Boi duong hoc sinh gioi hoa hoc 8+9 dao huu vinh p2

B a i giai

2.

PhSn tfng dien phan n6ng chay K C l :
2KC1

dpnc

2K + 2H2O

>

2K

+ cut

(A)

(B)

2 K O H + H2T
(D)

(1)

3.

FeO + CO2
-> A1(N03)3 + N 2 0 t +H2O
— > ZnS04 + H2ST +H2O

Z n + H2S04djc

5.

>
1r »

B a i giai

(C)

+ 8SO2

4FeS2+1102

IFejOi

2.

2 y A l + 3Fe,0y

3xFe + y A l 2 0 3

3.

Fe^Oy + (y - x ) C O

xFeO + (y - x)C02

4.

8A1 + 3OHNO3

V i theo cdc phan iJng ( 1 , 2, 3, 4) so mol H C l = so m o l K O H nen k h i cho het

5.

4Zn + 5H2S04d5c

D vao E dung dich trd thanh trung tinh va quy chuyen t r d l a i mau t i m .

Bai 4. Hon hdp A g o m AI2O3, M g O , Fe304, CuO. Cho luong H2 dif qua A nung

2HC1

>

(3)

dd H C l (E)
K h i cho quy t i m vao E, quy t i m chuyen thanh m ^ u do. K h i them D vao E xay
ra phan uTng:
KOH + HCl

> K C l + H2O

(4)

-> khong phan uTng

M g + CUSO4

-> MgS04 + Cui

M g + 2AgN03

-> Mg(N03)2 + 2 A g i

2A1 + 6HC1
2A1 + 2NaOH + 2H2O
2A1 + 3CUSO4
A l + 3AgN03

> 2AICI3 + 3H2T
> 2NaA102 + S H j t

8A1(N03)3 + 3 N 2 0 t + I5H2O
-> 4ZnS04 + H2ST +4H2O

-

B a i giai

1. CAc phan uTng:

Mg + NaOH

;3nt! = i i.nq -; ^

HNO3 loang (dir C O k h i duy nhat N O bay ra). V i e t cdc PTPU xay ra.

B a i giai

MgCl2 + H2t

Cdc phan tfug:

'UO

3Fe + 4H2O

Fe304 + 4H2
CuO + H2

-> Cu + H2O

Chat ran B g o m Fe, Cu, AI2O3, M g O :
T^c dung cua B v d i dung dich N a O H va vcli CO2

-

AI2O3 + 2NaOH

Al2(S04)3 + 3 C u i

> 2NaA102 + H2O

CO2 + 2H2O + NaA102

-> Al(N03)3 + 3Ag>l

Tic

Al(0H)3i +NaHC03

dung v d i dung dich HNO3:

FeCl2 + H 2 t

M g O + 2HNO3

Mg(N03)2 + H2O

Fe + N a O H

-> khong phan ufng

Fe + 4HNO3

Fe(N03)3 + 2H20 + N O t

Fe + CUSO4

-> FeS04 + Cui

3Cu + 8HNO3

3Cu(N03)2 + 4H2O + 2 N 0 1

Fe(N03)2 + 2 A g i

N e u d i r A g N 0 3 thi:
Fe(N03)2 + A g N 0 3

1-

Chosddobienhod:
-> A ,

-> Fe(N03)3 + A g ^

B a i 3. Can blng cac phuTcfng tnnh phan iJng:
FeS2 + O2

•ad:

Tdc dung v d i H2:

Fe + 2HC1

Fe + 2 A g N 0 3

aii^r

v^ chat rin D. Sue k h i CO2 dU vao dung dich C va hoa tan D bhng dung dich

N a O H , C U S O 4 . A g N O j . V i e t cac phiTdng trinh phan ufng xay ra.

Mg + 2HC1

j^nti

^ht>^

n6ng diTdc chat r ^ n B . Hoa tan B vao dung dich N a O H diT dtfdc dung dich C

B a i 2. Cho cac k i m loai M g , A l , Fe Ian liTdt tdc dung v d i cac dung dich HCl,

102

Fe^Oy + C O

1.

H2 + CI2

1.

Fe + AI2O3

A l + HNO3

4.
(2)

A l + Fe.Oy

-> FezOj + SO2

+x -> A2

+Y

-> A :
-> Fe(OH)3

|Fe(0H)3
B,

+z

^ B2

+T

B
103


Cty TNHH MTV DWH Khang Viet
Tim cong thuTc cua cac cha't vCng vdi cac chat A,, B,... Vie't cdc phifdng trin},

SO3 + H 2 O

phan tfng theo sd do do.
2. Cho sd do b i e n hoa:
^

A,

+Y

+x ^ A2

CaCO.,

CaC03
+z

B,

+T

> B

phan ufng theo sd do.

H

H2 + CI2

2HC1

2Fe + 3CI2

> 2FeCl3

(8)

>

2FeCl3 + 3H20

(2)

3 A g C U +Fe(N03)3

(3)

2H2O + Ba

Ba(OH)2 + H 2 t

(4)

1.

Fe203 + CO

B a ( 0 H ) 2 + Na2C03

BaCOji

(5)

2.

X2 + X 3

(6)

3.

X2 + X 4

4.

X5 + X ,

-

5.

X7 + Xs

-

6.

X y + X|()

Fe(N03)3 + 3 N a O H

-> F e ( 0 H ) 3 ; + 3 N a N 0 3

G h i c h u : Co the su" dung cac phan ufng khac, axit khac.

-> CaO + C02T

CaCOj

(1)

CaO + H2O

-> C a ( 0 H ) 2

(2)

C a ( 0 H ) 2 + 2HCI

-> CaCl2 + 2H2O

(3)

2CO2 + B a ( 0 H ) 2

-> Ba(HC03)2

2 N a O H + Ba(HC03)2

> B a C O j ^ + Na2C03 + 2H2O

C a ( 0 H ) 2 + Ba(HC03)2

> C a C 0 3 i + B a C 0 3 i + 2H2O

C a C l j + Na2C03

B a i 6. Tuf cac nguyen lieu ban dau la khoang vat pirit, m u o i an, nU'dc, cac chai

7.

xuc tac va thiet b i can thiet, vie't cac phU'dng trinh phan tfng dieu che' Fe.

8.
9.

B a i giai
10.

4FeS, + 1 1 0 ,
2SO2 + O2

2Fc203 + 8 S 0 2 t
V2O5

-> 2SO3

(1)
(2)

(10)

Bai 7. T i m cac chat X , , X2, X 3 . . . ihich hdp va hoan thanh cac PTPlJ sau:

Co the silr dung cac phan iJng khac, chat khac.

Cac phan iJng can thiet

NaHS04 + H C l t

(9)

Co the siir dung cac sd do khac.

r

-> Fe,0> + X i
>

—>•
>

B a S 0 4 i +Na2S04 + C 0 2 t + H2O
B a S 0 4 i + Na2S04 + CO2T + H2O
A g 2 0 i + K N O 3 + H2O
Ca(H2P04)2
Fe2(S04)3 + S02T + H 2 O

Xy

+ X|()

>

Fe2(S04)3 + S 0 2 t

+H2O

Xy

+ X|0

>

Fe2(S04)3 + S 0 2 t

+H2O

Xy

+ Xlo

>

Fe2(S04)3 + S 0 2 t + H2O

Xy

+ XlO

—— >

Fe2(S04)3 + S 0 2 t

+H2O

Xy

+ X||)

— >

Fe2(S04)3 + S 0 2 t

+H2O

CaC03^ +2NaCl

FeClj, F e ( 0 H ) 3 , NaHS04.

IPeCh)

> F e ( 0 H ) 3 i + 3NaCl

NaCl,,i„, + H2S04dac — ^

FczO, + 6HC1

+2NaOH

(6)
(7)

(1)

-

.^

> FeClj + H21

Fe203 + 3 H 2 0 t

+ 3AgN03

•

(4)

Fe + 2 H C I

2Fe(OH)3

FCCI3

+ CI21 + 2 N a O H

> 2Fe + 3 H 2 O

'"

FeCl3 + 3 N a O H

1. Cac phan ufng:

(3)
1

2

(hoac 2FeCl2 + CI2

B a i giai

104

2 N a C l + 2H2O —

Fe203 + 3H2

T i m cong ihtfc cua cac cha't uTng v d i cac chaft A i , B, ... Vie't cac phiTdng trinh

2.

> H2SO4

X|ii + X |

1

Ag2S04 + S 0 2 t

+H2O

BaC03i+H20

X,2

— >

X3 + X,3

-— >

BaCOsi + C a C 0 3 i + HjO

X,; + X , 4

-— >

FC(N03)2 + X , 5

X3 +

Bai giai
Hoan chinh cac phU'dng trinh phan iJng:
XFC2O3 + ( 3 x - 2 y ) C 0

> 2Fc,0y + ( 3 x - 2y)C02


2.

2NaHS04 + Ba(HC03)2

B a S 0 4 i + Na2S04 + 2 C 0 2 t + 2H2O

3.

2NaHS04 + BaCO,

—> B a S 0 4 i +Na2S04 + C 0 2 t + H2O

4.

2AgN03 + 2KC(ti

—> AgzOJ' + 2KNO3 + H2O

5.

Ca3(P04)2 + 4H3PO4

6.

2Fe

+

6H2S04dac

2FeO + 4H2S04dac

4H2S04dac

CUSO4 + 2 K 0 H

-> Fe2(S04)3 + S02T +6H2O

4H2S04dac

Fe2(S04)3 + S 0 2 t +4H2O

8.

Ba(HC03)2 + Ba(0H)2

9.

Ba(HC03)2 + Ca(0H)2

10.

Fe + Cu(N03)2
Chu de 3.

3Fe2(S04)3 + S 0 2 t + IOH2O

2FeO +

NaHS04 + BaCl2

^ B a S 0 4 i + NaCl + HCl

K O H + BaCl2

-> khong

NaHS04 + KOH

NaCl khong c6 phan iJng vdi 4 chat c6n lai.
Tren cd sd cac phan tfng ta c6 the nhan biet:
-

Cha't tao 2 ket tua trang la BaCl2

-> 2BaC03i +2H2O

- Chat tao I ket tua trang la NaHS04

BaCOai +CaC03i + 2H2O
-> Fe(N03)2 + Cu4'

B a i giai

Tri/dc het cho vao moi dung dich mot mieng kem (hoSc s^t, magie), ndi nao
CO khi thoat ra la HCl:
> ZnCl2 + H 2 t

Sau do trpn dung djch HCl vdi dung dich NaCl dtfdc dung dich A, va trgn
dung dich HCl vdi dung dich NaN03 dUdc dung djch B.
Cho hai mieng Cu vao 2 dung dich A va B, ndi nao c6 khi khong mau thoai
ra, ho^ nau trong khong khi la hon hdp HCl va NaNOs. Cac phan iJng xay ra:
3Cu + 8HC1 + 8NaN03

> 3Cu(N03)2 + 2N0 T + 8NaCl + 4H2O

3Cu + 8HC1 + 2NaN03

> SCuCls + 2 N 0 T + 2NaCl + 4H2O

> 2NO2

- Chat tao 1 ket tua trang, 1 ket tua xanh la CUSO4
- Cha't tao 1 ket tua xanh la KOH.
Bai 3. Cho 4 mau kim loai A, B, C, D mau sang bac giong nhau Ian Itfdt tac
dung vdi cac dung dich HNO3 dac, ngupi, dung dich HCl va dung dich
NaOH. Ke't qua thu diTdc nhi/ sau:
axit

A

HNO3

-

HCl

+

NaOH

-

B

C

D

+
+
+

+
-

+
+

-

-

Hoi A, B, C, D la nhffng kim loai nao trong so ci.c kim loai sau: nhom, sat,
kem, magie, bac?
B a i giai

Cho cac kim loai Ian liTdt tac dung vdi HCl, HNO3, NaOH ta c6 cac ket qua sau:
A l + HNO3

dac. nguoi

2A1 + 6HC1

>

khong phan iJng

> 2AICI3 + 3H21

2A1 + 2NaOH + 2H2O

> 2NaA102 + 3H21

V$y A l khong ihuoc A, B, C, D.

B a i 2. Co 5 dung dich CUSO4, BaClz, NaHS04, KOH, NaCl. Cho cac dung djcli

Fe + HNO3

d6 tac dung vdi nhau tirng doi mot. Viet cac PTPLf xay ra. Tren cd sd ca^

Fe + 2HC1

106

> NaKS04 + H2O khong c6 hifn tUdng dSc triftig

-> Ag2S04 + SO2T + 2H2O

khong: NaCl, HCl, NaN03.

2NO + O2

C u ( 0 H ) 2 i +K2SO4

Chat khong c6 phan iJng gi la NaCl.

B a i 1. Chi diTdc dung kim loai c6 the phan biet diTdc cac dung dich sau day hay

Hoac

-> khong

.

N h S n bi^'t - T a c h hon h
Zn + 2HC1

B a S 0 4 i +CUCI2

CUSO4 + BaCl2

->Fe2(S04)3 + S 0 2 t +4H2O

Fe2(S04)3 + 9 S 0 2 T + IOH2O

+ 2H2S04dac

Cic phanuTng:
CUSO4 + NaHS04

10H2SO4dac

2Ag

B a i giai

-> Fe2(S04)3 + 3 S 0 2 t +6H2O

2FeS +

7.

piat nhan.

3Ca(H2P04)2

2Fe304+ 10H2SO4 3ac
2Fe(OH)2 +

ph^n
ph5 i?ng d6 c6 nh$n biS't drfdc cdc dung dich khong, ne'u cdc dung dich bi

dac, nguoi

>

khong phan uTng

> FcCh + H2 T
107


Fe + N a O H

-)• khong phan tfng

V a y F e la kim loai A

Zn + 4HNO3

3.

^ Zn(N03)2 + 2NO2T + 2H2O

Zn + 2HC1

ZnCl2 + H 2 t

Zn + 2NaOH

NazZnOj + H z t

V a y Zn la kim loai B .

Mg + 4HNO3

4.

-> Mg(N03)2 + 2NO2T + 2H2O

Mg + 2HC1
Mg + 2NaOH

-> khong phan ufng

A g N 0 3 + N 0 2 t +H2O

Ag + H C l

khong phan uTng

L a y mot phan dung dich B, them mot liTdng d\i HNO3 va sau 66 A g N 0 3 tha'y
j{6't tua trang: nhan bie't dtfdc NaCl.
AgN03 + NaCl

>AgCU+NaN03.

pai 5. M u o i N a C l b i Ian cac tap chat Na2S04, MgCl2, CaCl2, C aS04, NaBr.
Trinh bay phu'dng phap hoa hoc de thu dtfdc N a C l nguyen cha't.
Bai giai
Hoa tan hoan toan N a C l vao niTdc (CaS04 v i liTdng i t nen tan het) sau do

Na2S04 + BaCl2

> B a S 0 4 i + 2NaCl

Na2C03 + MgCl2

> M g C 0 3 i + 2NaCl

Na2C03 + CaCl2

> C a C O j i + 2NaCl

'

-> khong phan uTng

V a y Ag la kim loai C .
B a i 4. Trinh bay phi/dng phap hoa hoc ngan gpn nhat de nhan bie't tiifng muoi
natri trong mot dung dich gom: Na2C03, NaHC03, Na2S03, Na2S04, NaCl,
NaNOj.
B a i giai
Tru-dc het cho lu-png du" dung dich Ba(CH3COO)2 vao dung dich ban dau thu
diTdc ke't tua A va dung djch B .
Ket tua A gom BaS04, BaS03, B a C O j .

.

> 3CuCl2 + 2 N O T + 2 N a C l + 4H2O.

th6m luTdng du" BaCl2 de ke't tua het muo i sunfat:

Ag + 2HNO3

Ag + N a O H

3Cu + 8HC1 + 2NaN03

MgCl2 + H 2 t

V a y Mg la kim loai D.
5.

