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Lời giải đề thi học sinh giỏi Hóa học lớp 9 phần 1

DE

SO 1

DE THI HOC S1I\IH GiOl HOA HOC LOfP 9, TP. HQ CHI MINH NAM HOC 1998
Cdu

I.

Cdu

I I Viet phuang

Cdu

Viet 3 phuang

trlnli

khdc


trinh phdn

> Fe(0H)2

FeCh

> Fe(0H)3

mang

(NHJ.SO,;

dugc

NaoCOs;

de dicu

die muoi

ling de bieu dlen

FeCls

III. Co 6 ong nghiem

NaOH;

nhau

chiiSi

ZnCl2.
bien

^ FeSO,

PbiNO^)^;

so ndo dUng chat ndo? Viet phdn


CaCl..

icng minh

Fe(N0s)2

> Fe.

so tii 1 den 6 chica

Ba(N03)2;

hoa sau:
>

> FesOs

ddnh

- 1999

cdc dung

dicli:

Hay cho bict

6'ng

hga. Biet

rdng:

a) Dung

dich

(2) cho ket tila trdng

vai cdc dung

dich

(1), (3), (4). ''•

b) Dung

dich

(5) cho ket tua trdng

vai cdc dung

dich

(1), (3), (4).

c) Dung

dich

(2) khong

tgo ket tila vai dung

d) Dung

dich

(1) khong

tgo ket tua vai cdc dung

dich

e) Dung

dich

(G) khong

phdn

(5).

f) Dung

dich

(5) bi trung

i(ng vai dung

hoa bdi dung

dich

(5).

dich

(3), (4).

dich HCl.

g) Dung dich (3) tgo ket tua trdng vai HCl, khi dun nong ket tua nay se tan.

»

CduIV.
a) Nong

Cau



do dung

Tinh

dich

bdo hoa KCl d 40°C la

do tan cua dung

) Xdc dinh
AgNOs

lugng

dich KCl a ciing

AgNOs

tdch

28,57%.

nhiet

do.

ra khi Idm Ignh

bdo hoa a 60°C xuong

10°C. Cho biet

2500

gam dung

dich

do tan cua AgNOs

o

60°C Id 525 gam, a 10°C Id 170 gam.
V. (A) la dung
Trgn
Lay
them

dich H2SO4;

(B) la dung

0,3 lit (B) vai 0,2 lit (A) dugc
20 nd (C), them
tic tit dung

dich

dich

NaOH.

0,5 lit (C).

mot it quy tim vdo thdy
HCl 0,05M

c6 mdu xanli.

tai khi quy tim ddi thdnh

Sau do
mdu tim

thdy het 40 ??il axit.
Trgn

0,2 lit (B) vai 0,3 lit (A) dugc

Lay

20 ml dung

Sau

do them

dich

(D), them

tii tii: dung

dich

Tim

dich

do mol/l dung

I r(i n i A i n c T U I u n r C I M M n i f i i H H A H f i n Q

mot it quy tim vdo thdy c6 mdu do.

dich NaOH

tim thdy het 80 ml dung
nong

0,5 lit (D).

0,1M tai khi quy ddi thdnh

NaOH.
(A) vd (B).

mdu


Cdu

VI. Xdc dinh

cong

thiJtc ciia hai oxit sdt A vd B, biet



23,2 gam (A) tan vica dil trong



32 gam (B) khi k/iii bdng Ho tqo thdnh

r&iig:

' ^ ) ^ 60"C, t r o n g 525 + 100 b a n g 625 g a m d u n g d i c h c6 525 g a m AgNOg
v a 100 g a m H2O. T r o n g 2500 g a m d u n g d i c h c6 x g a m AgNOa v a

0,8 lit HCl IM.
sdt vd 10,8 gam IhO.

y g a m nxidc:

L6\I

Cdu I.

Cdu 11.

* Zn + CI2

ZnCla

>

.

Z n O + 2HC1



Z n S 0 4 + BaCl2

X

Fe + 2IIC1

+ BaS04i

> FeCl2 + H 2 t

FeCl2 + 2 N a O H

> Fe(0H)2^ + 2 N a C l
(trdng

Fe(0H)2 + H2SO4

2Fe + 3CI2

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

FeCla + 3NaOH

> FeCOIDai + 3NaCl
(ndu do) •

= 2100

(gam)

400 g a m ni/dfc.
CJ 10°C, CLf 100 g a m ntfcJc h 6 a t a n 170 g a m AgNOa
400 g a m niJcJc h o a t a n z g a m AgNOa
z =

Do do khoi

400x170
100

liTgng AgNOa

= 680 (gam) AgNOa.

ket t i n h k h i lam lanh:

2100 - 680 = 1420 (gam)
Cdu V.

Phan ufng:

— > Fe203 + 3H2O

Fe203 + 3 C 0

625

& 60°C t r o n g 2500 g a m d u n g d i c h c6 2 1 0 0 g a m A g N O s v a

xanh)

> 2FeCl3

2Fe(OH)3

Vay

> FeS04 + 2 H 2 O

FeS04 + Ba(N03)2

2500 X 525

y = 2500 - 2100 = 4 0 0 g (nxidc)

> ZnCl2 + H2O
> ZnCh

=

H2SO4 + 2 N a O H

> Na2S04 + 2 H 2 O

L a n thuf n h a t q u y t i m c6 m a u x a n h chufng t o diT N a O H . T h e m a x i t

— > 2 F e + 3CO2T

Cdu I I I . Theo cac duf k i e n de b a i neu r a , cac lo d i i n g cac h o a c h a t sau:

H C l de t r u n g h o a N a O H dif:
HCl + NaOH

f ) => L o so 5 d i i n g : Na2C03

> N a C l + H2O

(2)

g)

L o so 3 d i i n g : Pb(N03)2

L a n 2 l a m quy t i m c6 m a u do chufng t o H2SO4 dir.

d)

L o so 1 chufa: Ba(N03)2, lo 4 chuTa CaCl2.

T h e m N a O H de t r u n g h 5 a H2SO4 d i / .
2 N a O H + H2SO4

a ) => L o 2 chijfa: (NH4)2S04.

(NH4)2S04 + Ba(N03)2

> B a S 0 4 i + 2NH4NO3

(NH4)2S04 + Pb(N03)2

> P b S 0 4 i + 2NH4NO3

(NH4)2S04 + CaCl2

(3)

Theo cac p h a n ufng (1), (2), (3) t a c6 phiiofng t r i n h :
0,3y - 2

> C a S 0 4 i + 2NH4CI '
0,3x -

Cac p h a n i l n g c o n l a i hoc s i n h tii vie't.
Cdu rv.

X

0,2x = 0,05

0,2y

0,1

x

0,08

X

2

0,04

2

x

500
20
500
20

= 0,05
= 0,1

(a)
(b)

G i a i h e p h u a n g t r i n h (a), (b) => x = 0 , 7 M v a y = 1,1M.

a ) G o i S l a do t a n cua K C l d 40°C.

Cdu VI.

K h o i lUcfng d u n g d i c h t h u dUgfc: (S + 100) g a m
N o n g do p h a n t r a m cua c h a t t a n t r o n g d u n g d i c h b a o h o a :
S + 100
S + 100 - 3,5S »

> Na2S04 + 2H2O

G o i X , y l a n l u g t l a n o n g do cua H2SO4 v a N a O H .

e) =^ L o 6 chijfa: N a O H .
P h a n irng:

(1)

S + 100

G o i c o n g thufc o x i t s a t l a FexOy.
Fe^Oy + 2 y H C l
(mol)

0,4

0,8

S = 40 (gam).
I fi\l n £ T u i u n r s l u u Rini Hni Hnr. a

^^IfilAlflg THI HOC SINH GIOI
GIO HOA HOC 9

> x FeClayx + y H 2 0

(1)


Ta c6: I I H C I = 0,8 X 1 = 0,8 (mol)

PHAN B. Bai toan

Theo de:

Bai

23,2 = — (56x + 16y)
y

<=>

56x = 42y

<=>

X _

3

fx = 3

y

4

[y = 4

FexOy + yH2

vd K2SO4

dugc trgn Idn theo tl Ic

Hoa tan hon hgp vdo

102 gam nUac thi thu dugc dung dich A. Cho 1664 gam dung dich

BaCL

10% vdo dung dich A. Lgc ket tua, them H2SO4 du vdo tiudc vUa Igc tin
thdy tgo ra 46,6 gam ket tua. Xdc dinh nong do phdn tram cua NosSOj

> xFe + y H 2 0

0,6

(mol)

Co mot hon hgp gom Na2S04

I : 2 ve so mol (1 mol Na2S04 yd.2 mol K2SO4).

Vay oxit sSt (A): Fe304 (sdt tii oxit)


1-

(2)

vd K2SO4 trong dung dich ddu. •
Bai 2.

0,6

Tinh

CM cua dung dich H2SO4

vd NaOH,

biet rdng 10 nd dung

dich H2SO4 tdc dung vUa du vdi 30 ml dung dich NaOH. Neu lay 20 ud
10,8

Ta c6:

18

dung dich H2SO4

= 0,6 (mol)

lugng du ndy tdc dung vita du vdi 10 nd NaOH.

'

TheodI:

^ ( 5 6 x + 16y] = 3 2 «
y ^

x = 2

LdlGIAI

- = -=>•
y 3
[y = 3

PHAN

=> Cong thufc oxit (B): FeaOs-

Cdu

DE SO 2

NaOH + N H 4 C I

2 N a O H + FeCl2

PHAN A. Li thuyet
Chi dugc diing thuoc thii de nhgn biet cdc muoi sau: NH4CI,
MgCl2,

NaCl, AICI3

(Gidi thich vd viet phuang

trinh phdn

ling

2Fe(OH)2 + - O2 + H2O

> Fe2(S04)3

b) MgCOs

> Fe(0H)3

> MgS04
> MgCOs

> FcsOa

> FeCls
> MgCl2
> CO2

> Mg(0H)2
> Ca(HC03)2

III. Chi dugc dung guy tini vd dung dich AgN03

IV. Sdt nguyen chdt trong khong khi thi khong

CO tap chdt de Idu ngdy trong khong
hien tugng
a

3 N a O H + FeCls

>Fe

>FeCl2

phdn biet cdc dung dich: NaOH, NaCl, HCl, H2S,
Cdu

> 2Fe(OH)3

Mau CO ket tija nau do la FeCls:

trinh phdn ling bieu dien cdc bien hoa sau:

> FeClz

Cdu

,

2

II. Viet phuang
a) Fe

> Fe(0H)2i + 2NaCl

trang xanh

FeClo,

neu c6).
Cdu

— > NaCl + N H g t + H^O

Mau tao ket tiia trSng xanh, hoa nau do trong khong k h i la FeC]2:

TP. HO CHJ MINH NAM HOC 1998 - 1S99

FeCls,

Cho dung dich NaOH dii Ian liicft vao cac mau thif t r e n va dun nhe;
Mau thuf nao c6 k h i m i i i khai bay ra la N H 4 C I :

DE THI CHOIV flOl TUYEN KQC SINH GIOIHOA HOC 9, qUAAl TAN BINH

I.

A. Li thuyet

I. - T r i c h m6i lo mot i t lam mt\ thuf.
-

Cdu

cho tdc dung vdi 2,5 gam CaCOs thi axit con du vd

^
> Fe(0H)3i + 3NaCl

nau do

> AgCl.

Mau tao ket tua keo trSng sau d6 tan trong NaOH dif la AICI3:

> MgO

3 N a O H + AICI3

> CaCOs

> A l ( 0 H ) 3 t + 3NaCl

keo trdng

c6 sdn, neu each
A 1 ( 0 H ) 3 + NaOH

H2SO4.

> NaAlOa + 2H2O

Mau tao ket tua trSng la MgCl2:

bi han gl, nhUng sdl

khi Igi bi han gl. Hay gidi

'

thich

2 N a O H + MgCl2

nay.

^ Mg(0H)2 + 2NaCl

Mau khong c6 hien tJong gi la NaCl.
Lfll GIAI Oi THI HOC SINH GIOI HOA HOC 9

Lfll GIAI

T H I unr emu ni/\i un» unr a

^


Cdu

Cciu

II.
a)

2Fe + 6H2SO4 damdac

2Fe(OH)3

-

K h i t r o n g s a t b i I a n t a p c h a t , de l a u n g a y b i b a n g i do x a y r a s u
a n m o n k i m l o a i tufc l a bie'n sat t h a n h h o p c h a t cua sat.

— > FeaOs + 3H2O

Gidi

— > 2Fe + SCOat

l u o n g o x i n e n c h u y e n Fe -> Fe^\

F e + 2HC1

> FeCla + H a t

V a o x i hoa t a n t r o n g niidc theo qua t r i n h : O2 + 2H2O

2FeCl2 + CI2

> 2FeCl3

Sau do Fe^' k e t h o p vdri O H "

2FeCl3 + Fe

> SFeC^

M o t p h a n F e ( 0 H ) 2 b i o x i hoa t a o F e ( 0 H ) 3 (nau

MgCOg + H2SO4 loang

Mg(0H)2

> MgCl2 + B a S 0 4 i
> MgCOIDai + 2 N a C l

—>

M g O + CO2 —
MgCOa

+ CO^^t + HgO

> MgSOi

^

PHAN
Bai

1.

B. Bdi
rr.

-

T a co:

> C a C O a i + NaaCOg + 2H2O

hoac Ca(HC03)2 + C a ( 0 H ) 2

2 C a C 0 3 i + 2H2O

HI.

M a u k h o n g c6 h i e n t i i o n g l a N a C l .



Nhufng m a u l ^ m quy t i m h d a do: I I C l , H2S, H2SO4

Sau do cho dung dich AgNOa Ian li/gt vao cac m a u l a m quy t i m hoa do.
M a u nao tao k e t t u a t r S n g la H C l :
> A g C l i + HNO3


10

<-

X

1664

n

o

/

1^

= 0,8 ( m o l )

> BaS04i + 2NaCl

X

(mol)

2x

->

(1)

X

> B a S 0 4 i + 2KC1

2x

^

(2)

2x

V i k h i t h e m d u n g d i c h H2SO4 vao n L f d c loc l a i t a o k e t t u a n e n
t r o n g niidc loc con dtf BaCla.

