop gom NaOH 0,1M va Ba(0H)2 0,2M, sinh ra
m gam ket tua. Gia t r i cua m la.

A. 19,70 g.

B. 17,73 g.

C. 9,85 g.

D. 11,82 g.

Hiidng dan gidi
^Ca(0H)2

Dang

2:

^^^'^

T H I : n^^o^ - nc^co^

nNaoii = " N a * = " o H " " ^ ' ^ ^

T H 2 : n^o^ - 2.nc^,o„)^ -''^c^co^

I n^„. = 0,05 + 0,2 =0,25 mol;

B i e t n^o^,

nc^oH)^

t i m nc^co^

CO2 + OH" ^
X

o

W 2

^^CatOlllj

CO2

T H I : nco, = nc.co,
TH2

: ncacos

- •^^cMom,

~

"co^

^

BAI T A P V A N

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

= 0,02

mol

C O 3 ' " + H2O

y

x + y = 0,2

rx = 0,15

" x + 2y = 0,25

[ y = 0,05

0,05

0,05

Chon dap an C.

A. 0,02 mol va 0,04 mol

=

HCO3"

Khoi laong ket tua la: ra^^co, =0,05 .197 = 9,85 gam
^CMOMU

B a i 1. Hap thu toan bo x mol CO2 vho dung dich chiJa 0,03 mol Ca(0H)2
diroc 2 gam ket tua. gia t r i x?

nr,ro

2y

0,1

DUNG

2

M | = 0,2 mol

Ba'" + CO3'- ^ BaCOai

_ ,
"Ca(OIl)o ~ •

Neu n^o, ^ ric^co, =^

n^o,

X

+ 20H- ^

y

Dang 3: Biet nc^co^. "00, t i m
nCaC03

X

"^u^iom, = ^ 3 , . . = 0 , 1 mol

Bai 3: (BH A - 2009): Cho 0,448 l i t k h i CO2 (d dktc) hap thu het vao
100 ml dung dich chuTa hon hop NaOH 0,06M va Ba(0H)2 0,12M, thu
duoc m gam ket tua. Gia t r i cua m la.
A. 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;

I n ^ j j =0,006 + 0,012.2 = 0,03 mol;

HB^IOH),

nco, =

= n^^,, = 0,012mol
= 0,02mol


C02 + O f f

^

X

X

X

2y
x + y = 0,02

[ y = 0,01

Ba^^ + CO^-

BaCOai

0,012

0,01

0,01

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

K h o i laang k e t t u a l a : mg^co, = 0 , 0 1 .197 = 1,97 gam
Ca(0H)2 + CO2

Chon dap a n C.
2,5 l i t dung dich B a ( 0 H ) 2 nong do a mol/1, t h u difdc 15,76 gam k e t

gidi

CaCOsi + H2O

Ca(0H)2 + 2CO2 ^

tua. Gia t r i cua a l a .

Ca(HC03)2

0,08 < - 0,04

B. 0,048 M .

C. 0,06 M .

D. 0,04 M .

Ca(HC03)2 —

Hiiafng d&n gidi

^

0,04

2,688 ^ , , „ „ , _
_ 1 5 ^
= 0,12mol; ng^co, =
= 0,08mol

nco, > neaco,

dan

0,06 < - 0,06

B a i 4: (DH A - 2007;.- H a p t h u hoan t o a n 2,688 l i t k h i CO2 (or dktc) vao

=

= 2.0,15 - 0,1 = 0,2 m o l

B a i 6: T h o i V l i t (dktc) CO2 vao 100 m l dung dich Ca(0H)2 I M , t h u diroc
6 gam k e t tiia. Loc bo k e t t u a lay dung dich t h u ducfc dun n o n g l a i c6
4gam k e t t i i a nufa. Gia t r i V l a :

fx = 0,01

' x + 2y = 0,03 ^

n^f>

ntettua

Chon dap a n D .

y

A . 0,032 M .

-

Vco, = 0,2.22,4 = 4,48 l i t

CO2 + 2 0 H - - > COa^" + H 2 O

y

n^^^ = 2nb,,o

TriCdng hap 2:

HCO3"

CaCOai + H2O + C 0 2 t

<-

0,04

I n c o , =0,08 + 0,06 = 0,14 m o l => Vco,= 0,14 .22,4 = 3,136 l i t
Chon dap a n A.

chi xay r a t r i / d n g hap 2

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?

C„^,„„,

= - ^ = 0,04M

C. G i a m 16,8gam

D G i a m 6,8gam.
gidi

Be tinh khoi lixang chat sau phan i2ng tang hay giam

B a i 5: D a n V h t CO2 (dkc) vao 300ml dung dich Ca(0H)2 0,5 M . Sau

p h a n l i n g duac lOg k e t tua. Gia t r i cua V bSng:
B. 3,36 l i t
Hildng
TriCang hap 1:

B. T a n g 20gam

Hii&ng dan

Chon dap a n D.