Sau do them tie'p Cu vao thay dung dich c6 m a u xanh, c6 k h i khong mau
thoat ra hoa nau trong khong k h i : nhan bie't dtfdc NaNOs:

Na2S04 + Ba(CH3COO)2

> BaS04 i + 2CH3COONa

NasCOj + Ba(CH3COO)2

> B a C 0 3 i + 2CH3COONa

Na2S03 + Ba(CH3COO)2

)• BaS03 i + 2CH3COONa

Phan nirdc loc con l a i (gom NaCl, NaBr, Na2C03) cho tac dung v d i dung dich
HCl d i / , l u c d 6 :
Na2C03 + 2HC1

> 2NaCl + H2O + CO2 T

'

Trong dung dich c h i con lai NaCl, NaBr. Sue k h i CI2 t d i du" va c6 can dung
dich ta se c6 N a C l nguyen cha't.
2NaBr + Cl2

)• 2NaCl + Br2

Bai 6. Co hon hdp M chiJa cdc chat CaC03, Fe203, AI2O3, Si02. H a y t n n h bay
phiTdng phap hoa hoc de lay tiifng cha't rieng le nguyen cha't, k h o i liTdng
khong d d i .
B a i giai

'

Hoa tan ke't tua A bkng dung dich H C l diT, mot phan khong tan la BaS04:

Cho hon hdp vao niTdc r o i sue k h i CO2 drf vao. L u c do x a y ra phan iJng hoa

nhan bie't diTdc Na2S04.

tan C a C O j , con l a i chat r ^ n R.

Na2C03 + 2HC1

> 2NaCI + H2O + CO2 T

Na2S03 + 2HC1

> 2NaCl + H2O + S O 2 1

Nhan bie't hon hdp CO2, SO2 nhiT cac bai tren. Nhif vay ta bie't dufdc Na2C03,
Na2S03.
Phan dung djch B c6 NaHC03, N a C l , NaNOj, C H 3 C 0 0 N a , Ba(CH3COO)2

CO2 + H2O + CaC03

> Ca(HC03)2

Lay dung dich thu du'dc dem dun nong ta c6 C a C O s
Ca(HC03)2

— ^

CaC03 + H 2 0 + C02T

Si02 + H C l

> khong phan u-ng

ra: nhan bie't N a H C O j .

Fe203 + 6HC1

> 2FeCl3 + 3H2O

AI2O3 + 6HC1

> 2 A I C I 3 + 3H2O

108

> N a C l + H2O + C O 2 1

(2)

Hoa tan R bkng dung dich H C l diT, thu dUdc S iO j

La'y mot phan dung dich B cho tac dung vdi dung dich H C l diT, thay khi bay
NaHC03 + H C l

(1)

I
^

'' '
(3)
J

(4)

109


Lay dung djch thu dUcJc (gom FeCh, A I C I 3 , HCl du) cho tac dung vdi duiig
dich NaOH dU'. Cac phtfcJng trmh phan i?ng xay ra:

(Iri) tC

HCl + NaOH

> NaCl + H2O

FeCl3 + 3 N a O H

>

AlCl3 + 3 N a O H

> A1(0H)3

Al(OH)3 + NaOHdu

> NaAlOj + 2H2O

Fe(OH)3 4. + 3 N a C l

i + 3NaCl

(5)

Fe203 + 3 H 2 0 t

^ siu dong thi do vat bing nhom rat ben khong bi htfhong, trai lai cdc do vat
b^ng s^t, dong thi bi hoen ri.

(6)

(7)
(8)

Lay ket tua nung 5 nhiet do cao ta c6 Fe203
2Fe(OH)3

Tai sao nh6m hoat dong hdn s^t, dong nhuhg khi de cic d6 vat hkng nhom,

Bai giai

£)5 v$t b^ng bac de trong khong khi van giiir duTdc anh kim vi Ag khong tdc
dung vdi O2 cua khong khi thanh A g 2 0 ; nhiftig khi khong khi nhiem ban H2S
fhi bi den xam do phan tfng tao thanh Ag2S (den) sau:
2Ag + 0,502 + H2S

(9)

> Ag2S + H2O

2 Khi bi rdi vai thuy ngan ra't doc (do bay hdi), khong the thu gom, v l Hg bi

Sue khi C O 2 vao phan ni/dc loc (gom N a A 1 0 2 va NaOH) sau do nung d nhiet

phan tin thanh nhffng hat ra't nho, do 66 ngiTdi ta phai r^c liTu huynh, luc 66

do cao ta diTcfc AI2O3

t^o thanh H2S khong bay hdi, ta c6 the thu gom de dang:

> >

C O 2 + NaOH

> NaHC03

C O 2 + 2 H 2 O + NaAIO2

2A1(0H)3

> A1(0H)3 i + NaHC03

5 ^ Al203 + 3 H 2 0 t

(10)
(11)

(12)

Bai7.

Hg + S

> HgS

3. Nhom Ih kim loai hoat dong hdn s^t, dong, nhtfng cac do vat hkng nhom de
ISu trong khong khi khong bi hoen r i do nhom tic dung vdi O2 (cua khong
khi) tao thanh mot Idp mang raft mong bao ve cho A l phia trong khong phan
tfng vdi O2.

1. Co the tach s^t kim loai khoi hon hdp kim loai sat, dong, nhom, tach s^t kim
loai khoi s^t sunfua hlng each diing nam cham hay khong?
2. Mot loai dirdng kinh bi Ian mot it cat. Lam the nao de c6 dufcJng nguyen chat.
Co the lay diTdng tiT dung dich di/dng tan trong riTdu etylic dtfdc khong?
Bai giai
1. Co the dung nam cham de tach sat kim loai ra khoi hon hdp sat, nhom, dong vl

Bai 9.

-A

1. C6 hai dung dich loang FeCl2 v^ FeCl3 (gan nhiT khong mau). Ta c6 the diJng
dung djch NaOH ho|c nxXdc brom, hoac dong kim loai de phSn biet 2 dung
dich dd. Hay giai thich bling cic phiTdng trinh phan tfng.
2. Co 5 ong nghiem diTdc d^nh so thi? tiT 1, 2, 3, 4, 5. M o i ong diTng 1 trong 5
dung dich sau day: Na2C03, BaCh, HCl, H2SO4, NaCl. Neu lay ong 2 66 vko

nhom, dong kim loai khong bj nam cham hut, nhiftig khong the tach Fe ra khoi

ong 1 thay c6 ket tua; lay ong 2 do vao ong 3 tha'y c6 khi tho^t ra, lay ong 1

FeS VI day la hdp chaft, nam cham khong the pha v9 hen ket giiJa Fe va S.

do vao ong 5 thay cd ket tua. Hoi ong n^o difng dung dich gi?

2. TriTdc het hoa tan diTdng kinh vao nu'dc, loc (gan) bo phan cat khong tan, sau
d6 c6 can can than ta thu dufdc dtfdng kinh. Co the lay diTcJng tif dung dich
dirdng trong riTdu bang each chifng ca't de lay rifcJu, diTcJng khong bay hdi do
do ta thu diTdc nhffng tinh the dirdng tr^ng.
Bai 8.
1. Tai sao do vat b^ng bac de trong khong khi van giff difdc anh kim, nhi^ng
neu khong khi bi nhiem ban H2S thi do vat bang bac bi nhanh chong doi maU
thanh den.
2. Khi v6 tinh lam v3 nhiet ke thuy ngan, thuy ngan kim loai luc 66 rdi va'
kh^p nha thanh nhQ-ng hat nho l i ti khong thu gom het. V i thuy ngan rat dpc
nen ngu'di ta dung bien phap r^c bot lu\ huynh vao nhffng cho c6 thuy ng'"'
rdi vai. Tai sao?

Bai giai
1- C6 the dilng NaOH de phan biet FeCh

thknh Fe(0H)2 mau tr^ng dnh luc v^ Fe(0H)3 mau nau do.
C6 the dilng nurdc brom (mau nau do), vi FeCh lam ma't mau niTdc brom, c6n
PeCl3 khong phan iJng:
6 F e C l 2 + 3Br2

> 4FeCl3

+ 2FeBr3

Cd the dfing dong kim loai vi luc dd FeCh khSng phan tfng vdi Cu, nhiftig
f'eClj hoa tan di/dc Cu thanh CuCb mau xanh.
2FeCl3 + Cu

> CUCI2

+ 2FeCl2

• "^rong 5 dung dich: Na2C03, BaCb, HCl, H2SO4, NaCl ta nhan thay chi cd
^aClz tao thanh ket tua vdi Na2C03 va H2SO4:
BaCl2 + Na2C03

m

FeCU (rat loang) v i Itic d6 tao

• BaC03 >l + 2NaCI

iO)--

111

> B a S 0 4 i + 2HC1

BaCl2 + H2SO4

N h i f v3y ong 1 phSi 1^ B a C h ; ong 2 phii Ik NazCOs \i cho v ^ o ong 3 cjj
k h i bay ra, va ong 3 p M i Ih H C l
NazCOj + 2 H C I

> 2NaCl + H2O + CO21

> 3 B a S 0 4 i +2FeCl3

Fe2(S04)3 + 3BaCl2

'

^- O n g 4 m H2SO4 va ong 5 m N a C l .
Bai 10. B p t dong oxit b i Ian bpt than (hon hdp A ) .

2.

2FeCl3 + 3Ag2S04

6AgCl>l +Fe2(S04)3

Fe(OH)3 + 3HC1

FeCb + 3H2O

FeCl3 + 3 N a O H

->. Fe(OH)3^ + 3 N a C l ^

Cu + CI2

-> CUCI2
FeCl2 + Cu i

1. T r i n h b ^ y phiTdng phap vat l i de lay rieng CuO.

Fe + CUCI2

2. L a y m o t i t hon hdp A nung n6ng trong chan khong (khong c6 mat cua oxit)

(C6 the thay Fe h\ng Z n , Mg, v.v...
C6 the d i e n phan

t d i k h i cac phan uTng xay ra hoan toan. G i a i thich sif bie'n d o i m ^ u ciia hon
*

.

hdp b^ng cac phiTdng trinh phan tfng. N e u nung n6ng hon hdp A trong khong

Y

dp

CuCb

k h i t h i hien tifdng xay ra nhiT the nao?
(C6the

1. V i bpt CuO ra't nang, con bpt than ra't nhe nen ta c6 the dilng phiTdng phap
l ^ n g gan de tach lay CuO: cho hon hdp A vao coc, t h e m ni^dc vao, khuay

^ Cu + C l s t )

->

CU + 2H2S04da.

Bai giai

CuS04 + S 0 2 t +2H2O

Cu + HgS04

CUSO4 + H g ) .

a

CUSO4 + Fe

^ FeS04 + C u i

''"^

(Co the thay Fe bang Zn, Mg, v.v...)

deu r o i l i n g gan, bpt than nhe se troi theo niTdc ra ngoai, lap d i lap l a i vai ba
Ian ta c6 CuO sach.
2. K h i nung trong chan khong t h l xay ra phan tfng

i

-

i J j B + eOiS^

CUCI2 + Ag2S04

> 2AgCl i

CUSO4 + BaCl2

> B a S 0 4 i + CUCI2

+ CUSO4
(latO >JU X>?. W^' } .t.

N e u it than:

2CuO + C

Na'u nhieu than:

CuO + C

> Cu + CO

I . C6 the dieu chc Fe b^ng each khur s i t oxit theo cic phan ufng sau:

(hoac:

CO2 + C

> 2C0).

a)

Fe.Oa + CO

-> Fe + . . .

thanh m ^ u do v ^ n g cua C u ; con neu nhif diT C hoac dif CuO t h i h6n hdp c6

b)

FczOj + H2

^

m a u do Ian den (tuy t i le Cu va CuO hoac C).

c)

FejOj + A l

d)

Fc.Oy + A l

e)

FC2O3 + C

>

2Cu + CO2

N e u t i I9 so m o l C: C u O tuT 1 : 1 den 1 : 2 t h i mku den ciia hon hdp bien

N e u nung hon hdp U-ong khong k h i thi c6 the coi than chay he't thanh CO2.
con l a i C u O m a u den ( v i Cu b i oxi thanh CuO).

, ^ Chu de 4.

B d tuc p h a n tifng - D i e u ch6'

u2.

A

Fe + ...

%Fe$^!:^o

Fe + ...
Fe + ...

Hoan thanh cac PTPLf tren. Theo em phan ufng nao duTdc dung de san xua't
gang tiir cac quang oxit s i t .

Bai 1 . V i e t cdc phiTdng trinh phan uTng theo cac sd do bien hoa sau:
Fe2(S04)3 <

> Fe(0H)3

Cu <

2- Vie't cac P T P U theo sd do bien hoa

> CuCh

+ x
FeCb

to 1.1a
1.

112

CUSO4

Bai giai
Fe2(S04)3 + 6 N a O H

> 2Fe(OH)3 i

+ 3Na2S04

2Fe(OH)3 + 3H2SO4

> Fe2(S04)3 + 6H2O

A

Fe203

. +Y
FeCl2

B

•+ T

Trong do A, B, X, Y, Z, T la cac cha't khac nhau.
113


Bai giai

1. Hoan thanh cac phan uTng:
a) ,
FejOa + 3C0
— ^ 2Fe + SCOj
• ' - • •

b)
c)
d)
e)

If,

FeaOj + 3H2
— ^ 2Fe + 3H2O
FezOa + 2A1
2Fe + AI2O3
3FexOy + 2yAl
— ^ 3xFe + yAlzOj
FezOa + 3C
— ^ 2Fe + 3C0
phan iJng a) difcJc dung de san xua't gang

2. Cic phiTdng trinh phan tfng:

a)

FezOj + 3C0

b)

Fe + CUCI2

c)
FezOj + 6HCI

d)
2FeCl3 + Fe
Bai 3. Cho sd do bien hoa
+x. t"

— ^ 2Fe + 3CO2 (X c6 the' la H2, Al, C . )
> FeCh

+

Cui

> 2FeCl3 + 3H2O
> BFeClj (T c6 the \h Cu, KI)

' 3Fe304 + 8A1

>

Zn + CUCI2

> ZnCl2 +

Zn

> ZnCl2 + H 2 1

+ 2HC1

D

— C

— ^ 3Fe + 4H2O
9Fe + 4AI2O3

Fe + 2HC1

> FeCl2 +

(B)
2FeCl2 + Cl2
(E)
Fe304 + 8HC1

(D)
> 2FeCl3
(C)
> FeCl2 + 2FeCl3 + 4H2O

Hat

CuCh

^-i-lil tkm I/.