(mol)

> B a S 0 4 i + 2HC1

0,2
«o

<= ^

»

X

(3)

0,2

= 0,2 ( m o l )
233

T h e o de: ng^ci, = ^-^ ^n^^l)

ng^^i./d) + "BaCi,/(2) + ^Baa.,im

= ^'^

+ 2x + 0,2 = 0,8 => X = 0,2

«

mdung dich = 0,2

C % N a so

n a o CO ke't t u a den l a H2S:
2 A g N 0 3 + H2S

10

=

K h o i l a g n g d u n g d i c h A l a : nidungdich = m N a . s o ^ + " ^ K , S O , + "^H^O

irdng
MSU

m,,

=1:2

B^so,/,3,





X

M a u thuf n a o l ^ m quy t i m h o a x a n h l a N a O H .

AgNOs + H C l

X

BaCl2 + H2SO4

C h o quy t i m t a m ni/dc I a n \\iat vao cac mau thijf t r e n :



C%

BaCl2 + K2SO4

T r i c h m 6 i lo m o t i t l a m m a u thtf.

-

n3,,,^ =

(mol)

C a C 0 3 i + H2O + C 0 2 t

hoac Ca(HC03)2 + 2 N a O I I



toun

BaCl2 + Na2S04

> Ca(HC03)2
—>

4Fe(OH)3

Cac p h a n ufng x a y r a :

> M g O + COat

Ca(HC03)2

do)

G o i X l a so m o l c i i a Na2S04 => Hj^^g^^ = 2x ( m o l )

MgCOg

2CO2 + C a ( 0 H ) 2

->

xanh)

Do do s a t b i h a n g i c6 m a u n a u do.

"Na^so, • " K ^ S O ,

M g O + H2O

>• 4 0 1 1 "

> F e ( 0 H ) 2 (trdng

4 F e ( 0 H ) 2 + O2 + 2H2O

> Fe(N03)2 + 2 A g C U

MgCl2 + 2 N a O H

-

T r e n be m a t k i m l o a i c6 I d p nUdrc a m , d a h o a t a n m o t

thidi:

FeaOg + SCO

M g S 0 4 + BaCla

Cau

b e n t r o n g k h o n g k h i a n h i e t do t h u 6 n g .

> 2 F e ( O I I ) 3 i + 3Na2S04

FeCla + 2 A g N 0 3
b)

1^02(804)3 + 3SO2T + 6II2O

>

Fe2(S04)3 + 6 N a O H

IV.
- s a t n g u y e n c h a t k h o n g b i b a n g i v i n6 dugc bao ve bori I d p Fe203

^ ^ 2 ^ ° "

> kgiSi

^

^

^

1 74

+ 2HNO3

M d u k h o n g c6 h i e n t u o n g g i l a H2SO4.
i f i l R I 4 I f i F TH] H f i r SlUH f;inr u n A u n r n

=

X 142

Ml GIAI DE THI HOC SINH GIOI HOA HOC 9

^ ^
200
X

+ 0,4
X

x 174

+ 102

= 200

(gam)

100 = 14,2%

04
11


Bai 2.

T a c6:

2.5
= 0,025 (mol)
100

nC a C O ,

4. Tafco
them

P h i i n ufng:
H2SO4 + CaCOa (mol)

0,025

(mol)

a

2NaOH

->

Ong 3: dung dich bac

Na2S04 + 2H2O

20

5. Mot hon hap NaCl vd MgCl2 them nuac vao hon hap ta c6 dung dich A.

= 0,0125 (mol)

Them dung dich AgNOs vao dung dich A, khi phdn iing ket thuc

30 m l d u n g d i c h N a O H chufa y m o l N a O H
2a

30

X

10

Chia B lam 2 phdn: a vd b.
Phdn a: Sau khi c6 can tiep tuc dun nong thl diigc mot hon Jigp khi

= 6a (mol)

C. Cho hon hap khi nay qua binh dUng KOH.

H2SO4 + 2 N a O H
(mol)

Na2S04 + 2H2O

Phdn b: Cho vdo lugng dU dung dich HCl thi thu dugc ket tua trdng D.

0,0125 - > 0,025

Viet cdc phuang
V i p h a n ufng t r u n g hoa n e n 6a = ^'

"2SO4

"

0,01

logi

bo chat ket tua trdng, phdn dung dich con Igi la dung dich D.

T r o n g 10 m l d u n g d i c h N a O H chijfa 2a m o l N a O H

y =

Em hay cho biet

dng ndo con chat ket tua. Gidi thich vd viet phiXang trinh phdn icng.

10 m l d u n g d i c h H2SO4 chtfa x m o l H2SO4
=

nitrat

Sau do cho them axit nitric vao ca ba ong nghiem.

T r o n g 20 m l d u n g d i c h H2SO4 chufa 0,025 m o l H0SO4

X

vao:

Ong 2: dung dicli natri cacbonat

2a

10 X 0,025

ngUdi ta cho

diig l:~dung dich kali cacbonat

-> CaS04 + C O a t + H2O

< - 0,025

H2SO4 +

3 6'ng nghi&m deu dung dung dich bari clorua,

(mol)

6

trinh phdn

dng.

6. Lam the ndo de phdn biet cdc Ig hoa chat diidi day md khong
dung theni hoa chat ndo khdc: MgCl2, H2SO4, NaCl, CuSO^,

= 1,25M

dugc

NaOH.

7. Mot hon hap gom: CuO, FeO, AI2O3. Lam cdch ndo de tdch chiing ra

0,025
^6

2,5

0,03

khoi

M

nhau.

Cdu II. Bai

todn

Cho 26,Ig MnOo tdc dung vdi dung dich HCl c6 20 gam HCl. Cho hct

OE SO 3

khi do qua mot lit dung dich NaOH lodng dU.

DE THI HOC SINH GiOl HOA HOC 9 QUAN 3 TP. HQ CHJ MINH NAM HOC 1998 - 1999

a) Lugng HCl nay c6 du de phdn dng het vdi Mn02

Cdu 1. Li thuyet

b) Tinh nong do mol 11 cua muoi thu dUgc trong phdn I'ing giOa do vd NaOH.

1. Tit 7 Ig hoa chat sau, em c6 the dieu chc nhUng chat khi nuo?
Axit sunfuric;
natri hidroxit; amoni
sunfit; sdt sunfua vd kim logi keni.
2. Ta H2SO4 CO may cdch de dieu che

nitrat;

canxi

cacbonat;

c) Nung qugng pyrit sdt 'de tgo ra SO2. Cho khi SO2 sue vdo dung
natri

chda 2 muoi

khoi

lugng pyrit

dung dich

dich

tren.

Sau do them vdo mot lugng du Ba(N03)2.

CaSOJ

3. Viet cong thUc vd ten ggi 3 muoi diing trong nong nghiep (phdn dam,
phdn Idn vd plidn kali). Hay gidi thich tgi sao ngUdi ta khong trgn tro
bep vdi phdn dam de ban rugng?
19

khong?

can dung.

Tim khoi lugng ket tua va

Biet rdng lugng SO2 tdc dung

vUa du

muoi.

(Cho Mr, = 55; O = 16; H = 1; CI = 35,5; S = 32; Fe = 56; Ba = 137)

^^F:.

,

13


L d l GIAI

Cdu I. Li

dng

thuyet
1000°C

Zn

+

-> CaO + C O a t

FeS + H2SO4

—>



NH4NO3

> Na2S04 + S O s t + H2O

MgCl2 + 2AgN03

Phdn

2AgN03

—>

> CaS04 + 2H2O

Phdn

2 M g O + 4 N 0 2 t + O-^t

—>

b: K h i cho H C l d u vao d u n g d i c h B t h u diJOc k e t t u a . D i e u

H C l + AgNOs

> A g C l i + HNO3
(D)

> CaS04 + C O a t + H2O

Chu y: Con nliieu cdch klidc, xin nJiuang ban doc!
3. C o n g thufc ba m u o i d u n g t r o n g n o n g n g h i e p :

6, T a n h a n t h a y : t r o n g t a t ca cac d u n g d i c h t r e n t h i c h i c6 m o t d u n g
d i c h CO m a u x a n h l a : CUSO4, cac d u n g d i c h con l a i l a k h o n g m a u .



CO(NH2)2 ( 4 6 % n i t a ) : U r e ( p h a n d a m )

T r i c h m o i lo m o t i t l a m m a u thijf.



Ca(H2P04)2: Supe p h o t p h a t k e p ( p h a n I a n )

-



KRO'i.

[Ca(H2P04)2



m a n g t i n h k i e m n e n n g i / d i t a k h o n g t r o n p h a n d a m vcfi t r o bep.

Ong 2:

BaCl2 + K2CO3

M a u thCf cho ke't t i i a m a u x a n h la N a O H :
CUSO4 + 2 N a O H

kali nitrat (phan kali)

hoac K C l : k a h clorua.

4. Ong 1:

Cho d u n g d i c h CUSO4 I a n lifot vao cac m a u thuf t r e n :


2CaS0,

P h a n d a m de b i p h a n h i i y t r o n g m o i t r t f d n g k i e m , v i t r o bep

-

> B a ( N 0 3 ) 2 + C 0 2 t + H2O

BaCIa + Na^CO^

> BaCOgi + 2NaCl

2 H N O 3 + BaCOa

-> Ba(N03)2 + C02t + H2O

> C u ( 0 H ) 2 i + Na2S04

Cac m l u con l a i k h o n g c6 h i e n tu'ong.

D u n g N a O H l a m thuoc thiif, cho I a n liiot vao cac m a u ihxi con l a i :


Mau

t h L f CO

k e t tua mau t r S n g la MgClg:

2 N a O H + MgCla

> B a C O g i + 2KC1

2 H N O 3 + BaCOg

K N O 3 + K N O 2 + H2O

n a y cho t a t h a y r a n g t r o n g d u n g d i c h con A g N O s dif.

Cdch 4: Tac d u n g vcri muo'i cija c a n x i :

hoac supe pho't p h a t dcfn

-02t
2

— > 2Ag + 2 N 0 2 t + 0 2 t

2NO2 + 2 K 0 H

CaS04 + II2O

Cdch 3: Tac d u n g v6i bazcJ C a ( 0 H ) 2 :

H2SO4 + CaC03

NaNOa +

> CaS04 + H a t

H2SO4 + CaO

H2SO4 + Ca(0H)2

—>

2Mg(N03)2

2. D i e u che CaS04 t i f H2SO4:

Cdch 2: Tac d u n g vdri CaO:

Mg(N03)2

a: K h i d u n n o n g d u n g d i c h B :
NaNOs

) N 2 t - H ^ 0 2 t + 2H20
2

Cdch 1: Tac d u n g vdri c a n x i : H2SO4 + Ca

> 2 A g C U + Mg(N03)2

v a A g N O s dif.

> N a O t + 2H2O
>^°°°^

> A g C U + NaNOg

Sau k h i Ipc bo k e t t u a t h i d u n g d i c h (B) g o m : N a N O s ,

> N a N O s + N H a t + H2O

Na2S03 + H2SO4

hoac

5. P h a n ufng: N a C l + A g N 0 3

ZnS04 + S 0 2 t + 2H2O

-> FeS04 + H 2 S t

N a O H + NH4NO3

NH4NO3

mac dti l a a x i t H N O 3 v i t h e t r o n g o n g 3 con c h a t k e t t u a .

> ZnS04 + H 2 t

2H2S04dac

> Ba(N03)2 + 2 A g C l i

Trong 3 ong nghiem t h i ta thay ket tija AgCl k h o n g t a n trong axit

Zn + H2SO4 loang

hoac

BaCla + 2 A g N 0 3
A g C l + HNO3

1. T a d i e u che duac cac c h a t k h i sau:
CaCOa

3:



M a u t h i i tao d u n g d i c h t r o n g suot va toa n h i e t m a n h l a H2SO4.
2 N a O H + H2SO4



> Mg(0H)2 + 2NaCl

> Na2S04 + 2H2O

M a u k h o n g c6 h i e n t u g n g l a N a C l .


b ) T h e tich dung dich thu dUcfc chinh 1^ the tich cua N a O H tufc l a :

7. T a c h C u O , F e O , AI2O3 r a khoi nhau:
CuO
AI2O3

+C0,

^NaAlOg
N a O H da

CuO_

FeO

-> A l ( O H ) ,

+ HC1 du

FeO^

Fe(OH),

cliaii kliong

So' mol cua N a C l trong p h a n ufng (2): 0,1375 (mol)

+ N H . O H dil

FeCl,,

+HC1

[Cu(NH3)J(OH)2

CuCl^

Vduiigdich = 1 (lit)

->A1,03

->CuCL

+NaOH du

C^
-^Cu(OH),

-»CuO

AI2O3 + 2 N a O H

(mol)

> A l ( 0 H ) 3 i + NaHCOg

CuO + 2HC1

— > CUCI2 + H2O

F e O + 2HC1

-> FeCl2 + H2O

> Na2S04 + N a C l + H2O (2)
- > 0,1375

Ba(N03)2 + Na2S04

Txi (3)

^

^ ne^so, = ^i^.,so,

> [Cu(NH3)4](OH)2

(3)

+ 2NaN03

> Ba^O^i

0,1375

(mol)

(i)

0,1375

(mol) 0,1375 < - 0,1375

> Cu(0H)2i + 2 N H 4 C I .

Cu(0H)2 + 4 N H 3

^

2

SO2 + N a C l O + 2 N a O H

-> AI2O3 + 3H2O

CuClz + 2NH4OH —

> 2Fe203 + 8S02t

4FeS2 + I I O 2
> 2NaA102 + H2O

-

= 0,1375M

-^FeO

N a A l O z + CO2 + 2H2O
2A1(0H)3

=

c) P h a n ufng:

P h a n ufng:


=C„

0,1375

= 0.1375 (mol)

mg^so^ = 0,1375 x 233 = 32,04 (gam)

(xanh t i m , tan)
[Cu(NH3)4](OH)2 + 2HC1
CUCI2 + 2 N a O H
Cu(0H)2

Cdu

TO (1), (2)

+ 4 N H 3 T + H2O

> CuCOIDal + 2 N a C l

np^3

^so,
=^
2

_ 0,1375
=
(mol)
"
2
(gam).