A . 2,24 l i t

A. T a n g 13,2gam

C. 4,48 l i t
d&n gidi

n^o^ = nkettoa = 0,1 m o l
Vco = 0,1 .22,4 = 2,24 l i t

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^^^^

D. Ca A , C deu dung
J^COj -

2nbaza ~

nen

chi

xay

ra

TH2

^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 a n D.


B a i 8: Cho 5,6 h t h o n hdp X gom Ng va CO2 (dktc) d i cham qua 5 K t
dung dich Ca(0H)2 0,02M de p h a n ilng xay ra hoan toan t h u dUdc 5
gam k e t tua. T i n h t i k h o i h o i cua h 6 n hcfp X so v d i H2.
A. 18,8

B. 1,88

C. 37,6

Hiidng
= n^
Truang

hap 1:

dan

= O.lmol
n^o^ =

gidi
100

=3^
""'"^
2

j^>,15.44.0,1.28 ^
0,25

= 15,6

= 18,8

Chon dap a n A.
B a i 9: Sue CO2 vao 200 m l h 5 n 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 t h a y tao 23,6 gam k e t t u a .
C. 2,24 l i t

HU&ng dan
^on-

Nguyen

hap 1:

khi chuyen

hap 2:

= 0,2 m o l ;

Pt:

_ 23,6
-^BaCOj

~

CO2 + O H " -> HCO3"
X

CO2

+

y

X
20H-

2y

^

0 0 3 ' - +

=> X

(1)
(2)

2YCI3 + 3H2O +3CO2T

Theo 2 phuong t r i n h (1),(2) t a t h a y

•^muo'i clorua ~

^

Pt:

h h rauoi cacbonat

H2O

197

= 0,12mol

A C O 3 + H2SO4

-'^cOo

B2CO3 + H2SO4

^ | y = 0,12

"co, =0,12 + 0,26 = 0,38 mol

(loang) ^
(loang) ^

A S O 4 + CO2T + H 2 O
A 2 S O 4 + CO2T + H 2 O

Cach t i n h n h a n h cho m u o i 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 H C I thay thoat r a 0,4481 CO2 (dktc). K h o i liicfng muoi clorua tao r a
trong dung dich la:
A. 115,22g

B.151,22g

C. 116,22g

HUcfng dan

fx = 0,26

Vay Vco^ =0,38.22,4 = 8,512 l i t
Chon dap a n D.

XCO3 + 2HC1 -> X C I 2 + H2O + CO2T
Y2(C03)3 + 6HC1

y

y = 0,12
" x + 2y = 0,5

chdt.

Cach t i n h n h a n h dung cho tr&c n g h i e m .

ri^^mn, = ^ B a ^ * = ^' 1^"^°^

n^o^ = nkettta = 0,12 m o l

X

tii chat nay sang chdt khac Ung vdi 1 mol

lugng

.(71-60)
M u o i cacbonat tac dung v d i dung dich H 2 S O 4 loang

Vco.= 0,12 .22,4 = 2,688 l i t
Truang

D . 44,8

tdc cua phuang phdp nay Id dua vdo sU tang, gidm khoi

D. Ca A va B dung

gidi

I n ,' ,O, i, r. =0,2 +' 0,15.2 = 0,5 m o l ;'
Truang

C. 224

n c f = 2n^^,_ (ma M^,. = 35,5; M^^,. = 60)

B. 2,688 l i t

-

B.44,8 hoSc 224

Dang 6. MUQI CACBONAT TAG DUNG VO! DUNG D|CH
HCI, H 2 S 0 4 loang

T i n h Vco^ da dung d (dktc).

-

D . 16,745g

tuf cac b a i t r e n .

TMng

= 0,05 mol

hap 2: n^^^ = 2nb^,„ - nkeuaa = 2.0,1 - 0,05 = 0,15 m o l

n KOH

C. 9,85g

G i a i tiicfng t i i b a i t r e n

A. 44,8 hoSc 89,6

Khac dap an l o a i

A. 8,512 l i t

B. 14,775g

duac 0,2g k e t tua. Gia t r i V m l l a :

= 0,05mol

^
0,05.44 + 0,2.28
= 31,2 =^ d , / , , = ^
M =
0,25
'
"'"^
2
Truang

A. 23,64g

B a i 11: T h o i V m l (dktc) CO2 vao 300 m l dung dich Ca(0H)2 0,02M, t h u
5

n CRCOO

nkettua

D. 21

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 t o a n t h a y tao m
gam k e t tua. T i n h m .

D . 161,22g

gidi

C a c h g i a i 1:
A C O 3 + 2HC1 -

ACI2 + H^O + CO2 t

B 2 C O 3 + 2HC1

> 2BC1 + H2O + CO2 t

R 2 C O , + 2HC1 ^

2RC1 + H.,0 + CO,

t