Cu4'

'C '

Cic phan uTng trifc tie'p tao thanh kim loai tfif muo'i:
Zn + 2AgN03
> Zn(N03)2 + 2 A g i
'i
AgNOj
Ag + N O 2 1 + I / 2 O 2 1 ' •
CUCI2

+ Cl2t

^ ^ C u

2. C^c phan tfng tao thanh NaOH
+ 2H2O

> 2NaOH + H2 T

Na20 + H20

Biet ring A + HCl
> D + C + H2O
Tim cac chat tiTdng tfng vdi cac chuT cai A, B,... va viet cac PTPl/.
Bai giai
Nhin sd do ta thay A phai Ih oxit sat, va vi A + HCl tao ra hai loai muoi ncn
A phai Ih Fe304:
Fe304 + 4 C 0
— ^ 3Fe + 4CO2
J ^ Fe304 + 4H2

CU + CI2

2Na

> Fe

.. 5

pjay vie't 3 loai phan uTng tri/c tiep tao thanh muoi tit kim loai va 3 loai phan
^jig tT\ic tiep tao thanh kim loai ttf muoi. < ,
j^ay vie't 4 loai phan ufng tao thanh NaOH.
^
^
Bai giai
:
, •,
J Cac phan tfng trifc tiep tao thanh muoi tiT kim loai:

Ca(0H)2

'^^

> 2NaOH

+ Na2C03

NaHC03 + Ba(0H)2

> CaC03i + 2NaOH
>

v^adu hoacdu

BaCOs i + NaOH + H2O

v.v...
Bai 5. Cho 4 mau Na vao 4 dung dich sau: ZnCb, FeCl2, KCl, MgS04. Viet cac
PTPU xay ra.
Bai giai
Trirdc het Na tac dung vdi niTdc
'
2Na

+ 2H2O

> 2NaOH + H j t

'

' ••^ ' ;v /

-

Sau do: 2NaOH + ZnClj
> Zn(OH)2 i + 2NaCl
,
2 N a O H du- + Zn(0H)2
> Na2Zn02 + 2H2O
2NaOH + FeCl2 .
> Fe(0H)2 i + 2NaCl , ,,
Neu de trong khong khi:

4Fe(OH)2i

+2H2O + O2

KCl + NaOH
2NaOH + MgCl2

>

4Fe(OH)3i

> khong phan urng
> Mg(0H)2 i + 2NaCl

115


B a i 6. Tic cac chat A, B, C... thich hdp
B

> C

viS't cdc PTPLf theo sd do b i e n ho^.

FeS04 + Cu>l.

Fe + C U S O 4

(H)

> D
B

^

H

-> F e S i + N a 2 S 0 4

FeS04 + NazS

M

(M)

->. G
B i c t M la 1 hdp cha't cua 1 k i m loai

p h i kirn B , C, D, H 1^ cac hdp chf, f

chtfa liTu huynh, E, F, G, H la hdp chat cua s^t hoSc s^t k i m loai.

p a ' ^'
J V i ^ t l?i '^^"S ^^^^

N2H«C03, H2P20sCa, C4H,„06Ba

M phai la FeS. Cac PTPLf each 1 la:
> FciOi

(A)
>

thi chinh l a i cho dung.

(B)

2HBr +

— ^

+ 6H2S04dac

(F)

FeCl2

Fe2(S04)., + 3 S 0 2 t + 6H2O

H.T

(G)

FeS04 + NazS

(H)
FcSi

+Na2S04

(M)

CO

2FeO +

CO2

> Ca(H2P04)2

canxi dihidrophotphat

BaC4H,„06

> Ba(CH3C00)2.2H20 tinh the bari axetat ngam

4H2S04dic + 2FcO
(F)

FcO + H2

-> Fe2(S04)3 + S02T + 4 H 2 O
(B)

CuO
CUCO3)

> FeCh

2FeCl2 + CI2

> 2FeCl3

FeCl3 + 3 N a 0 H

+

Cui
(2)

> Fe(0H)3 i

+ 3NaCl

(3)

(CO the thay N a O H bang K O H , Ba(0H)2 v . v . . . )
Fe203 + 3 H 2 0 t

Fe203 + 3H2SO4

Chu de 5.

,i.

if!

(4)
(5)

> Fe2(S04)3 + 3H2O

(6)

> 3 B a S 0 4 i + 2Fe(N03)3

T i n h theo phi/cftig trinh p h a n ufng
.

^

1. De m i e n g A I nSng 5,4g trong khong k h i mot thdi gian thu di^dc cha't r ^ n

(G)
-> C U S O 4 +

Fe + C u C b

(1)

H i ^ u suS't phan ufng - Nong dp dung dich

- > Fc + H2O

(hoSc CO)

>FeCl2 + H2t

Fe2(S04)3 + 3Ba(N03)2

(F)

116

H2P20xCa

2Fe(OH)3

Cach21a:

(hoac Cu,

amoni cacbonat

Fe + 2HC1

2.

Hoac

2 A g C l i + FCSO4

H2SO4 +

> (NH4)2C03

hai phan tur H2O

Ag2S04 + F c C l :

(C)

N2HSCO3

(D)
FeCb +

FC2O3 +

. ;:i.n

1. Viet lai cong thuTc dung va goi ten:

> Ag2S04 + H2O

Fc + 2HC1

Fe(N03)3

B a i giai

2Fe + 3H2O

(B)

+ Ag20

FeO

Fe2(S04)3

(C)

H2SO4

Fe(0H)3 — ^

FeCh

(C)

(F)
2Fe

Fe

H2SO4

(hoac CI2, O 3 , H2O2...)
Fe203 + 3H2

2 Viet cac phiTdng trinh phan iJng triTc tiep theo sd d6 b i e n hoa, neu ndi n^o sai

+ 2SO2

(E)

SO2 + 2H2O + Br2

thanh phan cho dvldi day v£k goi ten c'liing,

^^^^

nd'u c6ng thi?c nao sai dUdc phep chinh l a i chi so' cua m p t nguyen to':

B a i giai

2FeS + 7/2O2

A

^- Hoh tan hoan toan A b^ng dung dich H C l tha'y bay ra 6,5856 l i t H2 (dktc).
H2O

Tinh kho'i lu^dng ran A va %A1 b i o x i hoa thanh oxit.

-

(D)
117


B a i giai

Khi

3/ a < b tiJc du3/

trong khong khi Al bi oxi hoi mot phan th^nh oxit:
> AI2O3

2A1+ - O 2
(1)
2
Cic phan ufng hoa tan:
2A1 + 6HC1
2AlCl3 + 3 H 2 t

D
t
Dung
dich X c6 a mol FeS04 va (b - a) mol C U S O 4 ; chat r^n Y c6 a m o l Cu.
pai 3- Nhung mot mieng nhom kim loai nSng 10 gam vao 500 ml dung dich
C U S O 4 0,4M. Sau mot I h d i gian lay mieng nhom ra, ruTa sach, say kho, can
nang 11,38 gam.
t * ! .>,;:. •
1 Tinh khoi l i / d n g dong tho^t ra bam vao micng nhom (gia suT ta't ca dong thoat
ra bam vao mieng nhom).
2 Tinh nong dp mol cua cac chat sau phan vlng (gia suT the tich dung dich van
500 ml).
^ B S

( 2 )

Cich 1:

AI2O3 +

6HC1

Khoi lifdng Al con:

(3)

2 A I C I 3 + 3H2O

nAiciin=

—^H-,

-

— X

3 2 3

Ttfc 0,196 X 27 = 5,292 g
Khoi liTdng AI2O3 lao thanh:
_1

CUSO4

^'^^^^ -0,196mol
22,4

.

" 2 " ' ^ '

Al p h a n iJng la

2x,

l i f d n g C U S O 4 p h a n iJng la

"5y

^

•J-, •

B a i giai

Ta

can

xet

cac

triTdng

> FeSO^ + Cui

(1)

hdp:

CuO + CO

— ^ C u + C02

C02 + Ba(OH)2 -

2- Tinh so mol cac chat:
_

2/

Trurdng hdp 1: n, = n2 hieu suat h% =

s^t.

Dung dich X c6 b mol FeS04; cha't r^n Y c6 b mol Cu va
118

(a

- b) mol Fe.

'

V
°2 ="1

a
"CuO ="2

.

(1)

^BaCOji+HjO

1/ a = b ttfc cac cha't tdc dung vdi nhau vijfa du.
Dung dich X chi c6 a mol FeS04; chat r^n Y chi c6 b mol Cu.
a > b ttfc du-

'

^- Cac phu'dng trinh phan uTng:

B a i giai

Fe + C U S O 4

3x.

Dya treh nguyen tic l a Al phan tfng thi kho'i liTdng mieng Al bi giam, con Cu
tao thanh bam vao mie'ng Al nen khoi l i f d n g tang len, ta c6 phu'dng trinh:
1 0 - 2x X 27 + 3x X 6 4 = l l , 3 8 g a m
Giai ra ta di/dc: x = 0,01 mol
Khoi l i / d n g Cu thoat ra bkng 3x x 64 = 3 x 0,01 x 64 = l,92g
2. Tinh nong do mol.
n
0,01
^
0,5x0,4-3x0,01 ^
^ '
CA12(SO4)3 =

= " ' ^ ^ ' ^ '^CuS04
= 0,34M
Bai 4. Cho V lit khi CO (dktc) di qua ong stf diTng a gam CuO nung nong. Sau
khi ket thiic thi nghiem cho khi di ra khoi oag si? hap thu vao dung dich
Ba(0H)2 tha'y tao thanh m gam ket tua.
1- Viet cac PTPl/xayra.
2. Tinh hieu suat ciia phan i?ng khur CuO theo V, a, m.

2 .

tfng:

gnrdv:

1. Phan tfng: 2A1 + 3CuS04
> Al2(S04)3 + 3Cu>l'
"f'^^ "f
'
Goi X la so mol Al2(S04)3 tao t h a n h , nhu* vay l i f d n g Cu tao t h a n h la 3x, luTdng

5,4-5,292 1
,
h„^ =
• X - = 0,002mol
"AI2O3
27
2
Turc 0,002 X 102 = 0,204g
Vay khoi liTdng A = 5,292 + 0,204 = 5,496g
5,4

Cach 2:
54 3
Ne'u khong bi oxi hoa thi 5,4 gam Al tao ra: —-—x — x22,4 = 6,72 l i t H 2
Theo (1, 2) so mol nguyen tuT oxi phan iJng bkng so mol H2
• = 0,006mol
6,72-6,5856
22,4
Vay khoi liTcJng oxi da tham gia phan iJng b^ng: 0,006 x 16 = 0,096 gam
Do 66 khoi liTdng A = 5,4 + 0,096 = 5,496g
Bai
Cho a mol bot Fe vao dung dich chtfa b mol C U S O 4 . Sau khi ket iho"^
phan iJng ta thu diTdc dung dich X va chat r^n Y. Hoi trong X, Y c6 nhiWr
chat gi? Bao nhieu p i o l ?
Phan

.,

B a i giai

(2)

m
i B a C O , = "3 =

^^^^% hoac

n2

f

' f

M

^^^^%
n,

119


.^
n, xlOO„
Trirdng hcJp 2: ni > 02 hi?u suat h% = —
%

n = 3r=>M = 97,5 loai
E)6'i chieu vdi de bai, kim loai R 1^ Zn (kem)
pai 6. Cho 27,4 gam bari vao 400 gam dung dich

Trirdng hdp 3: n, < nz hieu suat h% =

3,2% thu diTdc khi A,

Icd't tija B va dung dich C.

"i

Bai 5. Cho 13,44 gam bot d6ng kim loai vao 1 coc difng 500 ml dung dich
AgNOs 0,3M, khuay deu dung dich mot thcJi gian sau d6 dem Ipc ta thu diroe
22,56 gam chat r^n A va dung dich B.

1. Tinh nong do mol cua cha't tan trong dung dich B. Gia suf the tich ciia dung

J Tinh the tich khi A (dktc).
2 Nung ke't tua B d nhiet do cao den khoi liTdng khong doi thi thu diTdc bao
nhieu gam cha't ran?
3 Tinh nong do % cua cha't tan trong dung dich C.

dich khong thay ddi.

Bai giai

2. Nhiing mot thanh kim loai R nSng 15 gam vao dung dich B, khuay deu de

Cac phan iJng xay ra:

c^c phan iJng xay ra hoan toan, sau do lay thanh kim loai R ra khoi dung

Ba + 2H2O

djch, can nang 17,205 gam (gia suT ta't ca kim loai thoat ra deu bam vko thanh

Ba(0H)2 +

R). Hoi R la kim loai gi trong so cac kim loai cho sau: Na = 23, M g = 24,
A l = 27, Fe = 56, N i = 59, Cu = 64, Zn = 65, Ag = 108, Pb = 207.
Bai giai

CUSO4

BaS04

)

1. Phdn tfng giffa Cu va AgNOj
Cu + i A g N O j

CUSO4

-> Ba(OH)2 + H 2 t

(1)

-» BaS04i +Cu(OH)2 4.

(2)

^

BaS04

Cu(OH)2

CuO + H . O t

(3)

m
1
400x3,2 „ „ o
,
nga =

= 0,2mol; np,,<-o. =
= 0,08mol
137
100x160
27,4
•

> Cu(N03)2 + 2Ag ^

n 44
Tinh: ncu = —— = 0,21mol;
64

(1)

n^gNOj = 0'5 x 0.3 = 0,15 mol

Theo phan ufng(l): n^^ =nB,(0H)2

""BU

=0,2mol

Gpi X la so mol Cu tham gia phan uTng (1) (nhd r^ng Cu chiTa phan tfng het)

The tich H2: VA = 0,2 x 22,4 = 4,48 lit

ta CO phufdng trinh ve khoi IiTdng cha't r^n A:

Theo cac phan uTng (2, 3) chat ran gom BaS04 va CuO, vi Ba(0H)2 diT nen

(0,21 - X ) X 64 + 2x X 108 = 22,56
Rut ra

X

= 0,06 mol

i

-

0,15-2x0,06

5^

= 0,06M;

lifdng cac cha't ban dau truf lufdng H2 bay ra va li/dng ke't tiia.

0,06
CCU(NO3)2 = " ^

=

^'^2M

(0,2 - 0,08) X 171x100
Ta c6:

2. Goi n va M la hoa tri va KLNT cua kim loai R ta c6 cac phan iJng:
R + nAgNOj

> R(N03)n + nAg i

(2)

2R + nCu(N03)2

> 2R(N03)„ + nCu i

(3)

Kho'i lufdng thanh R tang = 17,205 - 15,00 = 2,205 g
Theo cac phan utag (2, 3) ta c6 phifdng trinh ve suf thay doi khoi liTdng thanh R:
M

2M

(108 - — ) X 0,03 + (64
n

) X 0,06 = 2,205g
n

Riit ra M = 32,5n
Khi

C%B,(OH)2

= 5,12%

(400 + 2 7 , 4 ) - 0 , 2 x 2 - 0 , 0 8 x 233-0,08 x98

7. A la mot oxit kim loai chiJa 70% kim loai, can diing bao nhieu ml dung
dich H2SO4 24,5% (d = 1,2 g/ml) de hoa tan vira du 8g oxit A. ^
Bai giai
Tim kim loai R. Goi cong iMc cua oxit la RxOy, ta c6 ti le % khoi Irfdng:
xR

70

16y

30

R=

112y

56

2y

3x

3

X

56
-—n

Trong do n la hoa tri cua kim loai
Khi n = 1 => R = — loai;

Khi n = 2 :

„
56x2 , .
R =
loai

n = 1 =^ M = 32,5 loai
n = 2 = > M = 65 d o l a Z n

120

i,v

Trong dung dich C chi con Ba(0H)2, khoi liTdng dung djch C b^ng tdng kho'i

Vay nong do mol cua:
CAEN03 =

'

nBaS04 =ncu(OH)2 = n c u O =0,08mol

Kho'i lircfng cha't r^n = 0,08 x 233 + 0,08 x 80 = 25,04 gam

121


o •

Khi n = 3

R=

do la Fe

Phan iJng hoa tan: FejOj + 3H2SO4

>

Fe2(S04)3

+ SHjO

8

(1)

'>r' >

Theophaniirng(l): nH2S04 = 3 x n F e 2 ( s o 4 ) 3 = 3 x — = 0,15mol
V X1 2 X 24 5
Goi V la so mol axit can diJng, ta c6:
'

' = 0,15 V = 50ml.
100 X yo
B a i 8. Cho biet do tan cua dong sunfat 6 5"C la 15 gam va d 80"C la 50 gam,
Hoi khi lam lanh 600 gam dung dich bao hoa dong sunfat d 80"C xuong 5"c
•Y thi CO bao nhieu gam tinh the C U S O 4 . 5 H 2 O thoat ra.
Baigiai