-> F e ( 0 H ) 2 i + 2 N a C l

chankhong

DE SO 4

FeO + H2O

I I . Bai todn
T a c6:

:^

0,1375
=> mpes, = - ^ - r — X 120 = 8,25

> CuO + H2O

FeCl2 + 2 N a O H
Fe(0H)2

> CuCh

OE THI HOC SiNH GIOI HOA HOC 9 , CAP TP. HO CHI M I N H N A M HOC 1999 - ZOOO

n MnO.,

26,1
87

= 0,3 (mol): n^j^j =

M n O a + 4HC1

CI2 + 2 N a O H

36,5

0,55 (mol)

Cdu
(1)

0,1375
-> N a C l + N a C l O + H2O

(mol) 0,1375

I . Khi cho kirn

loai

vdo dung

Cdu

I I . Viet cdc phuang

(2)

Fe

0,1375

trinh

> FesO^

TCr p h a n ufng (1) t a c6 t i l e :

1

= 0,3>

Cdu

n HCl
4

0,1375
4

V a y l i r a n g H C l n a y k h o n g d i i de p h a n ufng h e t liicfng M n 0 2 da cho
h a y I I C l p h a n ijfng h e t v a M n 0 2 con d i i .

phdn

mud'i c6 the xdy ra nhitng

phan

I I I . Mot nhd hoa hoc dieu

dang

ben ngoai

vCng theo chuSi bien hoa sau:
FeCh

> Fe(0H)3

> FezOs

FeCl2

> Fe(0H)2

> FeO

<^
^

a ) L i f g n g H C l n a y c6 du de p h a n lifng h e t vcfi M n 0 2 k h o n g ?
n MiiO,

dich

ling gi? Cho vi du minh hoa.

> M n C l a + Cl2t + 2H2O

0,55

(mol)

20

che dugc

3 mdu

(mdu sdc) va da tim dugc phuang

chong.

dng lay cdc mdu

NaOH,

ket qua do dugc ghi tron^^M^:^,^^"^

L f l l R l A i n £ T u i u n n S I U H R i n i HflA H t l R 9

kim log.i cho tdc dung



kirn

loai gidng

phdp

phdn

nhau ve

biet

vdi^axitjm_d^g

nhanh
dich

6SNH THUAN


_

.

J O

17


Thuoc
Axit

HCl

Axit

HNO3

Dung dich

thii

Kim loai I
-

Kim loai II

Kim loai III

K i m loai I :

A g (Bac)

+

+

K i m loai I I :

A l (Nhom)

+

-

+

K i m loai I I I :

Zn (Kem)

-

+

+

NaOH

Trong do ddu (+) de chi trudng hap kim loai hoa tan, ddu (-) chi

Cdu I I I T h e o de b a i t h i :



triiang

ciiu, viet phuang

HCl, HNO3 ddc, AgNOa, KCl,
Vii't cdc phuang

trinh phdn

ling vd gidi



-

dung

dich A. Cho 1664 gam dung dich BaClz 10% vao dung dich A, xudt

hien

ket tua. Loc bo ket tua, them H2SO4 dU vdo nilac Igc thi tlidy tgo ra
ket tua. Xdc dinh nong do phdn

trong dung dich A ban

l a HNO3.
Cu + 4HNO3 dac
-

+) M a u CO k e t t u a m a u x a n h l a K O H .
Cu(N03)2 + 2 K 0 H
-

N e u k i m l o a i cho vao k h a c k i m l o a i t r o n g m u o i t h i x a y r a p h a n

Fe + CUSO4

nhiet la H C l

N e u k i m loai t r u n g v 6 i k i m loai t r o n g muoi t h i xay r a p h a n tfng tiT oxi hoa.
> SFeClg

Cdu I I . P h a n ufng:
3 F e + 202



Fe304 + 8HC1

> Fe304

> K C l + H2O

KOH + HCl
M a u con l a i l a K C l .

Cdu V. G o i k i m l o a i hoa t r i I I l a A va c6 a m o l => o x i t l a : A O .

A O + H2SO4
(mol)
Theode:

a ^

2Fe(OH)3

a

« , , „ , ^ . = ^
" ^ d d Hi,S04

a ( A + 96)100
C%ddASO,

> FeCOIDgi + 3 N a C l
—>

FeCla + N a O H

FezOs + SHsO
> Fe(0H)2i + 2NaCl

TT-^rt
chan khong

> FeO + H2O

> ASO4 + H2O

a

> FeCla + 2FeCl3 + 4H2O

FeCl3 + 3 N a O H



Cho d u n g d i c h K O H vao 2 m a u con l a i , m a u nao c6 p h a n

uTng

> FeS04 + Cu

Fe + 2FeCl3

> 2KNO3 + C u ( 0 H ) 2 i

+) Cac m a u con l a i k h o n g c6 h i e n t u g n g .

K h i cho k i m l o a i vao d u n g d i c h m u o i c6 t h e x a y r a :

Fe(0H)2

> Cu(N03)2 + 2 N 0 2 t + 2II2O

Sau do cho d u n g d i c h viia t h u di/oc a t r e n I a n liicft vao cac m a u con l a i

ddu.

t h e - o x i h o a khijf.

-

> Cu(N03)2 + 2 A g

+) M a u thuf nao vifa t a o d u n g d i c h m a u x a n h va c6 k h i n a u do bay r u

tram cua A^a^SOj vd K2SO4

Ld\I
-

trUcfng

Cho b o t k i m l o a i Cu d t i I a n l u a t vao cac m a u t h t f t r e n :

Cu + 2 A g N 0 3

(Cho Na = 23; S = 32; K = 39; Ba = 137; CI = 35,5).
Cdu I .

A l k h o n g tac d u n g v d i HNO3 v i A l b i t h u d o n g t r o n g m o i

+) M a u t h i i nao d u n g d i c h id k h o n g m a u chuyen sang x a n h la AgNOs.

vd K2SO4 dilgc trgn Idn iheo ti le

I : 2 ve so mol. Hoa tan hon hap vdo 102 gam nUac thi thu dilgc

46,6 gam

v d i NaOPI v i k h o n g p h a i l a k i m

ufng

Cdu IV. T r i c h m o i lo m o t I t l a m m a u thijf.

KOH.

trinh phdn dug xdy ra trong qua trinh nhgn biet.

VI. Co mot hon hap gom Na2S04

hoat

HNO3 dac n g u o i .

day:

Cdu V. Hoa tan oxit cua kim logi hoa tri II trong mot lilgng vita dil dung
dich H2SO4 20% thi thu dicgc dung dich mudi c6 nSng do 22,69c.
Xdc dinh kim logi do.
Cdu

v d i H C l v i A g dufng sau h i d r o t r o n g day

l o a i iLfdng t i n h .

thich vi sao kim loai khong tdc dung vai cdc chat dd cho.
Cdu TV. Chi diing kim loai, hay nhgn biet cdc dung dich sau

uTng

dong k i m loai. A g k h o n g phan

hap kim loai khong tdc dung vai dung dich kicm hay axit.
Hay xdc dinh kim loai nghien

Ag khong phan

=


™ d d AS04

Ma:
<=>
=>

a(A + 9 6 ) x l 0 0
a x 98 x 100

= a(A + 16) +
22,6
20
A = 24: M a g i e (Mg).

ufng

toa


VI. T a c6:

Cdu

„,

0 3

Cdu

= 10 x 1664 ^ ^
100 X 208

^^^'2
'^NajSO,,

Gidi
Cdu

> BaS04i + 2NaCl

X

< -

X

- >

2x

<-

BaS04l

>

2x

+ 2KC1

(2)

2x

K l i i t h e m dung dich H2SO4 vao lo niidfc loc thi tao ket tua nCfa nen

BaCl2

+ H2SO4

>

BaS04i +

<-

2HC1

(3)

0,2

Theo de:

H

=

0,2

lugng

lugng

tinh

the



Tinh



Khi

trgn

50

ml

0,004M

Cdu

174

+

142 X 0,2
= —^——
200
NaaSO^
200
„„

=

174x0,4
^

m^^^^^^ + m^^^^^^ + m^^^
102

100 =

C % N a so

C%„

=

200

dich

gam

tdch

dich

dich

dung

ra

dich

khoi

20%

dcm

nung

10°C.

Tinh

khoi

II9SO4
den

dung

(a 20°C) vd khoi

dich.

Biet

do

tan

lugng

rieng

cua

dung

dung

dich

mol.
0,012M

CaCl2

thi c6 ket tila xudt

vai

hien

150

ml

khong?

10%.

trong

cua kim logi M bdng

mot

lugng

ling thu dugc dung

dich

A. Them

Sau phdn

lugng

vUa dii dung

dugc

so lieu

dich AgNOs

tUong

sau: ti so thdnh
20
—.

2 oxit do bdng

20%

iCng cila ciuig

phdn

vita

mot

% ve khoi

dii
vdo

thu dugc

Xdc dinli cong thvCc hidroxit

tich 2 oxit vd 2 hidroxit

hda hoc

X 100 = 3 4 , 8 % .

1 hidroxit

muoi c6 nong do 8,965%.

b) Khi phdn

14,2%

200

'^2^^i

dung

Iglml.

dung

todn

A mot

x

thi

V.

a) Hda tan hodn

md.ngdich =

0,4

trong

Id 0,2g

a 20°C

dich

+

HCl

iCng.

do tan cua CaSO^ theo nong do

dung

142

het

bdo hda coi bdng

HCl

X

dich

17,4g.

b) Clio biet do tan CaSO^

dich

0,2

tan

CUSO4.5H2O

dung

=

ket tila dugc chat rdn D. Cho H2

vila dii, sau do Idm ngugi

K h o i lugng

mdungdich

trinh phdn

CuO

TCr (1), (2), (3) => x + 2x + 0,2 = 0,8 =:> x = 0,2

=>

dagc

(mol)

ng^Q^ ban ddu = 0,8 (mol)

dung dich (A) l a :

thi thu

phdn.

mol

nong

NaoSO^
=

nBaso,„3,

vd CuSOj

dUgc chat rdn E. Hda tan E vdo dung

Viet phuang

0,25

dich

0,2

(mol)

nong

CUSO4 a 10°C Id

trong nude loc con dii B a C l 2 .

mud'i Al2(S04)3

IV.

a) Cho

X

BaCls + K2SO4
(mol)

(1)

thich.

dich

B vd ket tua C. Nung

thdy E tan mot

Cac p h a n ijfng:

(mol)

dich

di qua D nung

G o i X l a so' m o l cua Na2S04 => n^^g^^ = 2x ( m o l )

B a C l 2 + Na2S04

Na vdo 2 dung

khi A, dung

=• 1 • 2

• '^K2S04

I I I . Cho

dung

tren.
nguyen

lugng

ciia

to
0x1

,

2T
Tl

DE SO 5
I.

Cho chuoi

phuang

trinh plidn
+A,

ling
I ,

sau:

hidroxit

dung

dich

Cdu

II. Dung

mot
FeCh;

trinh

phdn
HCl;

nhau

vd khdc

ting.

kim loai de nhdn
FeCh;

khdc

BaCh;

CaCOs)-

dung
b) Neu

biet cdc lo dung
(NH^2S04;

dich

AlCh;

sau:
NH4CI

X tdc dung

hda V ml dung

CaCOs

cdc chat A, B, C, X, Y, Z (Id cdc chat

do bdng

khoi

lugng

. Hay xdc dinh

VI. Cho X, Y Id hai dung

a) Khi
Viet cdc phuang

% ve

ciia

nhdm

Iiidroxit

trorig

hai

nguyen

to' do.

135
Cdu

+B,

CaCOs

Tim

phdn

107

OE THI HOC SINH GlDl HOA HOC, QUAN 9 (VONG 2 ) TP. HCM NAM HOC 19S9 - 2000
Cdu

thdnh

so

trgn

dich

vai-AgNOs

Z. Tinh

dich

c6 nong

du tao thdnh

dich

dich

Tinh

35,876g

dich NaOH

Lay

V ml

ket tila. De

trung

nhau.

0,3M.
dich

Y thu dUgc 2 (lit^

Z.

X vd lay 100 nd dung

het vdi kim logi Fe thi lugng
(lit)unA(dktc).
u n ra

do khdc

X vol V (lit) dung

CM dung

lay 100 nd dung

LCII Riai niihau
c T u , „ „0,448
o

HCl

Y can 500 ml dung

V (lit) dung

dich

dich

hidro

thodt ra trong

CM dung

dich X, Y.

dich

Y cho tdc

hai trUdng

hgp

dung
lech
21


Cdu

VII.

Trgn

20 nd dung

Idn
dich

10 ml dung

dich

A de diCgc dung

dich

B can 8 ml dung

dich

tqo thdnh

dung

dich

dich NaOH

dugc

b) Dung
nhieu

CM dung
dich

l,365g

dich

1,N=

dich

vai 20 ml dung
dich A, pha
doi.

khan.

Neu

dich

HNO3

II2O

vdo

dung

hoa 25 nd

dung

Bern c6 can

dung

them

Trung

8% (D = l,25g/nd).

muoi

cho 40 tnl dung

dich BaCl-z thi thu dugc

axit ban

C chi'ia hon

ml dung

Cho: H=

HCl
dung

B c6 the tich gap

vai mot liCgng dU dung

a) Tinli

dich

thu dugc

H2SO4

dich

0,932 ket

vd

M & u nao cho k e t t u a n a u do l a FeCla:
3Ba(OH)2 + 2FeCl3

M a u CO k e t t i i a v a k h i m i i i k h a i bay r a l a ( N H 4 ) 2 S 0 4 :
B a ( 0 H ) 2 + (NH4)2S04

B tdc

tiia.

3Ba(OH)2 + 2AICI3

hap NaOH

0,8M

vd Ba(0H)2

hoa het 50 ml dung

0,2M.
dich

Can

bao

B?