Tinh tong khoi lifdng C U S O 4 (khan).
Ctf 100 + 50 = 150 gam dung dich M o hoa d 80"C c6 50 gam C U S O 4 .
Nhir vay trong 600 gam c6 ^ ^ ^ ^ ^ "^^^^ ^ " ^ ^ ^
~
^ "^^^^
5x18
Gpi X la so gam tinh the C U S O 4 . 5 H 2 O thoat ra trong 66 c6 -^j^ x x = 0,36x
gam H 2 O va 0,64x gam C U S O 4 . Vay liTdng H 2 O con lai trong dung dich la
400 - 0,36x va li/dng C U S O 4 con lai la 200 - 0,64x, do do ta c6 ti 16 (d 5"C):
1^9 = 1^
rut ra X = 238,9 gam.
400-0,36x 200-0,64x
B a i 9. Mot loai da X chiJa 10,2% A I 2 O 3 , 16% Fe203. c6n lai la hon hdp CaCO,,
MgCOj. Lay 100 gam X, nung mot thdi gian thay con lai 82 gam chat r^n Y
De hoa tan hoan toan 10 gam Y can 173 ml dung dich HCl 2M. Neu lay lOHg
X nung tdi phan huy hoan toan muoi cacbonat thi con lai bao nhieu gam cha!
r^nZ?
B a i gial

^.f, Cac phan i^ng nung da X:
- '
CaC03
— ^ C a O + C02t

MgC03
— ^ M g O + C02t
Cac phan tjTng hoa tan Y bkng dung dich HCl:
+ 2HC1

> CaCl2 + H2O

(3)

MgO + 2HC1
CaCOj + 2HC1
MgCOj + 2HC1

> MgCh + H2O

(4)
(5)
(6)

CaO

122

(1)
(2)

>

C a C l 2 + H2O + C O 2 1

> MgCl2 + H2O + C O 2 1

A I 2 O 3 + 6HC1

> 2 A I C I 3 + 3H2O

(7)

Fe203 + 6HC1

> 2FeCl3 + 3H2O

(8)

82
S6 mol HCl can hoa tan 82g Y bing: mHci - 0,173 x 2 x — = 2,8372mol,
$6 mol do cua HCl cung b^ng so mol HCl hoa tan 100 g X, vi mot mol
CaC03 dil da bi phan huy thanh CaO hay khong deu can 2 mol HCl, tufdng tif
d6'i vdi MgCOj (xem cac phiTdng trinh phan iJng 3, 4, 5, 6). Goi x, y \k so mol
CaC03, MgC03 trong lOOg X.
^ 10,2
16 ^ x6 = 2,8372mol
'HCl = (x + y)x2 +
102 + 160
"AI2O3
nFe203

Rut ra X + y = n c o 2 = 0,8186 mol.
' *'
Khoi lifdng chat tin Z blng: 100 - 0,8186 x 44 = 63,98g.

;-||
,_
B a i 10.
1. Cho 500 gam dung dich Na2S04 x% vao 400 ml dung dich BaCh 0,2M thay
tao thanh 10,485 gam ket tua. Tinh x?
2. Dung dich A chtfa 24,4 gam hon hdp hai muoi Na2C03 v^ K 2 C O 3 . Cho dung
djch A tac dung vdi 33,3 gam CaCh thay tao thanh 20 gam ket tua va dung
dich B. Tinh so gam moi muo'i trong dung dich A, B.
B a i giai

1. Phan ifng Na2S04 + BaCl2
> BaS04i + 2NaCl
So'molke'ttua:
nBaso4 = ^ ^ ^ ^ = 0.045mol

(1)

'T
n g n <> V

SS'molBaCb:
Ug^cia =0,4x0,2 = 0,08mol
Vi so' mol ket tua BaS04 nho hdn so mol BaCh nen so mol BaS04 phai b^ng
s6' mol cua Na2S04 = 0,045 mol.
Do do n6ng do cua dung dich Na2S04 b^ng: x = Q'Q45x 142x100 ^ ^^78%
500
2' Cac phan lirng:
= 5^
Na2C03 + CaCl2
> CaCOji + 2NaCl

(1)
,i
K2C03 + CaCl2
>CaC03i+2KCl
(2)
a'5 ul^ '
mol CaCl2 = — = 0,3 mol. So mol CaC03 = — = 0,2 mol.
Ill
100
so mol CaC03 it hdn so' mol CaCl2 chiJug to CaCb di^ 0,3 - 0,2 Ik so mol
ciia Na2C03 va K 2 C O 3 ta c6 he phiTdng trinh
123


1 97
Theo cic p h i n tfng (1, 2): H C = ncoj = nBacoj = 7197
^ = O.Olmol
, , .V
0,01x12x100 „
,;,
vay %C trong thep b^ng
—
= 0,8%
•'

flOex + nSy = 24,4
[x + y = 0,2
Giai he phifdng trinh ta c6: x = 0,1 mol

y = 0,1 mol

Kho'i lifcfng cdc muo'i trong dung dich A:
NazCOj: 0,1 x 106 = 10,6 gam; K2CO3: 0,1 x 138 = 13,8 gam
jf Kho'i lU'ctng cac muoi trong dung dich B:

^

C a C l 2 : 0 , l X 111 = 11,1 gam

. ^ v 5x1000x4

90

,

Tron quang A vdi quSng B theo ti le khoi lufdng mA : ma = 2 : 5 ta ductc
quang C. Hoi trong 1 tan quang C c6 bao nhieu kg s^t?

1. Via't cdc P T P U .

KC1:0,1 X 2 X 74,5 = 14,9 gam
Bai 11.
1. A la mot loai quSng sat chtfa 60% Fe203; B la mot loai quSng s^t khdc chiJa
69,6%

Fe304. Hoi trong 1 tan quang A hoac B c6 chtfa bao nhieu kg s^t?

1. V i trong 1 tan quang A c6 60% Fe203 va trong 160g Fe203 c6 2 x 56 = 112g
Fe nen trong 1 tan (1000kg) quang Ac6:
. s

*

nmi

fat) v
3. Tinh khoi lifdng cua muoi s^t trong B.
1. Cac phan tfng:
-> Fe2(S04)3 + 3S02t + 6H20 (1)

2Fe + 6H2S04dac

^

2. Tron 2 tan quang A vdi 5 tan quang B ta dtfdc 7 tan quang C, lifdng s^t c6
trong 7 tan quang C la: 2 x 0,420 + 5 x 0,504 = 3,36 tan Fe.

Fe2(S04)3 + SO21 + 4H2O

2FeO + 4H2S04dic

= 0,48 tan Fe.

(2)

-> 3Fe2(S04)3 + S O 2 T + IOH2O (3)

2Fe304+ 10H2SO4dac

Fe2(S04)3 + 3H2O

Fe203+3H2S04

(4)

2. Gpi X , y, z, t la so mol Fe, FeO, Fe304, Fe203 trong A, ta c6 cdc phifdng trinh:

B^i 12.

56x + 72y + 232x + 160t = 49,6

1. De xdc dinh ham lifdng cacbon trong thep (khong c6 lifu huynh), ngifdi ta
cho mot d6ng oxi dif di qua o'ng stf difng 15 gam thdp (dang bot) dot ndng va
cho khi di ra khoi ong sif hap thu hoan toan vao dung dich Ba(0H)2 dif thay
tao thanh 1,97 gam ke't tua. Tinh ham lifcJng % cua cacbon trong thep.
2. * De san xua't th6p tijf gang ngifdi ta cd the loai bdt cacbon cua gang bSng
Fe203 theo phan urng: Fe203 + 3C

> 2Fe + 3 C O t

Hoi muon loai bdt 90% lifdng cacbon cd trong 5 tan gang chtfa 4% cacbon thl
can bao nhieu kg Fe203?
Bai giai
Cac phan ifng:

|£M

Bai giai

^^^"^^0^^^^ ^
100x160

Tinh tiTdng tir, 1 tan quang B co: ^000^69,6x3x56 ^
100x232

Vay 1 tan quang C c6 ^

Kh6i lifdng C can phai loai bang

2. Tinh % khoi lifdng ciia oxi trong A.

Bai giai

#

A

(1)

—
x — = 180 kg '
100
100
Xheo phan urng (1) de loai 3 x 12 = 36 g cacbon can 160g Fe203.
Vay de loai 180kg cacbon can: ^ ^ ^ ^ ^ = 800 kg Fe203.
36
Bai 13. Hoa tan hoan toan 49,6 g hon hdp A gom Fe, FeO, Fe304, Fe203 bkng
dung dich H2SO4 dac ndng thu difdc 8,96 lit SO2 (dktc) va dung dich B chi
chtfa mot loai muo'i s^t.

.t,

NaCl: 0,1 X 2 X 58,5 = 11,7 gam

2.

2. Phan tfng khuT Fe203 bkng C: Fe203 + 3C — ^ 2Fe + 3COT

C + O2 — ^ CO21

(1)

CO2 + Ba(0H)2

(2)

> BaC03 4. + H2O

8,96
"802

"

22,4

.

.

.

.

.

.

(I)
.

.

= 0,4 = l,5x + 0,5y + 0,5z

S6' mol oxi trong A = y + 4z + 3t = no.

'

''I i>')sl'''

.isai'-"''

Ta (I, II) rut ra no = 0,65 mol
vay % kho'i hfdng oxi = ^ ' ^ ^ ^ ^ ^ ^ ^ ^ ^ = 20,96%
• ^
49,6
3- Khoi lifcJng cua Fe trong A = 49,6 - 0,65 x 16 (khoi lifdng oxi) = 39,2 gam
ttfc 0,7 mol va so mol Fe2(S04)3 = — = 0,35mol
2

Khoi lifdng Fe2(S04)3 trong B = 0,35 x 400 = 140 gam.
(Co the giai cau 2 theo phu'dng phap bao toan electron
125


"B<5/ amng hoc sittngwt nuu nut o, y - uuu

Bai 14. Cho 220 ml dung dich HNO3 nong do C mol/1 tdc dung vdi 5 gam hS^
gom NO va N2O c6 ti kho'i so vdi H2 bkng 16,75 dung dich A va c6n lai 2,0l3

dn gap 2,5 Ian nong dp mol cua FeS04. Them dung dich NaOH du v^o coc,
Ipc '.lay ket tua roi nung ngoai khong khi tdi kho'i liTpng khong doi thu diTpc
InC
14,5 gam cha't r^n.

gam kim loai.

Tinh khoi lu'dng Cu bam vao moi thanh va nong dp mol cua dung dich

hop Zn, A l . Sau khi ke't thuc cac phan iSng thu diTOc 0,896 lit hSn hdp khi X

CUSO4 ban dau.

1. Hoi CO can dung dich A thu diTdc bao nhieu gam muoi khan?

Bai giai

2. TinhC.
.

3Zn(N03)2 + 2 N 0 T +

4H2O

A l + 4HNO3

-> Al(N03)3 + N O t + 2 H 2 O

4Zn + IOHNO3

-> 4Zn(N03)2 + N 2 0 t

8A1 + 3OHNO3

-> 8A1(N03)3 + 3 N 2 O T + I 5 H 2 O

Tinh so mol hon hdp khi: n =
Tinh KLPTTB hlng:

0,896

+5H2O

-+ ZnS04 + Cu 4-

(1)

(1)

Fe + CUSO4

-> FeS04 + Cui

(2)

(2)

FeS04 + 2NaOH

-+ Fe(OH)2i +Na2S04

(3)

CuS04 + 2NaOH

-+ C u ( 0 H ) 2 i +Na2S04

(4)

ZnS04 + 2NaOH

-> Z n ( 0 H ) 2 i +Na2S04

(5)

Zn(0H)2 + 2NaOH

-> Na2Zn02 + 2H2O

(6)

2Fe203 + 4 H 2 0 t

(7)

(4)

= 0,04 mol

4Fe(OH)2 + O2

22,4

(Phan tfng (7) c6 the viet 2 giai doan

16,75 x 2 = 33.5

4Fe(OH)2 + O2 + 2H2O

x + y = 0,04
Goi X, y la so mol NO, N2O ta c6 he phiTdng trinh:

30x^^33^3

Cu(0H)2

Giai ra ta co x = 0,03 mol (NO); y = 0,01 mol (N2O)
Theo cac phan tfng ( 1 , 2 ) so mol goc - NO3 tao muoi nitrat luon b^ng 3 so
Theo cac phan uTng (4, 5) so mol goc - NO3 tao muoi luon b^ng 8 l^n s6' mol
N2O. Do do tong so mol goc - NO3 tao muoi b^ng:

3 + 0,01

X

Tong kho'i liTdng muoi = tong khoi liTdng kim loai + khoi liTdng goc nitrat:
m™,tikhan

= 2,987 + 0,17

X

62 = 13,527 gam.

0,17

+ 0,03 + 0,01x2 =0,22
^

—
'
phan t^o khi

0,22
-HNO3

| T a tha'y ciJ 1 mol CUSO4 chuyen th^nh 1 mol FeS04 khoi liTpng dung dich
gi^m (64 + 96) - (56 + 96) = 8gam;
Do do ta CO phiTPng trinh: 8x - 2,5x = 0,22 => x = 0,04 mol
Vay khoi luTpng Cu bdm vao sat = 64 x 0,04 = 2,56 gam
bam vao kem = 64

0,04

x

2,5 = 6,4 gam.

Kh6'i lirpng mpejoj = 160 x 0,02 = 3,2 gam
= IM

0,22

dich CUSO4. Sau mot thcJi gian lay 2 thanh kim loai ra khoi coc, liic d6 tat

T^ng so mol CUSO4 ban dau bang:
X

+ 2,5x +

11.3
80

Cu thoat ra deu bdm het vao 2 thanh kim loai va khoi liTdng dung djch trong
coc bi giam 0,22 gam. Trong dung dich sau phan iJng nong dp mol ciia ZnSO^

;Mii-+ !

Vay khoi liTpng CuO = 14,5 - 3,2 = 11,3 gam.

Bai 15. Nhiing 1 thanh sat va 1 thanh kem vao ciing mot coc chuTa 500 ml dung

126

x

Theo cic phan iJug (2, 3, 7):

2. Tong so mol HNO3 trong 220 ml dung dich HNO3 bing
%'—^—'
^
phan tao muoi

la so mol FeS04 tao thanh thi so mol ZnS04 Ik 2,5x; so mol Cu him

I TiWng tir doi vdi Zn tang (65 + 96 - (64 - 96) = 1 gam.

8 = 0,17 mol

Khoi liTdng kim loai tan trong axit b^ng: 5 - 2,013 = 2,987 gam
.

X

— ^ C u O + H20t

vho thanh s^t la x va so mol Cu bam vao thanh kem Ik 2,5x.

mol NO, khong phu thuoc kim loai nao.

X

I'Gpi

-> 4Fe(OH)3

-> Fe203 + 3 H 2 0 T )

2Fe(OH)3

0,04

0,03

it*.;

Zn + CUSO4

1. Cdc phan uTng hoa tan:
3Zn + 8HNO3

•m

Cic phan tfng xay ra:

Baigiai

= 0,04 + 0,1 +0,14125 = 0,28125

0,28125
CcuS04 '

0,5

= 0,5625M

:
127


B a i 16. Hoa tan hoan to^n m o t m i e n g bac k i m loai v ^ o m o t ItfcJng diTdung dici^
HNO3

15,75% thu diTcJc k h i duy nha't N O va a gam dung dich F trong do nong

3
3 5 94
5
J 896
Tinh H A =nH2 =2^A\-^ = O,33mol; ng = n c i 2 " 2 " " ^ 5 8 " " ° ' ^ ^ ™ ° ^

do C% cua AgNOs bang C% cua H N O 3 dir. T h e m a gam dung dich HC|

3
" c ="02 =2"'
1,46% vao dung dich F. H o i c6 bao nhieu % A g N O j tac dung v d i H C l .
Baigiai
PTPU:

> 3 A g N O j + N O t + 2H2O

3Ag + 4HN03

(1)

,

=2no2 + nci2 =^ 0,33 = 2 X 0,15 + 0,03

Vi

15,75% = 0,1575m

Dung dich D la dung dich H C l

'

Kh6'i lu-dng H2O tao thanh = 0,3 x 18 = 5,4 gam.

Va so gam H N 0 3 d i r = 0 , 1 5 7 5 m - 4 X 63.
K h o i lifring dung dich F la a = 3 x 108 + m - 30 = m + 294 (trong do 30

»y ' 1 - • "

K h o i lu-png H C l tao thanh = 0,06 x 36,5 = 2,19 gam.
Do do C % H c , = 1 4 ^ ^ - 2 8 , 8 5 %

k h o i liTcJng m o l cua N O )

5,4 + 2,19

Ta c6: C % A « N O , = ^ ^ ^ ' ^ ^ = 2.0528
^'^
1000x36,5
PTPLf:

12,25
5 "

Theo cac phan iJng (4, 5) thi cac chat phan iJng vuTa du v d i nhau.

Cach 1: Gpi m la so gam dung dich H N O 3 da dijng de hoa tan 3 m o l A g ,
so gam H N O 3 = itix

3

"2''T22

AgN03 + HCl

>AgCli+HN03

Bai 18. Nung hon hdp X g o m FeS2 va FeCOj trong khong k h i t d i phan iJng hoan
toan thu duTpc san p h a m g o m mot oxit s^t duy nhat va hon hdp hai k h i A , B .

(2)

1. V i e t P T P l / x a y ra.

2 0528

V a y % A g N 0 3 da phan lirng bang

: •

2. Neu cho ttfng khi A va B l o i tijf tijT qua dung dich Ca(0H)2 t d i diT thi c6 cac

— x 100 = 68,40%

Cach 2: Gia suT l a y 100 gam dung dich HNO3 15,75%, luc do so m o l Ag hi

hien tiTdng g i x a y ra. G i a i thich bang cac P T P l /
3. Trinh bay phifdng phap hoa hoc de nhan biet cac k h i A , B trong hon hdp cua

hoa tan la x. Ta t i m difdc x = 0,062 m o l , a = 106,076 gam
0 0424 X 1 0 0
nHci = 0,0424 m o l va %AgN03 phan uTng bang
—
= 68,40%
0,062

chung.
4. Cho biet 1 l i t hon hdp k h i A , B d dktc nang 2,1875 gam. T i n h % k h o i liTdng
moi chat trong hon hdp X .

>

B a i 17. Hoa tan h e t 5,94 gam A l bang dung dich N a O H diTdc k h i A . Cho dung
dich H C l dac, diT tac dung v d i 1,896 gam K M n 0 4 di/dc k h i B . Nhi$t phan

Baigiai

1. Cac phan i^ng;

' ~>

hoan toan 12,25 gam KCIO3 (xuc tac M n O j ) ihu diTdc k h i C. Cho 3 k h i A, B,
C vao binh k i n va tic'n hanh phan iJng no hoan loan, sau do l a m lanh binh
xuong nhiet dp thiTcfng, gia sijT luc do niTdc nguTng tu he't va chat tan het vao
nirdc diTdc dung dich D .

4FeS2 + 1102

—

^ 2Fe203 + 8 S 0 2 T

(1)

4FeC03 + 02

—

^ 2Fe203 + 4 C 0 2 t

(2)

2. Neu cho tuTng k h i A , B l o i qua nifdc v o i trong dau tien ta thay due (ke't tua),

1. V i e t cac PTPLf xay ra.

sau do dung dich l a i trong suot do cac phan i^ng:

2. T i n h nong do C % cua cha't tan trong D .
B a i giai
Cac phan ufng x a y ra:
..

;

'^''^ '

2A1 + 2 N a O H + 2 H 2 O - - - > 2NaA102 + 3H2

(1)

> 2KC1 + 2MnCl2 + 5 C l 2 t + 8H2O (2)

2 K M n 0 4 + 16HC1

C02 + Ca(OH)2

)-CaC03;+H20

'^"'^ '

C02d^ + H20 + CaC03

> Ca(HC03)2

^^'f^l

SO2 + Ca(0H)2

> C a S 0 3 i + H2O

S O 2 + H2O + CaSOj

> Ca(HS03)2

J
^

^ '
;
^- Cho hon hdp k h i Ian lu-dt qua binh 1 diTng ni/dc brom du" va binh 2 difng niTdc
2KCIO3

— ^ 2 K C l + 302t

(3)

SO2 + 2H2O + Br2

xt

128

Voi trong, dn tha'y mau nau cua ni/dc brom b i nhat do phan iJng:

2H2 + O2

—

^ 2H2O

(4)

H2 + CI2

—

^ 2HC1

(5)

> 2HBr + H2SO4

trong binh 2 xua't hien ket tua:
C02 + Ca(OH)2

?',,» rf

'

•

•

>CaC03^+H20
129

4. Khoi liTdng cua 1 mol hon hOp A, B = 2,1875 x 22,4 = 49 gam.

pjoa tan cha't r^n A b^ng axit HNO3 dSc thi c6 bao nhieu lit khi m^u nau
^' thoatra(c(dktc)

Goi X va 1 - X la so' mol S O 2 va C O 2 trong I mol hon hcJp, ta c6:

64x + 4 4 ( l - x ) = 49

Bai giai

Giai ra ta c6 x = 0,25 mol va 1 - x = 0,75 mol; tiJc ty le so mol
Tir do theo cac phan tfng (1,2) suy ra ti le so mol:

: nco^ = 1 : 3

Up^s^ : n F e C 0 3 ^ ^' ^

J C^c phan iJng:
2AgN03 + Fe

(Cach tim: nhan phiTcJng trinh (2) vofi 6 de cho so' mol CO2 gap 3 so mol SO^)
Vay % khoi liTdng cua: %FeS2 = ^^^^"^^Q^
• ^
1x120 + 3x116

Cu(N03)2 + Fe

Tinh:

25,64%

npe = ^

/ .sj-:;,

•
> Fe(N03)2 + 2Ag i

(1)

> Fe(N03)2 + Cu i

(2) ^ .^4. H U : .

= 0,04mol; n^gNO:, = 0,20x0,1 = 0,02mol;

,

ncu(N03)2 ==0'20>< 0.5 = 0, Imol

Va %FeC03 = 100 - 25,64 = 74,36%.
Bai 19. Tien hanh dien phan (dien cifc trd mang ngan xop) 500 ml dung dich

S6'mol Fe tham gia phan tfug (1) bSng: ^^n^gNo,^ =rnAg = - ^ ^ = 0,01mol

NaCl 4 M (d = 1,2 g/ml). Sau khi 75% NaCl bj dien phan thi dCfng lai.
1. Tinh nong do % cua cac chat trong dung djch sau dien phan.

So mol Fe tham gia phan tfng (2) bang: npc = 0,04 - 0,01 = 0,03 mol

2. La'y cac khi thoat ra cho tac dung het vdi nhau dufdc san pham A. Hoa tan A

Do do: ncu(N03)2 ' ^ ^ ^

vao 500 g nirdc dU'cJc dung dich A. Tinh nong do % cua dung dich A.

0,02 x 108 + 0,03

1. Phan iJng dien phan:
H21 + CI21 + 2NaOH

(1)

m.n.x
UNaci

b j d i p n phan

= nNaOH

75 •
= 0,5 x 4 x — = l,5mol

CFe(N03)2 = - ^ = 0'2M ;

25

= 0,5

X

4

X

1

1
— - = 0,5mol; nH2 =nci2 = - n N a O H = - x 1,5 = 0,75mol

Khoi lifdng dung dich sau dien phan bang:
500 X 1,2 - 0,75 X 2 - 0.75 X 71 = 545,25 g
w
XT r^u 1,5x40x100
0,5x58,5x100
. ,,„
Vay % NaOH =
=^11% ; %NaCl =
= 5,36%
545,25
545,25
2. Phan tfug: H2 + CI2
> 2HC1
(2)
Theo phan drug (2):

64 = 4,08 g

'

*

HHCI =

2

X

0,75 = 1,5 mol

CCU(NO3)2

=

;•

=

j^.j^j. ;

,

•

0.35M

3. Cdc phan ufng hoa tan:

lUO

nNacic6n

X

. ^^^j,^^, „ ^^.^ j ^ , ^ . .

2. Tinh nong do cac muoi trong dung dich B:

2NaCl + 2H2O
Theo phan u'ng(l):

= 0,1 - 0,03 = 0,07 mol

Chat ran A gom Ag, Cu c6 kho'i li/dng b^ng:

Bai giai

"

Ag + 2HN03

> AgNOj + NOzT

(3)

Cu + 4HNO3

> Cu(N03)2 + 2NO21 + 2H2O

+H2O

(4)

Theo phan tfng (3, 4): nN02 ^ " A g +2xncu =0,02 + 2 x 0 , 0 3 = 0,08mol
Vay the tich N O 2 (theo dktc) = 0,08 x 22,4 = 1,792 lit
Bai 21. X la quang hematit chiJa 60% Fe203. Y la quSng manhetit chtfa 69,6%
Fe304.
^- Hoi tir 1 tan quang X hoac Y c6 the dieu che diTcJc toi da bao nhieu kg sat
kim loai?
Can tron X va Y theo ti le khoi lu-dng nhu the nao de dirdc quang Z ma ttf 1

Vay nong do % cua dung dich A b^ng: %HC1 = ^'^^^6,5x100 ^ ^ ^^^^
500 + 1,5x36,5
Bai 20. Co 200 ml dung dich hon hdp AgNOj 0,1M va Cu(N03)2 0,5M. Them
2,24 gam bot s^t kim loai vao dung dich do. Khuay deu tdi phan uTng hoai^

tSn Z CO the dieu che dtfcJc 0,5 tan gang chiJa 4% cacbon.
Bai giai
C)»ra theo cong thuTc cua Fe203 va Fe304 thi tiT 1 tan X hoSc Y c6 the dieu che
nhiTng lUdng Fe nhir sau:

toan thu diTdc chat r^n A v^ dung dich B.
1. Tinh so gam chat ran A.
2. Tinh nong do mol cua cac muoi trong dung dich B, biet r^ng the tich dung
dich khong doi.
130

m p ^ , ^ , = l ^ x l l ^ = 0,42T = 420kg
^^^^^
100
160
^
1x69,6

168

_

"^'^^•<^^=^^^ = °''°''^ = '°''^

' '
.iTun;u-r

-

'


2. LrfcJng Fe c6 trong 1 ta'n quSng Z hKng:

mp^^2.) = ^ ' ^ ^ ^ ^ = 0,48T = 480kg
100
Goi X , y la so ta'n quang X va quSng Y can tron, la c6 he phiTdng trinh:

X'-a::;'-

Jx + y==l

Qlii chii: CO the g i a i theo phuTdng trinh bSo loan electron nghia la t^ng so'
electron cac chat khur cho bang long so electron cac cha't o x i hoa nhan, d day
pg la cha't khur, c6n O2 va N*^ la chat o x i ho^ nen ta c6 phu'dng trinh:
jn ^
56

" 0,42x + 0,504y = 0,48
2
5
G i a i he phufdng trinh ta difcJc: x = - va y = - ; nghla la phai tron X , Y the;oti

3 ^ i l z i ! ! X 4 + 0,1 X 3.
32

3 T6'ng k h o i lirdng muo'i: npe = npe(N03)3 =
podo:

1$ mx : my = 2 : 5

Rut ra m = 10,08g
= 0,18 m o l

^

mpg(fjo^)^ = 0,18 X 242 = 43,56 gam

B a i 22. De m gam bpt s^t nguyen chat trong khong khi mot thdi gian thu duoj,

pai 23- H a i mie'ng k e m c6 cung k h o i li/dng 100 gam. Mie'ng thi? nha't nhiing vao

chat r^n A nSng 12 gam gom Fe, FeO, Fe304 va F e j O j . Hoa tan hoan loan

100 m l dung dich CUSO4 diT, mie'ng thuT hai nhiing v^o 500 m l dung dich

A bkng dung dich HNO3 loang thay c6 2,24 lit khi N O duy nhat

AgN03 dir. Sau m o t thcfi gian la'y 2 mie'ng k e m k h o i dung dich nhan thay

cha't dn

(dktc) va dung dich B chi chtfa mot muoi nitrat sat duy nhat.

mieng thiJ nha't g i a m 0 , 1 % kho'i liTdng, nong do m o l cua cdc m u o i k e m trong

1. Viet cac PTPLf xay ra

hai dung dich bkng nhau. H o i k h o i liTdng mie'ng k e m thiir hai thay d o i nhuTthe

2. Tinh khoi liTdng m.

nao? Gia si( cac kirn loai thoat ra deu b a m vao mie'ng k e m .
B a i giai

3. Tinh khoi liTdng muoi trong dung dich B.
C i c phan

B a i giai
1. C a c phan uTng xay ra:
2Fe + O2

2FeO

(1)

3Fe + 2O2

-> Fe304

(2)

4Fe + 3O2

-> 2Fe203

(3)

Fe + 4HNO3

Fe(N03)3 + N O t +2H2O

(4)

3 F e O + IOHNO3

3Fe(N03)3 + N O T + SH.O

(5)

3Fe304 + 28HNO3

Fe203 + 6HNO3

-> 9Fe(N03)3 + N O t + I4H2O

(6)
(7)

2Fc(N03)3 + 3H2O

'

' '

Z n + CuS04

>ZnS04 + C u i

(1)

Zn + 2AgN03

)• Zn(N03)2 + 2Ag>l

(2)

< ^
t

t

X = nzn phin iJng = n Z n S 0 4

" "Cu

ta c6 bieu thiJc: 100 - 65x + 64x = 100 - —

= 99,9 gam. R i i t ra x = 0,1

100

^

V I the tich dung djch A g N 0 3 gap 5 Ian the tich dung dich CUSO4 nen so m o l
Zn(N03)2 gap 5 Ian so m o l CUSO4, ttJc b i n g 0,1 x 5 = 0,5 m o l . G o i a la k h o i
IiTcJng mie'ng k e m thi? hai sau phan iJng, ta c6 bieu thiJc:
100 - 0,5 X 65 + 2 X 0,5 x 108 = a. G i a i ra c6 a = 175,5 g
^ai 24. Cho m o t mie'ng nhom nSng 20 gam vao 400 m l dung dich C u C b 0 , 5 M .

56x + 72y + 2 3 3 z + 1 6 0 1 = 12

(a)

Khi nong do dung dich CUCI2 g i a m 25% thi lay mie'ng n h o m ra, ruTa sach, sa'y

y z
2,24
= 0,1
x + ^ + - = nNo =
22,4
3

(b)

kho, can nang bao nhieu gam? Gia suT dong bam het vao mie'ng nhom.
B a i giai

So m o l nguyen tijT o x i tham gia phan iJng b^ng: y + 4z + 3t =
T i r ( a , b ) t a c 6 : y + 4z + 3 t = ^
= 0.12
160
Tur (c, d) n i l ra m = 10,08g

132

tfug:

'

Gpi X la so' m o l Z n tham gia phan tfng. Theo phan lirng (1) Ihi:

2. G o i X , y, z, t la so m o l cua Fe, FcO, Fe304, Fe203 trong chat ran A . Thco dicn
k i e n bai loan va theo cac phiTcfng trinh phan tfng, ta c6 he phiTdng trinh:

'

12-m
16

(c)

Phan iJng:

2A1 + 3CuCl2 - — >

2AICI3 + 3Cu i

(1)

So mol C u C l j tham gia phan u"ng b^ng: 0,4 x 0,5 x

= 0,05 m o l
100

(d)

^ o i m lii k h o i lUdng mie'ng nhom sau phan iJng, theo phan tfug (1) ta c6 bieu
thu-c:

20 - - X 0,05 x 27 + 0,05 x 64 = 22,3 gam.
133


'64x + 27y = 100

X a c d|nh thanh p h ^ n h S n hqrp

Chu di 6.