B a ( 0 H ) 2 + 2NH4CI

B a ( 0 H ) 2 + 2HC1

GIAI

t-HCl

->Ca{OH),

lAl

'Cdu

->CaCl,

III. P h a n ufng:

(Bl

N a + 1120

CO,,

+ NH011

+CO.,*U.,0

-^Na^CO,

6 N a O H + Al2(S04)3
->NaHCO.,

CaC03

iooo"c

CaO + CO2T
(A)

CaO + I I 2 O

(X)

Ca(0H)2



CO2 + 2 N a O H

> Na2C03 +

> 2NaHC03
(Z)


-> C a C O y i + 2 N a N 0 3 + CO.T + II2O

Cho k i r n l o a i B a vao cac m a u ihii t r e n , dau t i e a c6 p h a n ufng:
B a ( 0 H ) 2 + H2T

M a u nao cho k e t t i i a t r f i n g x a n h la FeCl2:
Ba(01i)2 + FeCL —

-> CuO + H2O

^ — > Cu + H2O

AI2O3 + 6 H C 1

- > 2AICI3 + 3H2O

CuO + 2HC1 -

-> CuCla + H2O

CuO + H2SO4
(mol)

0,25 ->



> CUSO4 + H2O

0,25

0,25

K h o i iLfoag d u n g d i c h H2SO4 2 0 % :

II. T r l c h m o i d u n g d i c h m o t i t l a m m a u thuf.

B a + 2H2O



H2O

(Y)

2 N a I i C 0 3 + Ca(N03)2

Cu(0H)2

- > AI2O3 + 3H2O

V i E t a n 1 p h a n n e n E chufa: CuO, Cu, AI2O3

> CaCO^i + 2NaCl

NaaCOs + CO2 + I I 2 O

-

CuO + H2 -

> CaCl2 + 2H2O

CaCl2 + NaaCOa

2A1(0H)3

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

AI2O3 + H2

(B)
C a ( 0 H ) 2 + 2HC1

> 2 A l ( O H ) 3 i + 3Na2S04

2 N a O H + CUSO4 —

P h a n ijfng:

22

-> N a O H + ^ H a t

CaCO;,

CaCOg

Cdu

> BaCl2 + 2H2O

C o n l a i l a BaCla.

ThiTc h i e n chuoi p h a n ufng:
CaO

> BaCl2 + 2 N H 3 t + 2H2O

M a u cho d u n g d i c h t r o n g suo't v a toa n h i e t l a H C l :

Cu = 64, Ag = 108, Ba = 137
Cdu I.

> 2 A l ( O H ) 3 i + SBaCla

M a u c h i CO k h i m t i i k h a i bay r a l a NH4CI:

14, O = 16, Na = 23, S ^ 32; CI = 35,5, Co = 40, Fe = 56,
LCfl

> B a S 0 4 ^ + 2NH3T + 2H2O

M S U cho k e t t i i a keo t r a n g la AICI3:

ddu.

C de trung

> 2 F e ( O H ) 3 i + 3BaCl2

> FeCOIDai + 2BaCl2
I rii p i A i n c T u i u n r Qtwu n n i unA u n r Q

0, 25

X

98

X

100%

20

= 122,5

K h o i l u g n g CUSO4: 0,25 x 160 = 40

(gam)

(gam)

K h o i lu'gng d u n g d i c h sau p h a n ufng:
' ^ d u i i g d i c h s a u p h a n iMig —

niCuO +

= 0,25

X

'^^ddn^SO^

80 + 122,5 = 142,5

(gam)


K h i h a n h i e t do:

K h o i lucfng d u n g d i c h t r o n g (2):

• C U S O 4 + 5H2O
(mol)

— > CUSO4.5H2O

a

->

mdu„gdich2

_

T h e o de, n o n g do cua d u n g d i c h m u d i t h u dirge sau p h a n ufng (2) l a :

17,4

8,965% =

4 6 9 6 - 18784a = 2479,5 - 4 3 5 0 a

«

2216,5 = 1 4 4 3 4 a

Cong thuc oxit:

20°C: 100 g a m H2O + 0,2g CaS04 - > 100,2 g a m d u n g d i c h CaSO,

CO d = I g / m l

(mol)

M2OX,

V^^^^^ =

0,0006

"caci, =

o

16y

= 20 : 27

2 M + 16x • 2 M + 16y
x

M + 8y

20

M + 8x

y

27

2 7 M x + 216xy = 2 0 M y + 160xy
2 7 M x + 56xy = 2 0 M y

> CaS04 + 2 N a C l

17y

17x

M a t khac:

M + 17x • M + 17y

X 0,012 = 0,0006 ( m o l )

o

^Na^so^ = O'^'^ ^ 0'004 = 0,0006 ( m o l )

(1)

= 107 : 135

M + 17y

:

107

y

M + 17X

135

135xM + 2295xy = 1 0 7 M y + 1819xy

d 2Q°C t r o n g 1 l i t d u n g d i c h CaS04 bao h o a c6 0,015 m o l CaS04
t r o n g 0,2 l i t d u n g d i c h CaS04 bao h o a c6 0,003 m o l CaS04
V i 0,6.10"^ < 3.10"^ n e n k h o n g c6 h i e n t i / g n g k e t t i i a .
Cdu

M 2 0 y

16x

= 100,2 ( m l )

C^.
=
x — - — = 0,015M
Mc,so,
136
0,1002

0,0006

M = 39a

C o n g thilfc h i d r o x i t : M ( O H ) ^ , M ( O H ) y
T h e o de b a i , t a c6:

+ Na2S04



b ) Goi nguyen to can t i m la M :

2 5 0 a = 250 x 0,154 = 38,5 (gam)

CaCla

100%

X

D o do c o n g thufc c a n t i m l a K O H .

Vay k h o i lifgng CUSO4.5H2O k e t t i n h :

Khi tron:

M + 62a
M+1232a-143,5a

100%

=^ N g h i e m h o p l i : a = 1; R = 39: k a l i ( K )

a = 0,154 ( m o l )

V a y thi t i c h d u n g d i c h CaS04 Ik:

X

M +1232a-143,5a

'3'a



Suy r a :

M + 62a

C% M ( N 0 , ) ,

142,5 - 2 5 0 a " 100 + 17,4

h)*a

+ 382a + 170a : 2 0 % - 143,5a

= M + 1232a - 143,5a ( g a m )

K h o i l i i o n g d u n g d i c h c o n l a i : 142,5 - 2 5 0 a ( g a m )
40-160a

=

ra^i^ + ra^^j^^o,

a

Kho'i liicrng C U S O 4 c6n l a i t r o n g d u n g d i c h : 40 - 1 6 0 a ( g a m )

.

=

135xM + 476xy = 107My

<=>

(2)

T i f (1) v a (2), t a c6:
5(20My - 56xy) + 476xy = 107My

V.
a ) G o i c o n g thufc cua h i d r o x i t k i m l o a i M : M ( O H ) a
M(OH)a + aHCl

> M C l a + aHaO

(I)

MCla + aAgNOg

> aAgCll + M(N03)a

(2)

<=>

lOOMy - 2 8 0 x y + 4 7 6 x y = 1 0 7 M y

<=>

7 M y = 196xy <=> M = 28x

Bang bien luan:

K l i o ' i l u g n g d u n g d i c h t r o n g (1)

X

1

2

M

28

56

3 "
84

V a y M = 56 (Fe).
Vay
= M + 17a + 36,5a x 10 = M + 382a ( g a m )
.

I rli n i l n c T U I u n n c i u u n i n i unA unp n

c o n g thijfc o x i t :

FeO; Fe203

cong thiic h i d r o x i t : F e ( 0 H ) 2 ; F e ( 0 H ) 3
25


Cdu

Trifffng hap 2: Cx < Cy

VI.

H C l + AgNOa

a)
(mol)

—> A g C U + HNO3

0,25

0,25

HCl + NaOH
(mol)

(1)

^ N a C l + H2O

(2)

0,15 <- 0,15
35,876

T a c6: n ,

143,5

= 0,25 (mol) wk n^^oH = 0'^ x 0,3 = 0,15 (mol)

TCf p h a n ijfng (1) v a (2), t a c6:
0,25 + 0,15

0,4

v +v

(mol)

= 0,2M (Vi V + V = 2)

(mol)

FeCl2 + H a t

0,025

0,025

V

2V

(3)

V

(mol)

2V' •

- n^^^.,^

= n^^^^^ = Cx-V = 0 , 2 5 ( m o l )

-• nHci/,2)

= "NaOH

= CY-V

0 , 1 h t d u n g d i c h H C l t h i c6

= 0,15

(mol)

molHCl.

0 , 1 l i t d u n g d i c h H C l c6

(mol)

<=>

V.V

<=>

3V - 10 + 5V

<^

8V - 8V - 10 = 0 =^

(mol)

+

0,005CB <-

8V^
V, = 1 , 7
(loai)

> N a C l + PI2O

HCl
0,0025CA

— > N a N O s + H2O

0,004

> Na2S04 + 2H2O

(3)

0,005Cc
> B a S 0 4 i + 2H2O

<-

(2)

0,005CB

0,005CB


(1)

0,0025CA

HNO3 -

0,01Cc <- 0,005Cc ->

(4)

0,004

Goi CA l a n o n g do m o l cua d u n g d i c h H C l .
So m o l cua H C l t r o n g 25 m l d u n g d i c h B :
0^.0,01x0,025

= 0,0025CA

G o i CB l a n o n g do m o l cua d u n g d i c h HNO3
c

^
V

V

=> So m o l ciia HNO3 t r o n g 25 m l d u n g d i c h B :

= 8

C B . 0 , 0 2 x b , 025
0,1

=> So m o l ciia H2SO4 t r o n g 25 m l d u n g d i c h B :

\5

Cg.0,02 X 0,025

V2 = 2,5 > 2 (loai)

I rtl n i f l l n p T H I H n r c i w u n i n i u n A u n p n

= 0,005CB

G o i Cc l a n o n g do m o l cua d u n g d i c h H2SO4

10 - 5V - 3V = 16V - SV^

0,1

V = 0,5 => V = 1,5 ^ Cx = 0,5M v a Cy = 0,1M

26

V:

T h e t i c h cua d u n g d i c h (B): 100 m l = 0,1 l i t

molHCl.

V

NaOH
0,0025CA <

= 8; V + V = 2 (V, V < 2)

8V^ - 24V + 10 - 0

16V

0,1

5(2 - V) - 3V = 8V(2 - V) «
«

3V - 5(2 - V) = 8V(2 - V)

= 0,02 (mol)

0,015^-^,32
2V

5V' - 3V

o

H2SO4 + BaCla

TriTdrng hdp 1: N e u Cx > Cy
0,025
2V

3V - 5V' = 8 W ' , V + V = 2

2 N a O H + H2SO4

T i i a n g tLf: T r o n g V l i t d u n g d i c h H C l c6 0 , 1 5 m q l H C l .

22,4

<=>

NaOH

T r o n g V l i t d u n g d j c h H C l t h i c6 0 , 2 5 m o l H C l .

T h e o de: n H„

V'

5

a ) Cac p h a n ufng:

(4)

0,015

0,448

2V

25
3
=
40
<
=
>
V
V

vn.

(mol)

-> FeCl2 + H 2 t

0,015

T h e o cau (a):

2V'

15

= 0,02

V i = 1,7 => V = 0,3 =^ C^ = 0,147 v a Cy = 0,5.
Cau

2HC1 + Fe

0,025

V2 = -0,72

b) P h a n ufng
2HC1 + Fe

0,015

"^LSIMI

THI HOC SINH GIOI HOA HOC 9

= 0,005Cc

27


Khoi

iLforng

d u n g d i c h N a O H = 8 x 1,25 = 10 ( g a m )

So m o l cua N a O H :

^
100

So m o l cua B a S 0 4 :

0 Q*^?
— — = 0,004 ( m o l )
233

X

40

G o i a, b , c I a n Ixxat Ik so m o l ciia H C l , HNO3, H2SO4 t h a m g i a p h a n

= 0,02 ( m o l )

L^ng (1), (2), (3).
a', b ' , c' I a n l i i a t l a so m o l cua H C l , HNO3, H2SO4 t h a m g i a p h a n

ang (4), (5), (6).
,
C A X 0,01x0,05
i i H c i = a + a' =

So' m o l cua H2SO4 t h a m g i a p h a n ufng (4):
X 0,02x0,04

= 0,004 =^ 0,008Cc = 0,004

0,1

^Cc

0 , 0 0 2 5 C A + 0 , 0 0 5 C B = 0,02 - O.OlCc =
2 5 C A + 5 0 C B = 150

«

CA+2CB = 6

Va:

5CA + I O C B =

0,015

0 , 1 3 2 5 C B = 0,1325 =^ C B = 1 (mol/1)

Suy r a :

= 6 - 2CB

CA

=

2

(0,8V - 2c - b) + a' + b + b ' + c' + c = 0,035

(1!

0,14625(6 - 2 C B ) + 0 , 4 2 5 C B = 1,01
o

^

0 , 8 V - c + (0,4V - 2c') + c' = 0,035

=^

1,2V - (c + c') = 0,035
1,2V - 0,005 = 0,035

=:>

4 (mol)

V - 0,03333 ( l i t ) = 33,33 m l .

DE SO 6

b ) Cac p h a n ufng t r u n g h o a :
NaOH
(mol)

a

+

<--

(mol)

+

2c

b

H2SO4

<-

NaOH

> N a C l + H2O

(1

a

2NaOH
(mol)

HCl

+

<-

> Na2S04 + 2H2O

(2

c

HNO3

> N a N O s + H2O

(3

> BaCla + 2H2O

(4

(mol)

b'


2

<-

c'

«-

> B a ( N 0 3 ) 2 + 2H2O

b'

B a ( 0 H ) 2 + H2SO4
(mol)

Cdu I.

Viet phan

c'

bien hoa

> Ca(0H)2

sau:

^
CaCOs

CaClo
Cdu

Ca(N03)2

11.
Tdch
phdp

hon

hap

hoa

Cdu in.
Hay

nhdn

gdm

BaCOs,

BaS04,

KCl,

MgCh

bdng

phuang

hoc.

b) Cho cac hoa

chat: Na, MgCh,

biet

Cho phuang
viet 4 phuang

xdy ra hodn
Ldl GlAl Oi THl HOC SINH Gidl HOA HOC 9

chuoi

CaCOs

hay
> B a S 0 4 + 2H2O

ling de hieu dien
CaO

a)
B a ( 0 H ) 2 + 2HNO3

(mol)

O E THl H O C SINK GIOI H O A H O C 9, C A P TP. H O CHJ M I N H N A M H O C 2 0 0 0 - 2 0 0 1

b

Ba(0H)2 + 2HC1

mol

= 0,01

2]n,^^ = a + a' + b + b ' + c + c' = 0,035

T h e ( I ) vao ( H ) t a c6:
0,8775 - 0 , 2 9 2 5 C B + 0 , 4 2 5 C B = 1,01

,

,
CcxO,02xO,05
,
= c + c =
-^-^1^
= 0,005 m o l

2

1,365 = 0 , 0 0 2 5 C A X 58,5 + 0 , 0 0 5 C B X 85 + 0,005Cc x 142

o



nH2S04



^ + K +c' = 0 , 2 V

(I

1,365 = 0 , 1 4 6 2 5 C A + 0 , 4 2 5 C B + 0,355

CB X 0,02x0,05

= b + b' =

a + b + 2c = 0 , 8 V ( v d i V l a t h e t i c h ciia d u n g d i c h C ) .