8 0 x + 1 0 2 x ^ = 1 0 0 x l , 3 5 = 135
2

B a i 1. H o a t a n h o a n t o a n 11,9 g a m h d p k i m Z n - A l b i n g d u n g d i c h H C l tj,
d t f d c 8,96 l i t H2 (cl d k t c ) . T i n h % k h o i I t f d n g m o i k i m l o a i t r o n g hcJp k i m .

^

G i a i h e phi/dng t r i n h ta c d x = 1,32

C a c p h a n iJng:

Zn + 2HC1

ZnCl2 + H j t

(1)

2A1 + 6HC1

>2AlCl3 + 3H2t

(2)

K h o i l i W n g c u a C u t r o n g X b a n g = 1,32 x 6 4 = 8 4 , 4 8 g

.<*>) d * ' >

vay % k h o i li/dng C u l a 8 4 , 4 8 % .

' ,.. ^

-j

'
^

p a i 3- D o n s n i t r a t b i p h a n h u y k h i n u n g t h e o p h a n iJng:

> C u O + N02t + O21

Cu(N03)2

G p i X v a y l a so m o l c u a Z n v a A l , ta c 6 h e phu'dng t r i n h :

'^J'

N u n g 15,04 g a m d o n g n i t r a t t h a y c o n l a i 8,56 g a m cha't r ^ n .

•65x + 2 7 y = l l , 9

J x i n h % dong nitrat b i phan huy.

3
8,96 . .
x +- y =
= 0,4
2
22,4

2 T i n h % k h o i liTdng m o i cha't t r o n g cha't r a n c o n l a i .

'

''»M.'I

•l'"*^'^*' •

3. T i n h t o n g t h e t i c h k h i d a t h o a t r a ( d k t c ) .

"

B a i giai

' V G i a i h e phU'dng t r i n h ta c d : x = 0,1 m o l ; y = 0,2 m o l .

,

Phan iJng p h a n h u y Cu(N03,2:

V a y % k h o i liTdng c u a Z n b ^ n g ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ = 5 4 , 6 %

C u O + 2NO21 + 0,5021

Cu(N03)2

V a c u a A l b ^ n g 100 - 5 4 , 6 = 4 5 , 4 %

1. T i n h :

B a i 2 . H o n h d p k i m l o a i g o m A l v a C u . C h o h o n h d p X v a o c o c d u n g d i c h HCl.

n c m N O j j j = 1 5 , 0 4 : 188 = 0,08 m o l

G g i X l a so' m o l Cu(N03)2 b j p h a n h u y , t h e o p h a n xing ( 1 ) ta c d phiTdng t r i n h

K h u a y d e u d u n g d j c h t d i k h i k h i n g i f n g t h o a t r a t h u du'dc c h a t r a n Y nang a

ve k h o i li/dng c h a t r a n c o n l a i : ( 0 , 0 8 - x ) x 188 + 8 0 x = 8,56

g a m . N u n g Y t r o n g k h o n g k h i t d i p h a n uTng h o a n t o a n t h u d t f d c 1,35a gam

G i a i r a ta c d : x = 0 , 0 6 m o l

o x i t . T i n h % k h o i liTdng C u t r o n g h o n h d p Y .
V a y % Cu(N03)2 b i p h a n h u y : " ' " ^ " " ^ ^ ^ = 7 5 %
"
0,08

B a i giai
P h a n iJng h o a t a n h o n h d p t r o n g H C l
2A1 + 6HC1

> 2AlCl3 + 3H2t

Cu + H C l

> khong

is

2 A 1 + 1,502

> CuO
AI2O3

3. T o n g so m o l k h i t h o a t r a :
n = 0,06

(2)

liTdng - 5 ^ ^ = — = 1,25, d i e u n a y k h o n g p h u h d p v d i d i e u k i e n c h o (1,35). D i '
64
m Cu

m 2AI

2x27

1 8 8 x 1 0 0 ^ ^^^^^^^
V

n = nN02 ~ " 0 2

2 + 0 , 0 6 x 0,5 = 0,15 m o l
"02

^^'^

^ V a y t o n g t h e t i c h k h i : V = 0,15 x 2 2 , 4 = 3 , 3 6 l i t
4. M o t l o a i d a v o i chiJa 8 0 % C a C O j , p h a n c o n l a i l a c a c cha't t r d . N u n g m
^ g a m da m o t t h d i g i a n t h u diTdc cha't r ^ n c d k h o i lufdng b a n g 0 , 7 8 m g a m .

^ T i n h h i e u sua't p h a n i^ng phaft h u y .
• T i n h % C a O t r o n g c h a t r ^ n sau k h i n u n g .

d d a x i t p h a i h e t v a A l c o n dir. T h e o p h a n ufng ( 3 ) t i l e t a n g k h o i liTdng c u a A l
= 1,89. D i e u d o p h i i h d p v d i d i e u k i e n b a i t o a n .

X

nN02

(3)

N e u A l h e t t h i cha't r ^ n Y c h i c o n C u , nhU' v a y theo p h a n iJng ( 2 ) t i 1^ tang kh*'

102

fflQ^-O.O^)^
8,56

K h i k h i ngijfng thoat r a c d t h e x a y r a h a i kha nang: hoac A l h e t hoac axit H C l het.

mAi203 _

%Cu(N0,)2 =

% C u 0 = 100 - 4 3 , 9 3 = 5 6 , 0 7 %

C a c p h a n uTng c d t h e c d k h i n u n g c h a t r ^ n Y t r o n g k h o n g k h i :
Cu + 0,502

2. T i n h % k h o ' i hTdng:

(D

B a i giai
1

o

• Phan li-ng p h a n h u y d a v o i : C a C O j

^0'}'''
• • •

'

'" > C a O + C 0 2 t

G i a sur k h o i liTdng Y l a lOOg

^ i a sur m l a lOOg, l u c d d t r o n g d a c d 8 0 g C a C O j tiJc 8 0 : 100 = 0,8 m o l

G p i X , y l a so m o l c u a C u v a A l t r o n g c h a t r ^ n Y , ta c d h e phuTdng t r i n h

K h o i lirang chat r ^ n b ^ n g 78 g a m .
135


Phan k h o i
Theo

ly

liTcJng

h u l di chinh bang k h o i

thuyet k h o i

liTdng

C O j toi da

H i e u suat phan uTng: h % = ^ ^ ^ ^ ^

C O j = 100 - 78 = 22g.

liTcfng

bang

0,8

x

Goi

44 = 35,2g.

X,

y

la so m o i

MgCO,, CaCOi

ta

c6

h e phiTdng trlnh:

i^^^ + lOOy - 8
[40x + 56y = 4,4

Giai

= ^2,5%

h e p h u - d n g t r i n h ta

,A,,0, =

2. So m o i C O 2 tao thanh b i n g so m o i CaO = 22 : 44 = 0,5 m o i .

c6 x = 0,0114

iiiSO = „ , , » ;

moi;

y = 0,0705

moi

, M g C O , = " • ° " ^ ' " " " < ' " ° = 10.6%

%CaC03 = 1 0 0 - 11,1 - 10,6 = 78,3%
78

J

B a i 5. Ngirdi ta c6 the dieu che CI2 bang each cho dung dich H C l dac tac dung

^

Cac phan i^ng hoa tan:
MgO

v d i K M n 0 4 hoac M n O j , K C I O , .

'' '

'

"

. *

> MgCl2 + H2O

+ 2HC1

1. Vie't cac phiTdng trinh phan i^ng.

CaO + 2HC1

> C a C l 2 + H2O

2. N e u lifcJng clo thu duTcJc trong 3 tri/dng hdp nhiT nhau t h i t i le k h o i li/cfng cua

A I 2 O 3 + 6HC1

> 2AICI3 + 3H2O

K M n 0 4 , M n 0 2 va K C I O 3 b^ng bao nhieu.

:

Be tinh nhanh, triTdc hct tinh liTdng H C l can de hoa tan 5,4 gam cha't A.

B a i giai

Tong so moi H C l = 2

x

UMgo

+ 2

x

ncao

+ 6

x

nAi203

1. Cac phiTdng trinh phan uTng:
2 K M n 0 4 + 16HC1
M n 0 2 + 4HC1

> M n C l 2 + C I 2 1 + 2H2O

(2)

K C I O 3 + 6HC1

> K C l + 3CI2 T + 3H2O

(3)

2. Gia suT luTdng CI2 bay ra trong moi tn/dng hdp la 1 moi, liic do theo cac phan iJrng
( 1 , 2. 3) can - moi K M n 0 4 , 1 moi M n O j va ^ moi KCIO3, tufc ty 1$ khoi lifdng:
5
3

m K M n 0 4 : m M „ 0 2 --^Kao,

=

| X 1 5 8 1 8 7 : i X 1 2 2 , 5 = 6 3 , 2 : 87 : 40,83.

B a i 6. M o t loai da chiJa M g C O j , CaCOj, A I 2 O 3 bang 1/8 tong liTdng hai muoi

cacbonat. Nung da d nhiet do cao t d i phan huy hoan toan hai m u o i cacbonat
thu du-dc chat ran A c6 k h o i lUdng bang 6 0 % khoi lu^dng da triTdc k h i nung.
1. T i n h % k h o i lifdng m o i chat trong da tru-ctc k h i nung.
dich H C l 0,5M?

13^

'X ?

Goi V la so lit H C l can dung: V x 0,5 = 0,0824 => V = 0,1648 lit = 164.8 m l .
Bai 7.

•

,

1. M o t loai dura co th^nh phan k h o i liTdng nhi/ sau: 9 4 % A l , 4 % Cu, 0,5% m o i
nguyen to M g , M n , Fe va Si. Neu co 1 tan A l nguyen cha't thi can lay cac
nguycn to kia, m o i nguyen to bao nhieu kg de luyen thanh dura co thanh
phan nhU'lrcn.
2. Hon hdp g o m cac k i m loai M g , A l va Cu.

»'

a) Oxi hoa hoan toan m gam hon hdp A thu dUdc 1,72 gam hon hdp 3 oxit.
Tinh % k h o i lifdng moi k i m loai trong hon hdp A.

ISyia:

MgCOj

MgO + COzt

(1)

CaC03

-> CaO + C 0 2 t

(2)

A1203

-> khong doi
gam

AI2O3;

AI2O3

mcu

=

^ " " " " 4• = 42,55kg
94

Con Fe, M g , M n , Si m o i nguyen to can lay

(3)
1

—
= 5,4 gam trong do c6 1 gam
100

CaO va M g O .

Vay de hoa tan 2 gam A can ^'^^^^""^ = 0,0824mol
5,4

Cu- 1 tan life lOOOkg A l ufng v d i 9 4 % ; do do lufdng cac nguyen to khac can

1. Cac phan i^ng phan huy cacbonat:

hdp

= 0,2226 moi.

B a i giai

B a i giai

Gia sur da ban dau c6 9 gam trong do co

102

b) Hoa tan m gam hon hdp A bang dung dich H C l diT thu du-dc 0,952m lit H2 (dktc).

2. M u o n hoa tan hoan toan 2 gam chaft ran A can toi thieu bao nhieu m l dung

A = 9x

= 2 X 0,0114 + 2 X 0,0705 + 6 x

> 2KC1 + 2MnCl2 + 5CI2 T + 8H2O (1)

1000x0,5
94

= 5,32kg

^' Ta C O the tif chon gia trj cua m la 100 gam
k h o i lufdng chat ra"^

va 5,4 - 1 = 4,4 gam 1"'"'

Cac phan ij-ng:
Mg+

0,502

->

MgO

(1)

2A1+

1,502

->

AI2O3

(2)
137


Cu + 0 , 5 0 2
b)

Mg

+ 2HC1

-> CuO
>MgCl2 + H 2 t

(3)
(1)

2A1 + 6HC1
>2AlCl3 + 3H2t
(2)
Gpi X, y, z la so mol cua Mg, Al, Cu trong 100 gam hon hdp A, ta c6 cac
phiTdng trinh: 24x + 27y + 64z = 100
Khoi liTrtng oxit = 40x + 102 I + 80z = 1,72 X m = 172
3
0,952xm 95,2 . . .
So mol hidro = x + - y =
=
= 4,25
2
22,4
22,4
Giai he phu^dng trinh ta c6: x = 1,25 mol; y = 2 mol; z = 0,25 mol
Vay % khoi lu'dng:
1,25x24x100
2x27x100
%Mg = —
— = 30%; %A1 =
= 54%;
^
100
100
^^^^^0,25x64x100^^^^^
100
(hoac 1 0 0 - 3 0 - 54= 16%.)
Bai 8. Nung 500 gam da voi chuTa 80% CaCO., (phan con lai la cac oxit nhom,
s^t (III) va silic), sau mot thdi gian thu diTdc chat r^n X va V lit khi Y.

1. Tinh khoi Ming chat ran X, biet hieu suat phan huy CaCOa la 75%.
2. Tinh % khoi lu-rfng CaO trong chat ran X.
3. Cho khi Y sue ra't tif lit vao 800 gam dung dich NaOH 2% thi thu dmc miioi
gi, nong do bao nhieu %?
Bai giai
1. Phan u-ng nung da voi: CaCOi — ^ CaO + C O j t
500x80 = 4mol
, ,
Tmh: nc.rn^
=
^"^"^ 100x100
phan huy = n c a o = nco2 = 4 X — = 3mol.
Khoi li/ctng chat dn bang khoi lu'dng ban dau triif khoi luTdng CO2 bay di
= 5 0 0 - 3 X 44 = 368 gam'
2. % C a O = l ^ ^ ^ ^ = 45,65%
368
3. Trirdc het tinh so mol NaOH = ^^^^ ^ ^ ^ 0,4mol
100x40
Vi so mol NaOH < so mol CO2 nen ta chi thu difdc muoi axit:
CO2 + NaOH
> NaHCOj
"cac03

=0,4mol
0,4x84x100 = 4,1%
Nong do %NaHC03 =
800 + 0,4x44
0ai 9*
^^"^ nhom kim loai trong khong khi mot thcJi gian thu diTcJc chat ran
/V CO kho'i lu'dng 2,802 gam. Hoa tan chat ran A bang dung djch HCl di/ thay

thoat ra 3,36 lit H2.
,
J Tinh % kho'i lu"dng ciaa Al va AI2O3 trong A.
2 Tinh % Al bi oxi hoa thanh AI2O3
3 Neu hoa tan hoan toan cha't ran A bkng axit nitric dSc nong thi c6 bao nhieu
lit khi mau nau duy nhat thoat ra.
Cho cac the tich khi do d dktc.
f
Bai giai
AI2O3
(1) . MuA)
1. Cac phan ifng:
2AlCl3 + 3H2t
2A1+ 1,502
(2)
^ 2AICI3 + 3H2O
(3)
2A1 + 6HC1
2
2 3 36
AI2O3 +
6HC1
TheoX phan
ufnggam.
(2) so mol nhom con lai = — n u , = —x-^— = 0,1 mol tuTc
0,1
27 = 2,7
va

nNaHCO.-, = n N a O H

3

22,4

Khoi liTdng AI2O3 = 2,802 - 2,7 = 0,102 gam.
2. Theo phan ijrng(l): nAihi.,xihoa = 2 x n^i^^^ = 2 x -0,102
^^zzO,002moI
•

,

H2

Vay % Al bi oxi hoa bang = 0,002x100 = 1,96%
'ts; '
0,1 + 0,002
3- Cac phan tfng hoa tan A bkng HNO3 dac:
(4)
AI2O3 + 6HNO3
> 2A1(N03)3 + 3H2O
(5)
Al + 6HNO3
> A1(N03)3 + 3NO21 + 3H2O
Theo phan urng (5) n N 0 2 =3nAi ==3xO,l = 0,3mol.
'
Vay the tich NO2 = 0,3 x 22,4 = 6,72 lit.
^ai 10. Chia hon hcJp kim loai Cu - Al thanh hai phan bang nhau.
Phan thu" nha't nung nong trong khong khi tdi phan tfng hoan toan thu difdc
18,2 gam hon hcJp hai oxit. Hoa tan hoan toan phan thu" hai b^ng dung dich

H2SO4 dac nong thay bay ra 8,96 lit SO2 (d dktc)
A;
',
^- Tinh so' mol moi kim loai trong hon hpp.
^' Ne'u hoa tan hoan toan 14,93 gam kim loai X b^ng dung djch H2SO4 dac
nong va thu diTdc mot Itfdng SO2 nhiT tren thi X la kim loai gi?
139


Bai giai
1. Cic

Cu + 0,502

p h a n uTng:

Kg'u

CuO

Cu+

0,502

2A1+

1,502

AI2O3

CuS04 + S 0 2 t

Cu + 2H2S04.1ac

+2H2O

Al2(S04)3 + 3 S 0 2 t + 6 H 2 O

2Al + 6H2S04dac

A l con:

'" > CuO

1,502

2A1+

(2)
,

> AI2O3

,

(3)

(1)

jChi khi H2 ngijrng thoat ra, c6 the A l het hoac HCl he't.