30

mniuo'i = n i N a C i + mNaNo, + ™Na,so,

<=>

, ,

nHN03

= 0,5 (mol/1)

So m o l cua N a O H t r u n g h 6 a t r o n g p h a n ufng 1 , 2, 3 1 ^ :
«

,

,
= 0,02 m o l

todn.

FeCls,

FeCk,

AICI3.

Chi dung

them

H2O

chung.
trinh
trinh

phan
phan

ling c6 dang

sau: BaCh

ling xdy ra. Biet

rdng

+ ? = NaCl
cac phan

+ ?
i2ng

deu


.au rV.d 25°C nguai ta da. hda tan 450 gam kali nitrat vao trong 500 gain
niiac cat (dung dich A). Biet rdng dp tan ciia nitrat kali la 32 gam „

Cau IIa) Scf do tach:
BaCOy

20°C. Hay xdc dinh khoi lugng kali nitrat tdch ra khoi dung dich khj

BaSO.

Idm lanh dung dich A den 20°C.
~lau y . Cho 3 gam

hdn hap hai kim loqi vun nguycn

Xdc dinh

BaSO^ +H;,o
KCl

cliudn.

MgCl^

thdnh phdn

phdn

tram

ve khoi lugng

ciia nlwm

vd magic

KCl

trong hdn hap.

Na + H2O

dich HCl thi sau kJii phdn ling ket thiic thu diigc 896 ml khi H2 (dicii

-

kiwi

hodn

M S U CO k e t tua mau trSng la MgCl2

-

> Mg(0H)2i + 2 N a C l

M a u nao c6 k e t tiia mau trang xanh, de lau hoa nau do la FeCl2.

I'ing vol axit thi Mij'

FeCla + N a O H

ling dcu xdy ru

> F e ( 0 H ) 2 i + 2NaCl
trdng xanh

todn).

Biet: H=1;N=

-> NaOH + - H a t
2

MgCl2 + 2NaOH

lugng cdc chat trong X, Y.

pJidn ling trade, het Mg mai den Fe. Cho biet cdc phdn

2Fe(OH)2 +

14; O ^16; Mg = 24, Al = 27; CI = 35,5; K = 39; Ca = 40; Fe = 56

-> 2 F e ( O H ) 3 i

+ H2O

ndu do

LCilGIAI
du I.

-

Chuoi bien hoa:

M a u cho k e t t u a m a u nau do l a F e C l s .
F e C l a + 3NaOH

CaCOs

^ - ^ ^ — > CaO + CO.T

CaCOa + 2HC1
CaO + 2HC1
CaO + H2O
Ca(0H)2 + CO2
CaCl2 + 2AgN03
Ca(0H)2 + 2HNO3
Ca(N03)2 + Na2C03

-

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

M a u cho k e t t u a keo trang la AICI3.
AICI3 + 3NaOH

> CaCh + COgt + H2O

-> A1(0H)3^ + 3NaCl

Neu N a O H dii thi k e t tiaa t a n dan:

> CaCl2 + H2O
> Ca(0H)2

->MgCl,

+) Cho dung dich N a O H I a n liigft vao cac mau thtf tren t h i :

kien chuan) vd c6 can dung dich thi dugc 6,68 gam clidt rdn Y.

ling vai nitdc vd khi phdn

+ HC1

+) Cho ni/orc Ian liJOt vao cac mau thiJf tren. Mau nao c6 k h i bay ra la natri.

Ncu cho hon hap gom a gam Fe vd b gam Mg vdo trong 400 ml dun/-

sii Mg khong phdn

'Mg(OH)^ i

b) Trich mSi chat 1 i t l a m mau thuf.

ling ket thuc dem c6 can dung dich thu dicgc 6,2 gam chat rdn X.

Tilth a, b, nong do phdn td gam ciia dung dich HCl vd thdnh phdn

KCl

+K011
vL/a du

MgCl^

^du VL Khi cho a gam Fe vdo trong 400 ml dung dich HCl, aau khi phun

(Gid

BaSO^ i

IdC

BaCO,,

chat Id nhom v,i

magie tdc dung hct vai H2SO4 lodng thi thu duqc 3,36 lit mot chat khi a
dieu kien ticu

BaCl. - ^ M loc
i_^BaCO. i

+HC1

Cdu

-> CaCOsi + IlaO
> Ca(N03)2 + 2AgCU

III,

A1(0H)3 + N a O H

> NaAlOa + 2H2O

BaCl2 + Na2C03

> B a C 0 3 i + 2NaCl

BaCl2 + Na2S04

— > B a S 0 4 i + 2NaCl

3BaCl2 + 2Na3P04
B a C l a + Na2S03

^ Ca(N03)2 + 2H2O

(hoac BaCl2 + NaaSiOa

^ CaC03>l + 2 N a N 0 3
LOi

r.iAi,

> Ba3(P04)2i + 6NaCl
> B a S O g i + 2NaCl
> BaSiOgi + 2NaCl


Cdu IV. Bap so: Kho'i Itfang kali n i t r a t t^ch ra k h o i dung dich:

Tif (3)

290 (gam) K N O 3 .

Cdu V. Goi a va b Ian liigt la so' mol cua M g
Phan ijfng:

Mg + H2SO4

(mol)

->•

(mol)

Giai he ta dtfOc: a = 0,05 va b =

Cdu

%mMg =

^'

3

x

mcLatrdn = 0,04

Theo de bai:

nichat rin

Phan Lfng:

M g + 2HC1

(mol)

= 0,15

(mol)

CO

dich t h i thu difgc 6,2 gam chat r ^ n . Neu Fe phan ijfng het tufc chat rSn
=

0M^9, (mol)

(*)

Vay:

niMg = 0,02
"^Fecij

Thi nghiem 2: Cho a gam Fe va b gam M g vao 400 m l dung dich H C l , c6
can dung dich t h i thu diigc 6,68 gam chat r ^ n va 0,896 l i t H2 (dktc).
Phan

Lfng:

M g + 2HC1

(mol)

X

> FeCla + H a t

y

mpedLT

^

nj,
"2

+y=

22,4

« x + y = 0,04

(2)

0,04

0,08

0,04

^0,04

=

6,68

|x + y = 0,04

fx = 0,02

[95x + 71y = 3,32

l y = 0,02

x 24

= 0,48

X 127

= 3,36 - 0,02

(gam)

= 2,54
x

(gam)

56 = 2,24 (gam) ^ b = 0,48 (gam).

Cdu 1.
^) Chi dung nuac hay nhdn biet 3 bgt kim loai: Ba, Al vd Ag.

(**)

^) Til cdc chat sau: Na^O, HCl, H2O, Al c6 the dieu che dugc n/ulng clid't
fnai ndo md kJiong dung them phiCang tien ndo khdc. Viet phan ling

0,04 (mol).
(mol)

=

+ ^Fed,

NAM HOC 2000 - 2001

so mol H2 cf t h i nghiem 1 la:

> FeCla + H g t

+ y = 0, 04

Ki THI CHON HOC SINK GIQIHOA HOC 9 TP. QUY NHdN, TJNH MW MW

han d t h i nghiem 1 nhifng so mol H2 thu diioc l a i i t hon. Do do trong

Fe + 2HC1

= X

DES0 7

So sanh (*) va (**) ta thay: K h o i lugng k i m loai d t h i nghiem 2 nhieu
t h i nghiem (1) t h i Fe di/ va H C l het

nj^

y

95x + 127y + (s, 36 - 56y) = 6,68

y

= X

y

(5)

+ y = 0,04

(1;

(Vdi X , y Ian liTOt 1^ so' mol cua M g va Fe tham gia phan iJtng tren).
TCr (1), (2)

X

> FeCla + Hat
->

X

Fe + 2HC1
(mol)

> MgCla + H2t

(4)

n^ran = n^MgCl, + ^FeCl,

Thi nghiem 1: K h i cho a gam Fe + 400 m l dung dich H C l , c6 can dung

= ^

(gam)

+ H2t

x

Tif (4), (5) ta CO he phtfcfng t r i n h :

= 60%.

= 7,16

> MgCh

y

+) Tinh a va thanh phan cua (X).

np^^i,

+ 3,36

->

X

n^^ =

56 = 2,24 (gam)

= 6,68 < 7,16 => M g khong du nen Fe da phan ijrng.

X

VL

Chi la FeCl2 ^

X 95

Fe + 2HC1


3
%mAi

x

6,2 - 5,08 = 1,12 (gam)

Neu k h o i lircfng M g dung du t h i :

(2)

24a + 27b = 3

100 = 40% va

0,04

uTng:

(gam)

+) Tinh b va thanh phan cua (Y)

2

a .

= 5,08

Vay: a = 2,24 + 1,12 = 3,36 (gam).

3b

->

Theo de bai, ta c6 he phuong t r i n h :

Vay:

(1)

> Al2(S04)3 + 3H2t

b

x 127

va khoi lircfng Fe diT:

a

2A1 + 3H2SO4

= 0,04

Khoi lugng Fe phan

Al.

> MgS04 + Hgt

a

ra^^cA.,

• minh hoa.
2. Viet cdc phuang trinh phan ling tlieo sa do sau:
A

(3)
i

-> B

+Y

D

+z,t°

A

^'•^'t C Id chat ket tua mdu do ndu vd A, B, C, D, X, Y, Z Id ki hieu I'Cng
cong thitc 1 chat.


Cdu

3. Bern nij gam hon hap ZnCOs, Zn dun nong ngodi khong
' i dc
plidn ling xdy ra hodn todn, thu dugc nig gam chat rdn.
Biet nil = mo. Tinh % khoi lugng ZnCOs trong hSn hap ddu.
Cuu 4. Bem dung dich chiia 0,1 mol sdt clorua tdc dung vai dung dich
NaOH du thu dugc 9,05 gam ket tua.
Xdc dinh cong thiCc sdt clorua vd tinh hieu sudt phdn ling.
Cdu 5. Bem 46,4 gain Fe^Oy tdc dung vdi Ho dun nong thu dugc chat rdn B
gom Fe vd Fe^Oy du. Bem chat rdn B tdc dung het vai dung dich HNO.
lodng du thu dUgc dung dich C c6 chUa 145,2 gam mud'i Fe(N03)3 va
a mol NO thodt ra. Tat cd phdn Ung xdy ra hodn todn.
a) Xdc dinh cong thUc Fe^Oy.
h) Biet a = 0,52, tinh khoi lUgng tiCng chat trong B.
Ldi

Cdu

. 2Zn + O2

(1)

2ZnO

(2)

Goi X , y Ian lircft la so mol ZnCOg va Zn trong h6n hop dau. V i m j = m2.
K h o i liiong CO2 thoat ra 0 (1) = khoi lugng oxi tham gia a (2)
=>
Vay

I

Cdu

44x = 16y ^

- = ~
y
11

% khoi iLforng ZnCOg = 41,15%.

4. Neu la FeCl2: FeCl2 + 2NaOH

> F e ( 0 H ) 2 i + 2NaCl

Khi phan ufng xay ra 100% thi khoi lugng ket tiia Fe(0H)2 = 9g < 9,05g

GIAI

1

=i> V6 11, vay do la FeCls.

a ) Cho 3 k i m loai vao 3 coc niicfc.
-

Tan

CO

FeCla + 3NaOH

bot k h i bay len la Ba:
Ba + 2 H 2 O

> Ba(0H)2 + H a t '

-

Kliong t a n la A l va Ag.

-

Cho 2 k i m loai A l va Ag Ian liicft vao hai coc chufa dung dich
Ba(0H)2: K i m loai nao tan c6 bot k h i bay len la A l :
2A1 + Ba(0H)2 + 2 H 2 O

-

=>
Cdu

NasO + H2O

145,2
> 2NaOH

NaAlOa + H C l + H2O

> A l ( 0 H ) 3 i + NaCl

A l + 6HC1

= 84,58%

56

= 33,6 gam

K h o i iLforng oxi trong 46,4 gam Fe^Oy = 46,4 - 33,6 = 12,8 gam
,

X
y

> NaCl + H2O
> 2AICI3 + 3H2T

33,6

X

16

^ 1 2 : ^

3
=

1

Fe + 4 H N O 3

b)

2. Phan ijfng:

,

FeCl, + 3NaOH
2Fe(OH)3
Fe203 + 3H2
B: FeCla;

-> 2FeCl3

^

> Fe(N03)3 + N O t + 2H2O
> 9Fe(N03)3 + NOT +

UlUO

Goi X , y Ian liigt la so mol ciia Fe va Fe304 trong hon hop.

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

V',-

So mol NO: x + ^ = 0,52
3

FeaOa + 3H2O

Bao toan Fe: 56x + 168y = 33,6

> 2Fe + 3H2O
C: Fe(0H)3;



- ^ " " ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^

3Fe304 + 28HNO3
2Fe + 3CI2

A: Fe;

X

242
> 2NaA102 + 3 H 2 t

NaOH + H C l

Hieu sua't (H) =

a ) K h o i liiang Fe c6 trong 145,2 gam FeCNOsJa:

> Ba(A102)2 + Sllat

2A1 + 2NaOH + 2 H 2 O

> F e ( 0 H ) 3 i + SNaCl

5.

Khong tan la Ag.

b)

Cdu

-> ZnO + CO2

ZnCOo

3.

D: Fe203;

=> X = 0,51 mol ^ m p e = 28,56 (gam)

Y: NaOH;
X: CI2;
Z: H2
A, B, D, X, Y, Z CO the khac, nhUng C phai la Fe(0H)3.

y = 0,03 mol ^
Wl

Rli, „S ....

m^^ ^

= 6,96 (gam).


L6\I

DE SO 8
flE T H I

HOC S I N H

GlDl H O A

HOC

9, C A P

T P . HQ CHI IVIINH N A M HOC 2 0 0 1 -

ZOOZ

Cdu 1: B o tuc v a c a n bkng:
3CaCl2 + 2Na3P04

Cdu 1: Bo tiic ud can bdng cdc phitang trinh phdn ling sau:
CaCh

+?

Ca3(P04)2^

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

+ ?

Ba(HC03)2 + ?

^ BaCOaJ-

+?

CaS03

+?

-> SO. f

+ ?+?

HCl

+ ?

-> NallCOs

+?

FeClo

+ ?

-> FeCh

FeCh

+ ?

FeCU

Cdu 2: Vict cdc pliUang trinh phdn I'tng de bleu dien chuoi bicn lioa sau:
FcS2 ^ SO2 -> SO3 ^ H2SO4 -> SO2

> C a 3 ( P 0 4 ) , i + 6NaCI
> 2 B a C 0 3 i + 2H2O

CaSOa + 2HC1

> S O a t + CaCl2 + H2O

H C l + NaaCOa

> NaHC03 + NaCl

2FeCl2 + CI2

> 2FeCl3

2FeCl3 + F e

> SFeClg

Cdu 2: V i e t phUcfng t r i n h

Na2S03 ^ BaSOa

4FeS2 + IIO2

2Fe203 + 8SO2

Cdu 3:
a) Cho 6 dung dich gom: NaCl, BaCU, CUSO4, NaOH,
KJioiig dung them hoa chat ndo khdc, hay nhan biet
b) Cho 3 dung

dich: BaCU, BafNOi).,

them mot hoa chat, hay nlidn bict

Ba(HC03)2.

MgCh,

2SO2 + O2

AgNO,.

chung.
SO3 + H2O

Chi diCgc si'i dung

cluing.

> CUSO4 + S02t + 2H2O
> Na2S03 + H2O

SO2 + 2 N a O H

rieng CO2.
Cdu 4: A vd B Id 2 loai clidt chi cliiia cdc nguyen to X, Y. Thdn.Ii

pJidn

Na2S03 + BaCl2

tram cua nguyen to X trong A vd B Idn liigt Id 30,4% vd 25,97/.

: i 2SO3

> H2SO4

2H2SO4 + C u

c) Cho hdn hap khi gom CO2, SOo. Bdng pliuang phdp hoa hoc, hay tdch

phdn

^

450*^ C

> BaSOsi + 2 NaCl

Neu cong thi'ic phdn til cua A la XY2, thi cong thiic phdn tii cua B Id gi?
Cdu 5: De gia tang nong do cila 50 gam dung dich CuSO^ 57c len gap luu
Idn, CO boil hoc sinh da thUc Jiien bdng bdn cdch khdc

nJiau:

Hoc sinh D: them 50 gam dung dich CuSO^ 157c vdo dung
Hoi hoc sinJi ndo da lam diing, gidi

nitac.

dich.

Hoc sinh C: them 4,63 gam tinh the CUSO4.5H2O vdo dung

dich.
dich.

thich.

Cdu 6: Hop chat A bi phdn hiiy a nhiet do coo thco phaang trinh plidn dug:
2A

B + 2D + 4E

San phdm tgo thdnh deu a the khi, khoi lugng mol trung blnh cua hon
hgp khi sau phdn ting Id 22,86 (gimol).

N h a n biet CUSO4: mau xanh
Cho mSu CUSO4 tac dung vdi cac mau c5n l a i c6 1 m a u cho ket tua

Hoc sinh A: dun nong dung dich de Idm bay hai phdn ni'ca lugng
Hoc sinh B: them 2,78 gam CuSO^ khan vdo dung

a)-

Tinh khoi lugng mol cua A.

Cho so lieu: H = 1; O = 16; S = 32; Cu = 6"-^.
Hoc sinh c6 the sii dung bdng do tan vd bdng he thong tudn hodn cdc nguyen to
hoa hoc.

x a n h l a m l a N a O H , mot mau cho ket tiia t r S n g l a BaCl2.
L a y mSu N a O H cho tac dung vdi cac mau con l a i : mau cho ket tua
t r S n g l a MgCla, mSu cho ket tua den 1^ AgNOa, mau khong tao ket
tua l a N a C l .
CUSO4 + 2 N a O H

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

CUSO4 + BaCl2 MgCl2 + 2 N a 0 H
AgNOg + N a O H
2AgOH

•> B a S 0 4 i + CuCla

-

> Agp

- > Mg(0H)2i + 2NaCl
^ AgOH + NaNO,
+

up


b) -

D u n nong m a u cho k e t tua la Ba(HC03)2

160

mp^so^

them

2,5 + 2 , 9 6 3 2 = 5 , 4 6 3 2 g a m

BaCl2 mSu con l a i la Ba(N03)2.

c) -

niduag
[ C u S 0djch
4 ] %= =5 0- 5+ ^ 4^, 6^3 X
= 100
5 4 , 6=3 g10%

> Ba(N03)2 + 2AgCU

54, 63

Cho qua dung dich thuoc t i m SO2 b i hap t h u , con CO2 bay ra (cho
qua dung dich Br2 chi c6 SO2 b i hap thu).

-

Co the dot chay SO2

™ C u S 0 4 them vAo

SO3 hap thu vao Ba(0H)2 sau do cho H C l

™cuS04

= 7,5 gam

-

dung dich sau =

2,5

+

7,5

-

Cong thufc A: XY2

n i d u n g dich =

-

Cong thufc B: X a Y b

[CuS04]% = 10%) ^ dung

-

Trong A:

-

% X = 30,4%
X

_ 30,4

2Y

69,6

% Y = 100% - 30,4% = 69,6%
_

bY
TCrd), (2)

X
Y

60,8
69,6

_

74,1
60,8

25,9b

69,6

74,1a

X

25,9b

Y

74,1a

Dat

5

50

+

50

100

10

gam

gam

B + 2D + 4E
a

2a (mol)

=

=

2a

4a

Neu lay 2a mol A nhiet phan se tao t h a n h 7a mol k h i .
D i n h luat bao to^n khoi liicfng cho:
M A X 2a = 22,86 x 7a

a _ 2
b
5

=>

M A = 80.

OE SO 9
BE THI HOC SINH GIOI HQA HOC 9, TP HAI PHONG (BANG A) NAM HOC Z001 - Z002

Cong thufc cua B la: X2Y5.
Cdu J.-

>

2A

Cdu 6:

%Y = 74,1%

%X = 25,9%

Trong B:

aX _ 25,9

2,5
.50 = 2,5 gam (n,„so_ = ^

Cdu I

= 0,015625 mol)

1. Cho bang phdn loai cdc chat:
^H,o

= 50 - 2,5 = 47,5 gam

Nong do can dat di/cfc la 10%.
Hoc sinh A lam bay hoi phan nijfa luong niidc
m^^Q

biioai =

m d u n g d . c h sau

= 23,75 = m^^o

Hoc sinh B:

cbn

X

100 = 9,52% ^ 10%

' 5,28 '
52,78 J

3

4

5

HI

NO

CO

O2

Fe

H2SO,

Na20

NO

SO2

H2S

CO2

CH4

6

7

8

Cu(0H)2

CH4

KOH

N2

KOH

C6H22O6

Ba(0H)2

Br2

NaOH

CCI4

NaOH

a ) Nhung dinh sdt da cqo sach gi vao dung dich

sai

Sue khi SO2 vao dung dich

sau:

CuSOj.

Ca(HC0a)2.

^) Dan khi etilen qua dung dich nUac brom.
^' Cho day chuyen hoa sau:

= 50 + 2,78 = 52,78 gam

[CuS04]% =

2

•2. Neu hien tUgng, viet phUpng trinh phdn ling cho cdc thi nghieni

mC U S O 4 = 2,5 + 2,78 = 5,28 gam
mdungdich

1

Hay cho blet vi tri (1), (2), (3), (4)., (5), (6), (7), (8) la cdc ti( gi.

= 50 - 23,75 = 26,25 gam

[CuS04]% = 126,25y
I T^M^rJ


diing

- ii_ .50

Hoc sinh D:

vao de giai phong CO2.
Cdu4:

gam

= 2,9632

250

I ^ C u S O . dung dich sau —

H a i mau con l a i cho tac dung vdri dung dich AgNOs. Mau cho i

BaCl2 + 2AgN03

4,63

X

=

^ B a C O g i + CO2T + H2O

Ba(HC03)2
-

Hoc sinh C:

Fe ->A

100 = 10% => dung

-^B

->C-^Fe

->E->-F-^D

dinh A, B, C, D, E, F. Viet cdc phiiang trinh phdn ring.
-.2.

11

I un/t u n r a

39


V Doi c/'idy hodn todn 1,344 lit (dktc) hon hgp 3 hidrocacbon

Cau II.
1. Dung dich Boocdo dung chong nam cho cay dugc pha theo tl le:
1kg CUSO4.5H2O

+ 10kg voi song (CaO) + 100 lit nuac

C H'>n + 2! CinH2m', CkH2k - 2- Sau phdn Ung, dan lion hgp sdn piidm Idn
lugt Q'^'^ H2SO4

fdgc),

dung dich NaOH

(dd^^) ^'^"^ ^'^^

Boocdo.

1 Tinh thdnh phdn % theo the tich hSn hgp 3 hidrocacbon,

2. Tie glucoza

va cdc chat v6 ca can thiet, viet cdc phuang

ling dieu che:

trinh phcn,

etylaxetat.

hidrocacbon

1. Ba khi A, B, C c6 phdn ti2 khoi bdng nhau vd bdng 28 dvC. A, B cu
the hi dot chdy trong khong khi, sdn phdm sinh ra deu c6 khi COo, li

CkH2k-2 trong hon hgp gap 3 Idn the tich

CO so nguyen
cua hidrocacon

cdc phuang trinh phdn iCng.
trong chuang trinh hoa hoc pho thong cap trung hoc ca sd. (P) vd (Ni
CO cung cong thvCc phdn tit.

con Igi.

Cdu I.
(1): axit;

(2): oxit;

(3): oxit khong tao muoi; (4); chat k h i ;

(5): dcfn chat;

(6): baza;

(7): chat hau ccf;

Fe +

a)

(M); klpt (X) = 3klpt (R) - Gklpt (Q)

(8): bazcf kiem.

- Co ket tiia cua. dong xuat hien.
SO2 + Ca(HC03)2

cJio the tich H2 Ian nhdt la 3,36 lit (dktc), chi c6 (R) phdn itng dugc

hay

vai dung dich NaOH. Tit (X) cd the dieu che ra (N), (R), (Q) c6 phdn
thich.

Cdu IV. Hoa tan mudi nitrat cHa mot kim logi hoa tri 2 vao nudc dugc 200 lu'i
dung dich (A). Cho vao dung dich (A) 200 ml dung dich K3PO4, phan iintxay ra vita du, thu dugc ket tua (B) vd dung dich (C). Klioi lugng ket tua (I'

2SO2 + Ca(HC03)2

> Ca(HS03)2 + 2CO2T

(c6 khi bay ra)
c)

'

CH2 = CH2 + Br2

> CH2Br-CH2Br

(mat mau ndu do dung dich nifdc brom)
3. Thiic hien day chuyen hoa:
Pe

y i _ > FeClg

vd khoi lugng mudi nitrat trong dung dich (A) khdc nhau 3,64 gam.
1. Tim nong do mol/lit cua dung dich (A) vd (C), gid thiet the tich dwni

> CaSOgl- + 2C02t + H2O

(cd ket tua va c6. khi)

Ve tinh chat: (M), (N), (R) cd phdn itng vai Na, tic 0,1 mol (M) c6 the

ling vai CI2 (chieu sang). Xdc dinh cong thitc cdu tgo cua (M), (N), (P'.

> FeS04 + C u i

CUSO4

- Dung dich mau xanh bi nhat dan.

b)

kien.

(Q), (R) vd cong thUc phdn tii cua (X). Gidi

tit cacbon

L d l GIAI

2,3 gam (N) hay 1,5 gam (Q) c6 the tinh bdng the tich cua 1,6 gam 0^
-

so nguyen

hidrocacbon

2. Viet phan ijfng:

Ve khoi lugng phdn tit (klpt):

cung dieu

CJl2„+2-

1. Ten goi cac v i t r i :

2. (M), (N), (P), (Q), (R), (X) Id nhUng hap chat hUu ca dugc biet den

klpt (N) = -klpt
2

biet the tich

biet rhng cd 2

tit cacbon bdng nhau vd bdng

CO the kill} dugc CuO a nhiet do coo, C la thdnh phdn quan trgiii:
trong phdn bon hoa hoc. Xdc dinh cong thilc phdn ti2 cua A, B, C, vii')

H2SO4

khoi lugng dung dich NaOH tang 7,04 gam.

2 Xdc dinh cong thiic phdn ti2 3 hidrocacbon,

Cdu III.

-

(du) thdy khoi lugng

Hay tinh thdnh phdn % theo khoi lUgng cdc chat c6 trong dung die/,
Viet cdc phiiang trinh phdn ling.

the khi:

:

> Fe(0H)3
Fe(0H)2

FeCl2

Fe203
> FeS04

Fe
> FeCl2.

Phan ling:

dich khong thay doi do pha trgn vd the tich ket tua khong ddng ke.
2. Cho dung dich NaOH (lay du) vao 100 ml dung dich (A) thu dugc kH
tua (D), Igc lay ket tua (D) roi dem nung den khoi lugng khong do'

1) 2Fe + 3CI2

2FeCl3

2) FeClg + 3NaOH

> Fe(OH)34 + 3NaCl

can dugc 2,4 gam chat rdn. Xdc dinli kim logi trong mudi nitrat.
40

1 ^ 1 niAi np T M i u n r c i u u mni un/t unr Q

nau do
THI HQC SINH Gini

HfiA

H n n

Q

"

41


Fe + 2HC1



> 2Fe + 3CO2T

4) FeaOg + SCO
5)

Cac p h a n ufng d i e u che:

-> FeaOa + SHgO

3 ) 2Fe(OH)3

.

FeCl2 + H2T

-

6) FeCl2 + 2NaOH



> F e ( 0 H ) 2 i + 2NaCl
tvang

7) Fe(0H)2 + H2SO4

xauh

m

> FeS04 + 2H2O

8) FeS04 + BaCla

10*
10
T h e o de: ncao = — x 1000 = - — (mol); n„
56
56

=

1000

10^

250

250

- C l a t h a n h p h a n q u a n t r o n g t r o n g p h a n b o n => (C): N2.
- ( A ) k h i c h ^ y tao CO2 ( M A = 28) => (A): C2H4: ( e t y l e n ) .