(2)

jsleu A l he't, chat ran con lai chi c6 Cu, va khi nung trong khong khi thu dU'dc

(3)

CuO: nhu- vay ti le khoi lu'dng tang bKng '"cuO
mcu

(4)

diTdc toi da 1,25a gam oxit. d day khoi lu'dng thu dU'dc la 1,36a chiJng to

Goi a va b la so mol Cu va A l ta c6 he phifdng trinh:

j 25.

a gam thu

64

trong X phai c6 A l v i ty le tang khoi lu'dng cua A l (theo phan vlng 3) bang:
•<•<

80a + 1 0 2 - = 18,2
2

• u' •

mjA,

3u
8,96
a + -b =
= 0,4
2

2x27

Chu de 7.

'

L a p cong thufc mpt c h i t

'

li

'

I

22,4
Bai 1. Them tijT tiif dung dich HCl vao 10 gam muo'i cacbonat kim loai hoa tri I I ,

Giai he phiTcfng trinh ta co: a = 0,1 mol va b = 0,2 mol

sau mot thdi gian thay lu'dng khi thoat ra da vu'dt qua 1,904 lit (dktc) va

2. Phan lirng hoa tan kim loai X hoa tri n:
2X + 2nH2S04 jac — ^ X2(S04)„ + nSOj t + 2nH20

lifdng muo'i tao thanh da vu'dt qua 8,585 gam. Hoi do la muo'i kim loai gi

(5)

trong so cac kim loai cho difdi day: Mg = 24, Ca = 40, Cu = 64, Ba = 137.

Theo phan iJng (5) va theo dieu kien cho ta c6 ty 1?:

Bai giai

- ^
= - - ^ = — = > r u t r a X = 18,66n
14,93
8^96
0,4

Phan li-ng hoa tan:

22,4

RCO3

+ 2HC1

>

RCI2 + H2O +

CO21

Khi khi thoat ra dung 1,904 lit thi:

Nghicm thich hdp: n = 3, X = 56 do la Fe
nco2 =nRCi2 =nRC03

Bai 11.
1. Hoa tan m, gam A l va m2 gam Zn bang dung dich HCl diT thu diTdc nhffng the

ThiTc tc so mol R C O 3 Mn hdn 0,085 mol, do do ta c6:

tich nhu^ nhau H2. Tinh ti le mi : m2.

—^>0,085
R + 60

2. Hoa tan hon hdp Al - Cu b^ng dung dich HCl cho tcfi khi khi ngiTng thoat ra ihav
con lai chat ran X. Lay a gam chat r^n X nung trong khong khi tdi phan iJng

Vay do la muoi canxi cacbonat CaC03.

> 2 A I C I 3 + 3H21
> ZnClj + H21

Zn + 2HC1

tu'cR<57,6

Ta CO 30 < R < 57,6.

Bai giai
2A1 + 6HC1

,.,

Mat khac khoi Itfdng muoi (R + 71) x 0,085 > 8,585 gam ttfc R > 30.

hoan loan thu dU'dc 1,36a gam oxit. Hoi Al bi hoa tan hoan toan hay khong?

1. Cac phan iJng:

= ^ ^ = 0,085mol

(0

^ai 2. Nung 25,28 gam hon hdp FeC03 va Fe.Oy trong khong khi ( O 2 dif) tdi

(2)

phan UTng hoan toan thu diTdc khi san pham A va 22,4 gam chat rj(n. Cho khi
A ha'p thu he't vao 400ml dung dich Ba(0H)2 0,I5M thay c6 7,88 gam ket tua

Theo cac phan i?ng (1,2), de c6 1 mol H2 bay ra can - mol A l hoac 1 m<''

tao thanh.
Zn. Vay ti le khoi liTdng:

^1
m-i

Vi^

_ 3
65

140

> 2AICI3 +

' ^

2- Tim C T P T cua Fe,0y

65

^

2. Cac phan uTng xay ra khi hoa tan va nung:
2A1 + 6HC1

cdc P T P U xay ra.

Bai giai

' Cac phu'dng trinh phan uTng xay ra:
3H21

.

p i i a ;v

,

- K a ^.

(1)
141


2FeC03 +
2Fe,Oy
•>¥

C02+

0,502

+

——> F e z O , + 2 C 0 2 t

> xFezOj

O2

>BaC03l+H20

Ba(OH)2

BaC03 + CO2 + H2O

> Ba(HC03)2

+ 2mHCl
> 2YCL + mH20
^heo phan tog ( l ) , t a c 6 tile:
X _ ^ n L 3 4 4 ^ ^ = > X = 32,5n

(D
(2)
(3)

3,9

'

pir(4)

Va npeco3 = n c o 2 =0'08mol
(*>€|$J
Do do khoi lircfng FeC03 = 0,08 x 116 = 9,28 gam.
K h o l li/dng O2 tham gia cac phan tfttg (1, 2) b^ng
25,28 - (22,4 + 0,08 x 44) = 0,64 gam
Kho'i liTdng O2 lham gia phan iJng (1) bhng
"FeCO^

4

'
'!

0,08

X 32 =

4

2 22,4

0,12

'

i^..-c,

.^

^^1,0 n = 1 X = 32,5
loai
n = 2 = > X = 65
XlaZn '
_ '
,
.
.
2Y
+

16m
2m
^,
Theo phan iirng (2), ta co ti le: — —
= ^ " ^ g ^ => Y = 18,67m

(4)

7 88
2. Tinh:
nQ^^Oj, = ^ ^ = 0,04mol; nBa(OH)2 = 0,4x0,15 = 0,06mol
Trirdng hrtp CO2 diT: CO xay ra phan tfng (4)
Cich 1: Tong so mol CO2 = 0,06 + (0,06-0,04) = 0,08mol
P^O)

(2)
..v.u......

Y20m

X 32 = 0,64gam

Do do O2 k h o n g tham gia p h a n uTng (2) chuTng to FCjOy p h a i la FeaOs vi FeO
ho5c Fe304 thi bj oxi h o a thanh Fe203.
Cach 2: Khoi liTdng Fe203 d phan uTng (1) bing ^ y ^ x 160 = 6,4gam
Do do khoi liTcJng F e 2 0 3 d phan Ung (2) b^ng 22,4 - 6,4 = 16 gam
Khoi liTdng FeC03 ban dau bang 9,28 g do do kho'i liTcJng Fe^Oy ban dau
bang 25,28 - 9,28 = 16 gam. Dieu do chufng to Fe^Oy ban dau chinh la Fe203.
Trirdng hdp CO2 thieu: khong c6 phan iJng (4)
Luc do n^oj = i^BaC03 ~ 0'04r"ol

TriTcfng hdp nay loai vi lufdng Fe^Oy Idn hdn Fe203 d phan tfng (2).
Bai 3. De hoa tan 3,9 gam kim loai X can diing V ml dung dich HCl va c6 LB^''
• lit H2 bay ra (d dktc). Mat khac de hoa tan 3,2 gam oxit cua kim loai Y cung
can diing V ml dung dich HCl d Iren. Hoi X, Y la cac kim loai gi?
'
Baigiai
Cac phan iJng hoa tan kim loai X hoa tri n va oxit kim loai Y hoa tri m.
>:
X + nHCl
> XC1„ + ^ H21
(1)
142

fjghiem thich hdp: m = 3, Y = 56 (Y la Fe). Cdc kim loai X, Y la Zn va Fe.
Bai 4.
J C6 the coi sat tijf oxit la hon hdp c6 cilng so' mol ciia FeO va Fe203 dufdc
khong,tai sao?
2 De san xuat mot lifdng gang nhU' nhau ngifcfi ta da diing mi tan quSng
hematit chuTa 60% FczO^ va m2 ta'n quSng manhetit chiJa 69,6% Fe304. Tinh
ti le m, : m2
Baigiai
,
1. Ve hinh thtfc cong thtfc Fe304 c6 the viet thanh FeO.Fe203, nhiftig kh6ng the
xem Fe304 nhiT hon hdp 2 oxit FeO va Fe203 cilng so' mol. Vi hon hdp FeO
va Fe203 khong c6 cac tinh chat ciia Fe304 nhiT khong c6 til tinh...
2. Vi san xua't mot liTcJng gang nhuT nhau nen li/dng Fe cung nhu* nhau. Gia suf de
c6 1 tan Fe thi can bao nhieu tan quSng?
Theo cong thuTc Fe203 va theo thanh phan quang hematit, de c6 1 tan Fe can:
J Y ^ x ^ = 2,38 tafn quSng hematit

j^,,i„^M riftii ?irf; n

TirOng tif doi vdi quang manhetit Fe304:
.j ,55
232 100 , ^„
,
X
= 1,98 tan quSng manhetit
168 69,6
^
*,
Vay ti le m, : m2 = 2,38 : 1,98 = 1,2
,a, ^< ,
' .>
B&iS.
^- Hoa tan hoan toan 6,66 gam tinh th^ Al2(S04)3. nHjO v^o nvCdc thanh dung
dich A. Lay 1/10 dung dich A cho tac dung vdi dung dich BaCb dU" thi thu diTcJc
0.699 gam ket tua. Xac dinh cong thtfc cua tinh the muoi sunfat cua nhom.
^- Hoa tan 24,4 gam B a C l 2 . xHjO vao 175,6 gam niTdtc thu dircfc dung dich 10,4%.
• C6 can ra't txi tuf 200ml dung dich CUSO4 0,2M thu diTdc 10 gam tinh thi
CUSO4. P H 2 O . Tinh p.
143


B a i giai

1. Hoa tan tinh the Al2(S04)3. nH20 v^o nvCdc ta thu diTdc dung dich Al2(S0<,)
Khi cho BaCl2 vao xay ra phan v?ng:
Al2(S04)3 + 3BaCl2
> 3BaS04>|. + 2AICI3

(1)
Theo phan uTng (1) ta thay cvt 1 mol tinh the, tvtc (342 + 18n) gam tinh the th^
dtfdc: 3 X 233 = 699 gam ket tua BaS04. Theo dieu kien bai toan cho thv
= 0,666 g tinh the thu diTdc 0,699 gam ket tua, nen ta c6 ti le:
342.18n^_699_^^P^^
0,666
0,699
rT.^.^.
.
666-342
Tilfdorutra: n = 18
18
Vay c6ng thuTc tinh the nh6m sunfat ngam nifdc la Al2(S04)3. I8H2O
2. KhoiliTcfng BaCl2 nguyen chat
= (24,4 + 1 7 5 , 6 ) X24,4-20,8
^ = 20,8gam, tfng vdi — = O,lmol
'
Dod6 n H 2 o = 0 . 1 x x =100
— ^ - r ^ : — ^ = 0,2mol
208
>i
18
Rut ra X = 2. Cong thtfc ciia tinh the 1^ BaCl2. 2H2O
3. Tinh ncuso4 = 0 , 2 x 0 , 2 = 0,04mol
>^ U H J O = nr^A
Do d6
0.04p = 10-0,04x160 = 0,2mol,
18
Rut ra p = 5. Cong thuTc ciaa tinh the la CUSO4. SHjO
C6 the tinh M i„h ,hd =

= 250. Do do p =

=5

Bai 6. Hoa tan hoan toan 14,2 gam hon hdp A gom MgCOs va mot muoi cacbonat
kim loai R b^ng mot liTOng vfiTa du dung dich HCI 7,3% thu diTdc dung dich D
va 3,36 lit CO2 (6 dktc). Nong do MgClj trong dung djch D la 6,028%.
1. Xac dinh kim loai R va tinh % khoi liTdng moi chat trong A.
2. Cho dung dich NaOH duT vao D roi lay ket tua nung ngoai khong khi cte"
khoi liTdng khong doi thi thu dUdc bao nhieu gam cha't ran?
B a i giai

1. Cac phan i^ng xay ra:

MgCOs + 2HC1
> MgClz + H2O + CO21
R2(C03)„ + 2nHCI
> 2RCI„ + nC021 + nHzO
144

(1)
(2)

Tinh: UHCI = 2. n^o^ = 2 x

= 0,3mol

x x ^ , 7,3%
^ = 0,3x36,5x100

g^hoi Wdng dung dich. HCI
—
= 150gam
KhS'i liTdng dung dich D b^ng
jnp =
mHci + mco2 = 14,2 + 150 - 0,15 x 44 = 157,6 gam.
fflA

+

S6'molC02C[phanu'ng(l)= n^gco.^ =nMgCi2 " ^ ^ ^ j ^ ^ ^ J ^ =
Do do khoi Itfdng MgCOs = 0,1 x 84 = 8,4 gam
S6' mol CO2 d phan uTng (2) = 0,15 - 0,1 = 0,05 mol, do do so mol
0,05 (2R + 60n) = 14,2 - 8,4 - 5,8gam.
R2(C03)n = 0,05
n
n

,0?, • •

Rdt ra R = 28n. Chi c6 n = 2 va R = 56 la dung (Fe).
Vay %MgC03 = ^ ' ^14,2
^ ^ ^ ^ = 59,155%; %FeC03 = 100 - 59,155 = 40,845%.
MgCl2 + H20 + C 0 2 t
(1)
MgCOj + 2HC1
2.
•^FeCl2 + H20 + C 0 2 t
FeCOj + 2HC1
(2) (

Mg(0H)2>l,
+2NaCl
MgCl2 + 2NaOH
(3)
->Fe(0H)2>l,
+2NaCl
FeCb + 2NaOH
(4)
Mg(0H)2
-> MgO + H s O t
(5)
4Fe(OH)2 + O2
(6)
2Fe203 + 4 H 2 0 t
Chat ran gom MgO va Fe203, khoi lifdng chat r^n b^ng:
0,1x40 + — x i x 160 = 8gam
^^TM^ 116_^ .
KL Fe203

Bai 7. Oxi hoa hoan toan p gam kim loai X thu dtfdc l,889p gam oxit, ho^ tan
mu6'i cacbonat kim loai Y (hoa tri II) bang mot liJdng vijfa du dung dich
H2SO4 9,8% thu diTdc dung dich muoi sunfat 14,18%.
!• Hoi X, Y la kim loai gi?
^ «,,1 •
2- A \n hdp kim loai X, Y.
^) Hoa tan hoan toan 3,61 gam A bang dung dich HCI thu diTdc 2,128 lit H2
(dktc). Tinh so' mol moi kim loai trong 3,61 gam A.
Cho 3,61 gam A vao 200 ml dung dich hon hdp AgN03 va Cu(N03)2. Sau khi
kS't tua cac phan tfng thu dtfdc chat ran B nSng 8,12 gam chiJa 3 kim loai.
Hoa tan B bang dung dich HCI diTthay bay ra 0,672 lit H2 (dktc).

Tinh nong dp mol cua AgNOs va Cu(N03)2 trong dung dich ban dau.
145


Bki giai
TimX:

1.

Giai he phiTdng trinh ta c6 x = 0,03 mol; y = 0,05 mol.

X20„

2X+ - O 2
2

(1)

Theo (1) ta C O ti le khoi luTdng cua kim loai

oxit:

2M,

Y C O 3 + H2SO4

=^

= 0.15M va

(2)

> Y S O 4 + H 2 O + CO21

(Mv+96)xl00
( M y + 6 0 ) + 1000-44

•

'

-

difdc mot dung dich chi chiJa mot muoi tan. Hay xac dinh kim loai M va so
mol muoi nitrat cua no trong dung dich.
2. Hoa tan hoan toan 9,9 gam hon hpp kim loai A hoa trj n va kim loai B hoa tri
m bang dung dich HNO3 loang thu diTpc dung dich X va 6,72 lit khi duy nhat
NO (dktc). Tinh tong khoi lufPng muoi nitrat c6 trong dung dich X.
Bai giai

2A1 + 6HC1

> 2 A I C I 3 + 3H21

(3)

Fe + 2HC1

> FeCh + H21

(4)

Gpi a, b la so mol A l , Fe trong 3,61gam A, ta c6

OH? +

.>i«i»*f

M + nAgN03

> M(N03)„ + n A g i

(1)

2M + nCu(N03)2

> 2M(N03)n + nCu 4

(2)

ncu(N03)2

Tinh:

= l a + b = ^ ^ = 0,095
2
22,4

?« o; lii* u

Gpi n la hoa tri cua kim loai, ta c6 cac phan i?ng:

rODo^

27a + 56b = 3,61

=nAgN03

= 0 . 5 x 0 , 4 = 0,2mol^;^^_,.^ ,^

Vay chat ran chiJa 3 kim loai chiJng to dir M va cac phan iJng ( 1 , 2) xay ra
ho^n toan. Theo khoi lu'png tang d cac phan iJng ( 1 , 2) ta cd phu^dng trinh ve

Giai ra ta difdc a = 0,03 mol va b = 0,05 mol.
b) V i A l hoat dpng hdn Fe va vi AgNOs tinh oxi hod manh hdn Cu(N03)2 n c n
phan tfug so'mpt la:

tong kho'i lu'dng tang sau day:
(108

- A

Al + 3AgN03

= 0,25M

ij-ng xay ra hoan toan, ta Ipc difdc (a + 27,2) gam chat r^n gom 3 kim loai va

2. a) Cac phan ufng hoa tan kim loai trong HCl:

"2

=^

hpp gom Cu(N03)2 va AgNOj deu c6 nong dp 0,4 mol/1. Sau khi cdc phan

Giai ra My = 56, do la Fe.

n„

€^(^03)2

J Cho a gam bpt kim loai M c6 hoa tri khong doi vao 500 ml dung dich hon

Theo (2) ciJ 1 mol Y C O 3 can 98g H2SO4 tiJc can ^ ^ ^ ^ = lOOOg dung dich
9,8
H2SO4 va lam thoat ra 44 gam C O 2 . Do do theo cong thtfc nong dp C% ta CO
14 18=

Q , N O ,

l,889p

Rut ra Mx = 9n, chi c6 n = 3; Mx = 27 (Al) la diing.
T i m Y:

Va

2 M x + 16n

> Al(N03)3 + 3 A g i

M
n

)

X

2M
0,2 + (64 - — )
n

X

0,2 = 27,2

'R'V

n •^'>

(5)
Giai ra ta cd: —
n

Sau phan uTng (5) neu A l het (hoac viTa du) thi nAg = 0,03 x 3 = 0,09

= 36 hay M = 12n.

-^'^

M

tifc khoi lirpng Ag = 0,09 x 108 = 9,72 gam trai vdi dieu kien bai toan cho,
Nghiem thich hdp n = 2 => M = 24 Mg. Vay M la kim loai Mg. ^'f'^^

^ do do sau phan iJng (5) se xay ra cac phan uTng:
4 ,

2A1 + 3Cu(N03)2

> 2A1(N03)3 + 3Cu >!'

(6)

Fe + Cu(N03)2

> Fe(N03)2 + Cu J'

(7)

> FeClj + H21

(8)

Phan u-ng hoa tan B: Fe + 2HC1

Gpi X , y la so mol AgN03, Cu(N03)2 trong dung dich dau ta c6 cac phu'cfng trinh:
Khoi lirpng B = 108x + 64y + 56 X

-

0 ^
oi672
nH2=-:::r^=0,03
22.4

.:

So'mol A l tac dung vdi AgN03, Cu(N03)2
2r

n A , = 0,03= - + - y - ( 0 , 0 5 - 0 , 0 3 ) '

=8,12

So mol muoi Mg(N03)2 bang - so mol AgNOj + so mol Cu(N03)2
0 2
= — + 0,2 = 0,3mol
^- Cac phan iJng hoa tan:
3A + 4nHN03

> 3A(N03)n + nNO t + 2nH20

(1)

3B + 4mHN03
> 3B(N03)„ + mNO t + 2mH20
(2)
^^ng khoi liTdng muoi nitrat bang tong khoi lu'dng kim loai + tong khoi lifdng
nitrat.

6

n
146

•

72

_ = 3 X nNo = 3 x - ^ — = 0,9mol
NO3
22,4

^•fysif''-''.

i'J. •••

147


Cfy TNHH MTV DWH

Khoi iMdng goc: - N O 3 = 0,9 x 62 = 55,8 gam.
Vay long khoi liTOng mu6'i = 9,9 + 55,8 = 65,7 gam.
Bai 9. Cho cac kim loai X hoa trj I , Y hoa tri I I va Z hoa tri 11, K L N T tiTdng iJng
la Mx, M Y va Mz.. Nhiing 2 thanh kim loai Z c6 cilng khoi liTcJng v^o 2 dung
dich muoi nitrat cua X va Y. Ngifdi ta nhan thay khi so mol muoi nitrat cua 2
f^: trong 2 dung dich hlng nhau thi khoi liTdng thanh thtf nhat tang a% c6n thatih
jf^, thtf 2 tang b%. Gia suf tat ca kim loai X, Y bam het vao thanh kim loai Z
•g^ Lap bieu thtfc tinh Mz theo Mx, My, a, b.

ax X I = nH2 =

= 0,105mol

Chia v e t h e o v d ' l v ^ I I t a c d :
^

— =

Khang

Viit

(H) ^
/

2

j ' T i
. * ^ t o Sf.i''.; .-fr;,:^ tji-54

t>

IChi n = 1 => - = - tifc M3O2 luc d6 theo (I) hoac (II) ta c6 a = 0,07 mol v^
8 12

laPT cua

Baigiai

Cdc phifdng trinh phan iJng:

M3O2 bing

= 116 ttfc 3M + 32 = 116 hay M = 28 loai

,

Z + 3XNO3

> Z(N03)3 + 3 X i

(1)

3
Khi n = 2 => - = - ttfc M3O4 luc dd a = 0,035 mol v£k KLPT cua M3O4 bling

2Z + 3Y(N03)2

> 2Z(N03)3 + 3 Y i

(2)

^

X

= 232 ttfc 3M + 32 = 232 hay M = 56, d6 la Fe va cong thtfc oxit la

Gia suf so mol muoi Z(N03)3 tao th^nh trong 2 dung dich, ta c6:

Fe304 (s^t tuf oxit)
3nMx-nMz = (3Mx-Mz)n = p x - ^
100

(I)

3n
f
3
^
h
• M Y - n M 2 = — M y - M z n = px
100
2
\

(ID

Trong do p la khoi liTdng ban dau cua 2 thanh kim loai Z.

Bai 11. Hoa tan 3,2 gam oxit M20„, trong mot ItfcJng vifa du dung dich H2SO4
10% thu duTcfc dung dich muoi sunfat 12,9%. C6 can dung dich muoi roi lam
lanh dung dich thay thodt ra 7,868 gam tinh the muoi sunfat vdi hieu suat
mu6'i ket tinh la 70%. Hay xdc dinh cong thtfc cua tinh the.
Bai giai
Theo phan iJug M20n + nH2S04

Chia (I), (II) ve theo vS", ta c6: |M2LZMz_ = i
^My-Mz
^

C % = 12,9 =

2M + 96n
2 M + 16n + 980n

R i J t r a M = 18,66n.
Nghiem thich hdp: n = 3 => M = 56 (Fe). Vay cong thtfc cua oxit Ik FCiOj"Fe203 = np,2(so4)3 = ^

= 0,02mol.

VI hipu suSft ket tinh chi 70% nen

•BaC03 4 . + H 2 0

(2)

M + nHCl

>MCl„ + n/2H2t

(3)

Goi a la so mol oxit, theo cdc phan urng ( 1 , 2, 3) ta c6:
ay = nco2 = ne.coj =

= 0,14mol

2

mol mu6i b i n g :

'^

0,02 X 0,70 = 0,014 mol

(D

C02 + Ba(OH)2

= 980n

(gam) dung dich H2SO4 10% va tao ra 1 mol muoi sunfat, n6n theo c6ng thtfc
n6ng dp C% ta c6.

C^c phan iJng xay ra:
_ ! l - > x M + yC02

(1)

Ci? 1 mol M20n c i n 98n gam H2SO4 nguyen cha't ttfc c i n

r^-. A , ^ J./,
6bMx-3aMY
Giai ra ta diTdc Mz =
2(b-a)
Bai 10. Cho dong khi CO di qua 6'ng stf nung nong, difng 8,12 gam m6t oxit cua
kim loai M khuT het o x i thslnh kim loai. Khi di ra khoi ong stf cho Ipi tir t^^ qua
binh difng liTcfng dir dung dich Ba(0H)2 thS'y tao th^nh 27,58 gam ket tua
tr^ng. Lay kim loai thu diTdc hoa tan hoan toin b^ng dung dich HCl thay bay

ra 2,352 lit H2 (d dktc). Hay x i c djnh kim loai M v^ cong thtfc ciia oxit.
Bk\i

MxOy + y C O

> M2(S04)„ + nH20

c6ng thtfc ciia muoi 1^: Fe2(S04)3. XH2O, ta c6:
(400 + 18x) X 0,014 = 7,868. Giii ra x = 9.
^.

c6ng thu-c cua muoi la Fe2(S04)3. 9 H 2 O .

'

"

'

' ^2. Hoa tan hoan toan a gam kim loai R c6 hoa tri khong ddi n v£lo b gam
(D

j^""g dich HCl diTdc dung dich D. Them 240 gam dung dich NaHCOa 7% vao
^hi vtra du tac dung het vdi HCl dU thu difdc dung dich E trong do nong do

148


J 50 sanh hoa tri cua R trong 2 muo'i clorua va nitrat.

% cua NaCl la 2,5% va cua muoi RC1„ la 8,12%. Them tiep liTdng dir du^
dich N a O H vao E, sau d6 loc lay ke't tua roi nung den khoi ItfOng khong
thi thu diTdc 16 gam cha't r^n.
1. V i e t cac PTPLf xay ra.

2 R la kim loai gi? Biet khoi Itfdng muoi nitrat tao thanh gap 1,905 Ian khoi
liTdng muo'i clorua.

>

Bai giai

2. H o i R la k i m loai gi?
Bai giai
1. Goi hod tri khong ddi cua R la n, ta c6 cac phan iJng:
>RC1„+-H2t

(1)

2

i

2. Theo (2)

KUij

lo-i

2R(OH)„ — ^

RjOn + nHaO

nNaHco3

=

(2)
(3)
(4)

jourji

>RCl„+|H2t

(1)

3R + 4mHN03

> 3R(N03)„, + mNO t + 2mH20

(2)

) >•

.

Theo phan tfng (1)

.

Theophaniirng(2)nNo=-a
3
Rutra m = l,5n

' ^Ifff

240 X 7
= "NaCi

R + nHCl

Goi a la so' mol cua R ta c6:

> NaCI + H2O + CO21
)• R(OH)„ + nNaCl

HCld^ + NaHCOj
RCln + nNaOH

'

j Vi khong biet kim loai R chi c6 1 hod tri duy nhat nen ta goi n va m la hoa tri
' c^a R khi hoa tan trong HCl va trong HNO3. Cac PTPLf: j ^ j ^ , q ^ .j,.,

3. Tinh C% cua dung dich HCl da dilng.

R + nHCl

-••ive:,

„ . = 0,2mol
100x84

^ ,_

mm:-

1
^ - U T T
0,2x58,5x100
^
Khoi liTdng dung dich E =
—
= 468,Ogam

=^a

'
•uiifc..o;.u^'
'+x-ii;*

x,i,,ftO,0.

sOrty .'si:

; ih 1

2. V i so mol 2 muoi nhiT nhau (deu bing a) nen ta co:

nii. •

j^w'y,^

j j '^i-^^j.' ^1'''

M + 62m = 1,905 ( M + 35,5n), trong do M la KLNT cua R.

„. . ^ , ,

Thay m = 1,5n ta rut ra M = 28n
Kho'i lifdng RC1„ = "^^^^^^'^^ = 38gam
Theo cic phiTdng trinh (3, 4) ta c6 ti le:

Chi CO n = 2 ; M = 56 la dung. Vay R la Fe.

38

mot cho cac coc diTng lifdng dtf dung dich CUSO4, sau khi phan iJng hoan

16

^

R u t r a M = 12n
Chi CO n = 2, M = 24 (Mg) la diing.
3. Xac dinh C% cua HCl.
Theo (1, 2, 4): nMg = nwgo =

Bai 14. Chia 8,64 gam hon hdp Fe, FeO va Fe203 thanh 2 phan b i n g nhau. Phan

^^JlLlil = 2R + 16n

•

loan tha'y trong coc co 4,4 gam cha't ran. Hoa tan het phan hai blng dung

u^,.

dich HNO3 loang, thu di/dc dung dich A va 0,448 lit khi NO duy nha't (dktc).
Co can tii tiS dung dich A thu diTdc 24,24 gam mot muo'i sat duy nhat B.
1. Tinh % khoi luTdng moi chat trong hon hdp ban dau.

= 0,4mol

HCI - 0,4 X 2 = 8,8 +

1HCI

HCI - 0,2 X 44 + 240
CO2

io mol H(
Rut ra mdd HCI = 228 gam. Tong so
HCl = 0,4 x 2 + 0,2 = 1,0 mol.
1x36,5x100
C%Hci =

•

= 16%

228
Bai 13. Hoa tan hoan to^n p gam kim loai R bing dung dich HCl thu diTcJc V li'
H2 (dktc). Mat khac hoa tan hoan toan p gam kim loai R bang dung di'^''
HNO3 loang thu diTdc V lit khi NO duy nhat.
150

iO-!
Bai giai

" H J = "Mg = 0,4mol; Uco^ - nN.ci = 0,2mol
Mat k h i c mjd E = 468 = 8,8 +

•

2. xac dinh CTPT cua muoi B.

Do d6 khoi liTdng Mg 1^ a = 0,4 x 24 = 9,6 gam.
Khoi Itfdng dung dich D = 9,6 +

f , i ^ « l ^ T J ^-

I

cac phiWng trinh phan liTng:

"^^'l'

Fe + CUSO4

> FeS04 + Cui

FeO + CUSO4

> khong xay ra

Fe203 + CUSO4 '

> khong xay ra

Fe + 4HNO3

> Fe(N03)3 + NO t + 2H2O

(2)

3FeO + IOHNO3

> 3Fe(N03)3 + N O t + 5H2O

(3)

Fe203 + 6HNO3

> 2Fe(N03)3 + 3H2O

(4)

^^^^^^

*

(1)

Goi X , y, z la so mol cua Fe, FeO, Fe203 ta c6 cac phiTdng trinh:
151