C"(o")2

250

(1)

n

m

250

X

56

250

10*

10^^

56

CalOH)^ d i i

100 = 0,49%

%m„

250
12918

Ca(OH).^dLJ
^

=

1 0 0 %

2,3
0,05
1,5
0, 05

= 46

(gam)

= 3 0 (gam)

7 4 *

(0,353%

3,36 l i t H2

3,36
22,4

= 0,15

(mol)

= 2 : 3 => ( M ) l a ri/gu 3 I a n r u g u (ri/gu 3 chiirc)

=^ ( M ) : CH„ - C H - C H ,
I
I
IOH
OH OH

(mol)

X

mol M + N a

> nM :

12918

P CO cong thufc C2H6O n h i f n g k h o n g p h a n ufng v d i N a v a N a O H .

(gam)

=^ (P): C H 3 - O - C H 3 (ete).
R viia p h a n ufng v d i N a v a viTa p h a n ufng v d i N a O H .

X

1 0 0 =

=> C o n g thufc (R): CH3COOH ( M R = 60g)

1 1 , 6 4 %

1 1 1 x 1 0 0 0
-

(mol)

T h e o de: p h a n ttf k h o i cua ( N ) = - k h o i liJgng p h a n t i i cua M
2
p h a n tijf k h o i ciia ( M ) l a : 2.46 = 92 (gam)
0,1

- 121 _ 1^

%m

p h a n ttf k h o i cua ( N ) :

= 0,05

(gam)

1 1 1 x 1 0 0 0

Ca(OH)^ d.r -

Do do:

32

M a t k h a c , ( N ) tac d u n g vdi N a n e n ( N ) l a : C2H5OH.



544
CaS04

n CO,,

( Q ) tac d u n g v d i CI2 chieu s a n g n e n ( Q ) l a : C2H6.

136 = 544

" 250

-> Cu + C O 2 T .

p h a n ttjf k h o i cua ( Q ) :

(mol)

10^

1,6

+ 2H2O

-> 2CO9T

T a c6:

(gam)

100 = 0,3539fc.

X1000

111

10'

CaSO,

%m

Tir(l)

0,392.10'^
Cu(OH),

2.

(mol)

X 98 = 0,392.10^

250

n CaSO,,
m

200,1

CO + CuO

> C u ( 0 H ) 2 i + CaS04
_ 10^

axetat)

III.

> Ca(0H)2

•^4

TCrd)

Cdu

(mol)

K h o i iLTcfng d u n g d i c h Boocdo t h u dirge l a : 100 + 1 + 10 = 111 ( k g )

.

> 2C2H5OH + 2C02t
len men giam
C2H5OH + O2
> CH3COOH + H2O
^ ^ ^
dam dac
C H 3 C O O H + C2H5OH
^ \
= =
^ CH3COOC2H5 + H2O

2 C 0 + O2 -

C a ( 0 H ) 2 + CUSO4

%m

riiau

P h a n ufng: C2H4 + 3O2

P h a n ijfng: CaO + H2O

m

len men

(etyl

(mol)

18



1. - A , B , C deu CO M = 28 dvC.
- B CO the k l i L f dagc CuO d n h i e t do cao v a k h i chay tao CO2 => (B): CO.

> FeCl2 + B a S 0 4 i

10^

,y

L6ni2*-'6

V a p h a n tuf k h o i cua (X) l a : M x = 3 M R = 3 X 6 0 = 180
+ 0,49%

+

11,64%)

=

87,517%.

(gam)

v a ( X ) d i e u che r a C2H5OH v a C H 3 C O O H => (X): C6H12OC (glucozcf).


Cdu

rv.

1. P h a n l i n g : 3 M ( N 0 3 ) 2 + 2K3PO4

> M 3 ( P 0 4 ) 2 i + 6KNO3

(li

Ta co:
B i n h d i i n g H2SO4 t a n g c h i n h la k h o i l u g n g cua I-I2O.

(Vdi M la k i m loai hoa t r i I I )
TU

m

p h a n ufng (1): Su k h a c n h a u ve kho'i l i j g n g la do t h a y 6 N ( )

(1)

ri ,

>

= —

372-190

M3(PO.,)2

^M(N03)^

=

^''M,(^O,).^

-

=

=

3

X 0,02

=

0,06

lirgng b a n g 2 I a n so m o l cua A .

(mol)

'KNOS

HA

= 0,3M

0,02
0,2

+

X

6
0,2

M(N03)2 + 2NaOH

(mol)

> M(0H)2l + 2NaN03

0,03

—>

0,03

Tii (2), (3)

=:>

nc = 0,03 ( m o l )

=>

nB = 0,06 - 0,04 = 0,02 ( m o l )

(2)

%VA =

Vay:

0,03

M(0H)2

= 0,01 ( m o l )

=

Vi la chat k h i nen % V = % n

2 . P h a n iJfng:

(mol)

M O + H2O

%VB

,

(3)

X 100 = 16,67%;

0,06
0,02

=

0, 06

X 100 = 33,33%

%Vc = 50%.
. T i m C T P T cua 3 h i d r o c a c b o n :

0,03

TCr (1), (2), (3) :=> ^ n^o^ = 0,01n + 0 , 0 2 m + 0,03k = 0,16

muo = 0,03 x ( M + 16) = 2,4
=> M = 64; D o n g (Cu).

n + 2 m + 3 k = 16

Cdu V.

T a c6: n hon hep 3 khi =

V i m , n , k la n g u y e n v a chSn n e n n g h i e m h g p l i l a : n = 2; m = 4; k = 2
^

" ^'^^

3 h i d r o c a c b o n c a n t i m l a : ( A ) : C2H6; (B): C4H8 v a (C): C2H2

1. T i n h p h a n t r a m the" t i c h m o i hidrocacbon.

DE SO 10

Cac p h a n ufng:
C„H2n.2 +

O2

\

'—^

CkH2k-2 + [ ^ ^ J

nC02 + ( n + D H g O

flE THI CHQN HOC SINH GIQ! HOA HOC 9, TJNH BJNH OjNH NAM HOC 2001 - 2002
(D

J

a
C„,H2:n + —
2

an

2.
(21

Sue a (mol)

Tinh

^

kC02 + ( k - D H s O

(3,*

3a

.

3ak

3a(k - 1)

bleu dien

® " 3. iV/ue^ phdn
ling xong

KNO3, BaSO^.

COo vdo dung

so mol CaCOs

Ve duang
O2

CO2 vd HoO, hay trinh

^ chdt rdn: K2CO3, BaCOj,

a ( n + 1)

mCOs + m l l a O

O2

1- Chi dugc dung

phdn
(mol)

(*)

Do la c h a t k h i n e n n , m , k < 4 va m , k > 2.

D a t : A : C„H2n+2 ; B : CniH2,n ; C: CkH2k;-2-

(mol)

= 0,16 ( m o l )

44

Theo de: nc = 3nA, t h i so m o l H2O g i a m v a n h o h a n so' m o l CO2 m o t
0,16-0,14

Vay:

7,04

= 7,04 g a m => n CO,

m.

0,02 ( m o l )

=

= 0,14 ( m o l )

18

B i n h d u n g N a O H t a n g c h i n h la k h o i l i f g n g ciia CO2.

( M = 372) b a n g 2PO4 ( M = 190).
^

2,52

= 2,52 g a m => n H . , 0

H,0

tgo thdnh

m 1 gam
ta thu dugc

I

Viet phdn

biet 4 Ig

iCng de niinh

chila

hga.

chi'ia 1 mol CafOWs.

-

ting vai gid tri a = 0; a = 1; a ^ 2.

so mol CaCOs

tgo thdnh

hSn hgp Mg, MgCOs
7712 ga77i

lugng Mg t7-o7ig hdn hap
: THI

dich

bay each phdn

ddu.

theo so mol CO2 da
ngodi

mot chat ra'n. Biet

khong

cho.

khi den

7711 = 77x2.

Tinh

khi
%


If) Chon 3 diem:

Cdu 4. Hoa tan hon hap Na20, NaHCOs, BaCl2, NH4CI c6 cung so mol vaa
nuac du, dun nong nlie thu dugc dung dich A vd ket tiia BaCOs- Hoi
dung dich A chiia gi? Viet phdn iCng ininh hga.
Cdu 5. Trgn 11,2 gam bgt Fe
khong CO khong khi de phdn
chat rdn tim dugc trong chen
HCl IM, thodt ra a (mol) hon

vd 4 gam bgt S trong chen sii dem ?iuiig
iCng xdy ra tao FeS vai hieu sudt 80%. Lay'
sii cho tdc dung viia du vai V lit dung dich
hap khi vd m (gam) chat rdn khong tan.

a) Viet tat cd phdn ling xdy ra.

C02

n C02

^' "^CaCO.j

= 0

•'

'•'CaCO;,

= 1

~ ^

^CaCOg

= 0

"

C&u3.
Phan ufng xay ra:

b) Ttnh gid tri V, a, m.
6. Dem hSn hgp gom

Cdu

lugng H2SO4

d4c,

nong

Mg+
0,1 mol Mg vd 0,2 mol Al tdc dung

viia du thu dugc hon hgp mudi,

0,075 mol S va

MgCOg

thdnh.

a
b

1
Lay moi lo mot it cho vao 4 co'c.

-

Che niidfc vao 4 co'c, phan diioc 2 nhom: nhom (I) t a n : dung dich

m •Mg
m^MgCOg
.

11.24 _ 264
4.84

NaaO + H2O

Sue k h i CO2 vao 2 coc nhom ( I I ) :

NaliCOg + NaOH

NH4CI + NaOH

Lay i t dung dich Ba(HC03)2 nho vao 2 coc cua nhom (I).
Neu coc nao tao ra ket tua trSng do la coc chufa K2CO3, phan ufng tao
raBaCOg.

Cdu

> NagCOg + H2O
-> BaCOgi + 2NaCl

-

NaCl + H2O + N H g t

(1)
(2)
(3)
(4)



FeS

(1)

Vi CO hieu suat 80% nen chat r a n gom: FeS, Fe di/, S diT

2.

FeS + 2HC1

a) T i n h so mol CaCOg: K h i a = 0 khong c6 phan ufng, so mol CaCOg = 0.
K h i a = 1 phan ufng xay ra:
> CaCOg + H2O

(D

=> So mol CaCOg = 1.

Fe + 2HC1

> FeCl2 + HaSt
> FeCl2 + H2t

(2)
(3)

^ ) Theo (1), neu hieu suat bkng 100% t h i S het, Fe dif.
K h i hieu suat 80% t h i : S d i i = 0,8 gam = a; Sd mol S phan ufng = Sd
oiol Fe phan ufng = So mol FeS sinh ra = 0,1 mol.

K h i a = 2 phan ufng xay ra:
2CO2 + Ca(0H)2

Klio'i liigng M g = 44%

Vi so mol cua 4 chat: NagO, BaClg, NaHCOg, NH4CI bSng nhau, nen
theo (1), (2), (3), (4) dung dich A chi chila NaCl.
5.

a) Phan ufng: Fe + S

> BaCOg + 2KHCO3

Coc khong ket tiia chufa KNOg.

CO2 + Ca(0H)2

~ 336

2NaOH

BaCl2 + NaaCOa -

> Ba(HC03)2

Coc khong tan chufa BaS04.

K2CO3 + Ba(HC0g)2

(2)

Cac phan ufng xay ra:

Neu tan la coc chtTa BaCOs, phan tfng xay ra tao ra dung dich BadlCOa)^:

Cdu

11

K2CO3, KNO3; nhom (II) khong tan; BaCOg, BaS04.

CO2 + H2O + BaCOg

MgO + C02t

Can 4.

-

-



Vi m i = m2 nen khoi lirgng O2 (1) = khoi liigng CO2 (2) =^ 16a = 44b

LCil GIAI

-

(1)

>oi a, b Ian iLTcft la so mol cua M g va MgCOg trong hSn hgp dau.

h) Tinh sd mol H2SO4 phdn iJCng viia du.
Cdu

MgO

vdi mot

0,175 mol SO2.
a) Tinh khdi lugng hdn hgp mudi tgo

^02

V a y s d m o l Fe = 0,1 mol.
> Ca(HC03)2

(2'

(2) va (3)

^

Sd mol HgS = Sd mol FeS = 0,1 mol.

=> So mol CaCOg = 0.

Sd mol H2 = Sd mol Fe = 0,1 mol =:> a = 0,2 mol, sd mol H C I dung la

Triicfng hop 2 c6 the viet phan ufng hoa tan het CaCOg.

0.4 mol

V = 0,4 l i t .


Co 6 to mat nhan diCng cdc dung dich khong

Cdu 6.

mdu Id: iVa^SOj (1);

MasCOs (2); BaCls (3); BafNOaJs (4); AgNOa (5); MgCls (6). Bdng

a) T i n h khS'i Itfgng hon hcfp muol:

phuang

CiJ 0,1 mol M g tao ra 0,1 mol M g S 0 4 (bao toan k h o i lirgng)

plidp hoa hoc vd khong dung them cdc hoa chat khdc hay trinh bay each

Cur 0,2 mol A l tao ra 0,1 mol Al2(S04)3 (bao toan k h o i lifgng)

nhan biet cdc dung dich tren, biet rdng chung deu c6 nong do du Ian de

Vay k h o i li/crng hon hgp muoi = (120.0,1 + 342.0,1) = 46,2 gam

cdc ket tua it tan cUng c6 the tao thdnh. (Khong

b) T i n h so mol H2SO4 dac n6ng da dung vCra du.

phdn

Phan ufng xay ra:

^°<^ dinh nong do cua cdc muoi NaHCOa vd Na2C03 trong mot

> 3MgS04 + S i + 4 H 2 O

(1)

dung dich

M g + 2H2SO4

> MgS04 + S 0 2 t + 2H2O

(2)

nghieni nhu sau:

2A1 + 4H2SO4

> Al2(S04)3 + S i + 4H2O

(3)

Thi nghiem

2A1 + 6 H 2 S O 4

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

(4)

dich HCl IM (du) dun nong hdn hgp, sau dd trung hoa lugng axit du

Theo (1) va (3)

^

^H^SO,

dung

= ^.ng = (0,075 x 4 ) mol = 0,3 (mol)

Theo (2) va (4) => n^^^gg^ ^^^^ = 2n^^^ = (0,175 x 2) mol = 0,35 (mol)
Vay so' mol H2SO4 da dung viTa du 1^ 0,65 (mol).

D E S 0 11
D E THI HOC SINH Gidl HOA HOC 9 (VONG 1), TJNH KHANH HOA N A M HOC 2001 - 2 0 0 2
Cdu I.

hdn hgp cua chung

(dung

dich A), nguai

1: Lay 25 ml dung dich A cho tdc dung

bdng lugng vita du la 14 ml dung dich NaOH

ta lam cdc thi

vai 100 nd dung

2M.

Thi nghiem 2: Lai lay 25 ml dung dich A, cho tdc dung vai lilang du dung
dich BaCl2. Loc bo ket tua mdi too thdnh, thu lay nUdc Igc vd nuac nla gap
Igi roi cho tdc dung vai lugng vica du la 26 ml dung dich HCl IM.
1. Viet cdc phuang

trinh phdn ling xay ra vd gidi thich vdn tdt.

2. Tinh nong do mol cua moi muoi trong dung dich A.
Cdu 4. Mot dung dich axit axetic CH3COOH

cd C% = 10%. Lay 300 gam

dung dich axit nay cho tdc dung vai 300 ml dung dich NaOH 2M tao ra

1. Hay thuc hien
phep tinh):

cdc phep tinh sau (yeu cdu ghi day du dan vi troii;;

a) Tinh so mol do trong 7,19 gam do;
b) Tinh

trinh

ling).

Cdu 3-

3Mg + 4H2SO4

can viet phuang

so mol O vd O2 trong

8 gam oxi;

dung dich A. Dung dich A cd tinh axit hay baza?
Tinh nong dd % cdc chat tan trong dung dich A, biet rdng dung
NaOH2M

cd d = 1,2

dich

giml.

c) Tinh khoi lugng cua 0,05 mol kem;

Ghi chu: Hoc sink dugc phep svt dung bdng tudn hodn, bdng tinh tan, gido vien

d) Tinh khoi lugng cua 0,75 mol nudc;

coi tin khong gidi thich gi them.

d) Tinh so nguyen

g) Tinh so phdn tvC CO2 trong 1,1 gam khi CO2;
h) Tinh so g cua 1 nguyen tii Na;
i) Tinh so g cua 1 phdn tii SO2.
2. Mot hdn hgp X gom FeO vd Fe203
hon hgp nay trong

LCll GIAI

tii C trong 0,02 gam C;

cd khoi lugng Id 30,4 gam. Nun.^

mot binh kin cd chica 22,4 lit CO (dktc). Kho'

lugng hon hgp khi thu dugc sau khi nung la 36 gam.
a) Hay xdc dinh thdnh phdn hon hgp khi. Biet rdng hdn hgp X b\
khv[ hodn toan thdnh Fe.
b) Tinh khoi lugng Fe thu dugc vd khoi lugng cua mSi oxit sdt trong X.

Cdu 1

7
1
1. a) nci = —-— (mol)
35,5

'

8
8
b) no = — (mol); n^^ = — (mol)
c ) mzn = 0,05 X 65,38 = 3,27 gam
d ) mj^^Q = 0,75 mol x 18 g/mol = 13,5 gam

iSlSi^li^ THI HOC SINH G|6'HOA

HQC 9


d ) nc =

Tii bang t r e n t a thay:

= 1,66.10^^ (mol)

D u n g dich n a o cho v a o t a o r a 4 I a n ket t u a l a d u n g dich Na2C03 va

So n g u y e n tuf C = 6,02.10^^ n g u y e n ti:f/mol.l,66.10^ m o l

• AgNOa (cap d u n g dich 1). D u n g dich n a o cho v a o t a o r a 3 I a n k e t t u a

= 9,993.10^° n g u y e n tuf.
g) n

la dung dich Na2S04 v a BaCla (cap dung dich 2). D u n g dich n a o cho

1,1
= 0,025 (mol)
44

CO2

vao t a o r a 2 I a n k e t t i i a l a dung dich M g C l s v a Ba(N03)2 (cap dung
dich 3).

So p h a n td CO2 = 0,025 x 6 , 0 2 . 1 0 ' ' = 1,505.10^^ p h a n tit.
h ) MNa = 23 g/mol

2,

niNa.-

MgQ = 64,06 g/mol =>

a)

Ta

nco

( b a n ddu) =

=

1 mol

=>

64,06
-,23

6,02.10'

mco

+) L a y m o t trong h a i chat a c&p dung dich 3 I a n lirgt cho v a o 2 dung

= 3,82.10"^^ g/nguyen t t f

6,02.10'23

i)

c6:

23

dich 6 cap 2, n e u c6 t a o r a k e t t i i a : thi chat cho v a o l a Ba(N03)2, con
l a i l a MgCl2.
C h a t t a o r a ket t u a a cap 2 l a Na2S04, c o n l a i l a BaCl2.

= 1 0 , 6 4 . 1 0 ' " ' g/nguyen tuf

+) L a y Ba(N03)2 d a tim diTOc a cap 3 cho vao h a i d u n g dich d cap 1,

( b a n d s u j = 28

neu CO ket t u a t h i : C h a t tao r a ket tiia vcfi Ba(N03)2 l a Na2C03 con

gam.

Do t a n g k h d i liiOng: 36 - 28 = 8 gam = mo => n^^ = 0,5 mol.

lai l a AgNOa.

V a y CO 0,5 mol C O ket h o p v d i 0,5 mol O cho r a 0,5 mol CO2.

Cdu 3

T h ^ n h p h a n h o n hap k h i l a : 0,5 mol C O v a 0,5 mol CO2.

Bat so mol Na2C03 v a NaHCOa trong 25 m l dung dich A I a n liTot l a x, y.

b ) T a c6: n^^ (c6 t r o n g FeO v a Fe203) = n^^ l a y r a ( d t r e n ) = 0,5 m o l .

Doi v(5i t h i nghiem 1, t a c6:
Na2C0a + 2HC1

G o i a = npeo v a b = np^^p^. V a y t r o n g X c6 (a + 3 b ) m o l O.
D o do:

a + 3 b = 0,5

(1)

Va:

72a + 160b = 30,4

(2)

G i a i (1) v a (2) dirge: a = 0,2 m o l FeO v a b = 0 , 1 m o l FeaOg.
Vay:

mpeo = 0,2 x 72 = 14,4 g a m
"^Fe.,03

= 0,1

X 160

=

16

(mol)

1^

( t h u dLToc)

nipe

( t h u duoc)

x

k

> N a C l + H2O

(3)

0 , 1 x 1 = 0 , 1 mol

So mol H C l da tac dung vdri dung dich A l a :
(4)

Boi vdfi t h i nghiem 2:

Cdu 2

BaCl2 + NaaCOa

1. L a y m o t d u n g d i c h b a t k i cho v a o 5 d u n g d i c h c o n l a i , t a c6 b a n g sau:

50

y

2x + y = 0,1 - 0,028 = 0,072

56 = 22,4 g a m .

Na2S04

Na2C03

BaCl2

Ba(N03)2

AgNOa

MgCl2

Na2S04

-

-

i

i

i

Na2C03

-

-

i

i

i

-

BaClz

i

i

-

-

i

-

Ba(N03)2

i

i

-

i

i

-

-

AgNOa

-

MgCl2

-

i

-

i

-

-

(2)

So mol H C l dir sau p h a n ufng (1) v a (2) l a : 0,014 x 2 = 0,028 mol.

= npe ( t r o n g FeO) + npe ( t r o n g Fe203)
= 0,4

> N a C l + CO.T + H2O

So mol H C l trong 100 m l dung dich l a :

= a + 2b = 0,4 mol.
Vay:

y

H C l (dii) + N a O H

H||

T h e o d i n h l u a t bao t o a n n g u y e n t o t h i :
npe

(mol)

(1)

2x

NaHCOa + H C l

I

gam.

X

> 2 N a C l + COgt + H2O

i

LOI GIAI at THI HOC SINH GIQI HOA HOC 9

> BaCOai + 2NaCl

(5)

Sau k h i loc bo ket tua, lay niidfc loc, ni/dc riia chufa N a H C O a cho tac dung
vefi dung dich H C l .
||^|^'

^



NaHCOa + H C l
(mol)

y

> N a C l + CO2 + H2O

y

^ i a i r a t a c6: y = 0,026 x 1,0 = 0,026 mol
x = 0,023 mol
nong do mol cua NaHCOa l a :

0,023 : 0,025 = 0 , 9 2 M .

nong do mol cua N a H C O g l a :

0,026 : 0,025 = 1,04M.

i J ^ i i l i l n f ....

(6)


Cdu

IL

4
300 g a m C H 3 C O O H

^

300 m l d u n g d i c h N a O H 2 M chufa

nNaOH

P h a n l i n g : CH3COOH + N a O H
0,5

= 0,3

x

icng dieu che

CaCOs.

vdi Mg thi khi dot chdy hap kim do, oxlt tqo nen c6 khoi

2 = 0,6 m o l .

doi khoi lugng

> C H a C O O N a + H2O

0,5

tri/ih phan

Cho 4 kim loai sau: Be, Na, Al, Ca. Hoi kim loai ndo khi tqo hap kim

\

So m o l C H 3 C O O H = 0,5mol.

(mol)

5 phuang
a) Viet
V

i

1 0 % chOra 30 g a m C H 3 C O O H

Cho

hgp kim ban ddu.
Kim

T i le p h a n lifng l a 1 : 1.

loai

Hda

V a y sau p h a n ufng con N a O H = 0,6 - 0,5 = 0 , 1 ( m o l )

tri.

Khoi

lugng

nguyen ti'i

Be

Na

Al

Ca

H

I

HI

H

9

23

27

40

d u n g d i c h A c6 t i n h bazcf.
c) Dot chdy
phdm

m ( d u n g d i c h A ) = m (dung d i c h CH3COOH) + m ( d u n g d i c h N a O l I
= 300

C% ( C H g C O O N a ) =

+ Vd,.>gdkh

0,5

X

82

X

X d = 300

100%

660

X 1,2

= 660

du

gam

Tinh

= 6,21%

Cdu

660

0,6%.

todn

tqo thdnh
khoi lugng

HI. Dot chdy

dich

5,6 lit (dktc)
dich

NaOH

inol trung

2M de tqo thdnh
cua mpt kim loqi

chi cd hai hda' tri 2 vd 3) trong

tqo nen c6 nong

Idng

Lam Iqnh

xud'ng

vd dung

muoi Id 23%. Xdc dinh cong thUc cua tinh the

DE THI HOC SINH GIQI HOA HOC 9, CAP TP. HO CHJ MINH NAM HOC 2002 - 2003

Cdu

muoi

trung

hda.

MeS (kim loqi

rV. Cho m gam hon hgp CaCOs

Me

oxi du. Hda tan chdt

viia du dung dich H2SO4 29,4%.

dp 34,5%.

the hidrat

khi nay tdc dung viCa

binh ciia hon hgp khi ban ddu.

ilng bdng mpt lugng

2,9 gam tinh

hdn hgp khi gom H2 vd H2S. sdn

la nuac vd mpt chdt khi. Lugng

1,76 gam sunfua

cdc hgp chat

rdn sau phan

DE SO 12
Cdu

hodn

vdi 100 ml dung

trong

= 2 i i L f ^ ,

«(NaOH)

+ 300

gap

biet:

0,5

T i n h k h o i lufOng d u n g d i c h A :

lugng

dung
dich

dich

Dung

thi thu

dugc

con Iqi cd nong

dp

hidrat.

vd FeS tdc dung

vUa du vdi V ml

dung dich HCl (D ^ 1,1 g/ml) thi thu dugc a lit hon hgp klii X (dktc) c6

I.

a) Cho biet dp tan mot loai muoi
neu lam Iqnh
thi phqm
nhiet

dung

c6 nong

dp % ciia muoi

dp theo bang sau
do trong

khodni^

do ndo?
dp

Dp tan (trong
b) Co 6 dung

lOOg

dich

nicac)

dugc ddnh

mpt chdt gSm: BaCh,

n,3g

10° C

20°C

30°C

40°C

15,2g

18,9g

24,8g

36,7g

so ngdu nhien
H2SO4,

luat thuc hien cdc thi nghiem

NaOH,

tic 1 den 6, moi dung c/?'
HCl, MgCl.,

NazCOa.

i

nghiem

1: Dung

dich (2) cho ket tua vai cdc dung dich (3) vd (4'i'

Thi

nghiem

2: Dung

dich (6) cho ket tua vai cdc dung dich (1) vd ^4'

Thi

nghiem

dung

dich

(4) cho khi bay len khi tdc dung

dich (3) vd (5).

Hay xdc dinh s6 cua cdc dung

dich.

vai

mol trung

binh

Id 40,67

(gam/mol)

vd dung

dich

Y cd khoi

ugng la b gam.
<.) Tinh

a theo m, V, b.

b) Ap dung:

Cho m = 1,44 gam; V = 400 ml; b = 440,83
mpt dU kien

^^0,67 (gam/mol),

La>-

ud dugc ket qua sau:

lugng

c) Neu chi sii dung

Thi

3: Dung

khoi

do Id 22% tie 50°Cl

vi de bdt ddu xudt Men ket tinh cua muoi

Nhiet

chiia

dich

bii'n doi theo nhiet

Cdu

khoi

hay tinh % theo khoi lugng

V. Mpt hon hgp A gom Na2C03

(iich

HCl du thu dugc

56 (gam/mol).
^"(0H)2du
Tinh

vd NaoSOs

hon hgp khi X cd klidi

C/io 0,224 tit (dktc)

•Sau thi nghiem

lugng

phdi

diing

gam. Tinh

mol trung

dich

ciia X la

cua hdn hgp ban ddu.
cho tdc dung
lugng

vdi

mol trung

khi X di qua 1 lit dung

50 ml dung

binh

a.

dich

dung

binh Id
Ba(0H)2-

HCl 0,2M de trung

hda

'

% so mol mdi khi trong

X vd % khoi

lugng

moi chdt

trong

hon

hgp A.
Tinh

nong

dp mol/lit

''^'10 sd lieu: H= 1,0=
VTHI HOC

S I N H Rini HOA

HOC

cua dung dich Ba(0H)2

ban ddu.

16, Na = 23. S = .32, CI = 35,5, Ca = 40, Fe = 56, Ba = 137
9

53


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