Boi duong hoc sinh gioi hoa hoc 8+9 dao huu vinh p3

mjd

sau phan ling =

rrijjCa(OH)2 + "^002

=

~

"^CaTO^

1000. 1,025 + 44. 0,075 - 100. (0,05 : 2) = 1025,8 g

^

n
n B,2=0,l.

162.M5
N o n g do m u o i CaCHCOj). trong dung dich la:
Cdch 2:

CO2
X

+Ca(0H)2

mol

2CO2

X

> CaCOjto + H2O

mol
>

y mol

nco2 = x + 2y = 0,075 1

. 100 = 0,4%.
v , „ . _ , . , , .

x mol

+ Ca(OH)2

2y m o l

^^^^^^

V a y A l a C2H4.

+

_

,^

0,05 m o l

0,05 mol

12 8 8
mc = - ~ 44

H2O

+

0,025 m o l

CaCOj

+ H2O

'

2 5 4

= 2,4 g; mn =

18

' *•

= 0,6 g; mo = 3 - 2,4 - 0,6

b) Dat cong thiJc A la C,Hy ta c6:

x : y =

mc
12

mH_
•

1

Ta CO ( 1 2 + 3 ) n < 40

0,025 m o l

(chii y ncaco3 = 0-0-'' - 0-025 = 0,025 mol)

Nghiem thich h d p n = 2

12

'

Cong thtfc p h a n tur A : C2H6.

d)

1 . D o t chay 12,6 gam hon hdp ba k h i CH4, C2H4 va C2H2, ta thu difdc 39,6 gam
CO2. N e u cho 12,6 gam hon hdp tren qua binh diTng dung dich brom thi c6 80

gam b r o m tham gia phan tfng. Xac dinh thanh phan phan t r a m k h o i Wdng
m o i hidrocacbon trong hon hdp.
2. D o t chay 11,2 l i t (do d dktc) hon hdp metan va etilen r o i cho san pham tac
dung v d i dung dich N a O H sau phan tfng ngi/di ta thu difdc 250 m l dung dicb
Na2C03 2,6M. Xac dinh thanh phan phan t r a m theo the tich m o i k h i trong
hon hdp.

T"

•

C H 3 - C H 3 + CI2

c 6 n g thQc p h a n ttf

' '

""""''"^ )

'iiu:imh:;.:r>-f./

CH3-CH2-CI

+HC1

3. K h i do't c h a y h o a n t o ^ n m o t t h e t i c h h i d r o c a c b o n A c a n 6 t h e t i c h o x i , s i n h r a
4 t h e t i c h k h i CO2.

Xac d i n h c o n g thtfc p h a n tuf c u a A. B i e t c a c t h e t i c h d o d

c u n g d i e u k i e n n h i e t d o v a a p sua't.
A l a h i d r o c a c b o n m a c h t h i n g . Hay v i e t c o n g thuTc cafu t a o c u a A.

Hi^dng d i n giai
O C l i n g d i e u k i e n v e n h i e t d o v a a p sua't, t i 16 t h e t i c h c a c c h a t k h i c u n g la t i

1$ so m o l .
Ne'u g p i c o n g thiJc c u a A l a C^Hy.

hpp. U p

~

0,6

n < 2,67

c) A k h o n g l a m m a t m a u d u n g d i c h Br2.

I I . Bal tap t i / g i i i

X ^ c djnh th^nh ph^n h6n

2,4

Cong thu-c A CO d a n g (CH3)„.

> Ca(HC03)2

T i e p tuc tinh m j j sau phan tfng va nong do Ca(HC03)2 nhiT each 1.

D a n g 4:

=0

V a y A chi chiJa 2 nguyen to C va H .

0,05 mol

0,025 m o l

-'^

d) V i e t phu'dng trinh hoa hoc cua A v d i clo k h i c6 anh sang.

phan iJng :
+

'
"

a) Tinh k h o i lu"dng cac nguyen to'co trong 3 g A

CO2 con du" 0,075 - 0,05 = 0,025 m o l l a i hoa tan CaCOs theo phuTdng trinh
CO2

•

3) Trong chaft hifu cd A c6 nhi?ng nguyen to nao ?

Hvtdng dan giai

y = 0,025; x = 0,025

> CaCOjio

-

c) Chat A CO l a m mat mau dung dich brom khong ?

Ca(HC03)2
y mol

Ca(OH)2

,o.u

2, D o t chay 3 gam chat hffu cd, thu duTdc 8,8 gam k h i CO2 va 5,4 gam H2O.

T i n h TUM sau phan tfng va nong do Ca(HC03)2 trong dung dich nhiTcach 1.
CO2

: ng^^ = 0,01 : 0,01 = 1 : 1

b) B i e t phan tuf k h o i cua A nho hdn 40. T i m cong thiJc phan tuT cija A.

'5

nca(0H)2 = x + y = 0,05

Cdch 3:

= 0 , 0 1 mol;

'

Tii phu'dng t r i n h p h a n vtng do't c h a y , v d i so' n g u y e n tuT c u a m o i n g u y e n to' 6
h a i ve' c u a phu'dng t r i n h p h a i b a n g n h a u , ta t i m diTdc g i a t r i c u a x , y .

I . Bdl tap CO Idfl giSl
1 . B i e t 0,01 m o l hidrocacbon A l a m mat mau vCfa dii 100 m l dung dich brom
0,1M. V a y A la hidrocacbon nao trong so cac chat sau :
a) CH4

b)C2H2

c)C2H4

d) CsHs.

CxHy + 602

>4C02 + y/2H20

So n g u y e n tijT C, O d h a i ve' c u a phiTdng t r i n h :
nc = x = 4 ; n o = 6 . 2 =

4.2 + y / 2 s u y r a y = 8.

'
•

'

-

'

Cong thiJc p h a n tuT c u a h i d r o c a c b o n A l a C4H8.
205


Cong thtfc cau tao ciia A c6 the la:
C H 3 - C H 2 - C H = C H 2 hoac la C H 3 - C H = C H - C H 3 .
4. Dot chay hoan toan mot liTctng hidrocacbon X, ngu-di ta thu dxidc 22 gam COj
va 13,5 gam HjO. Biet khoi Itfdng phan tur ciia X la 30 dvC. Tim cong thiJe
phan tuf cua X va viet cong thiJc cau tao cua no.
Hifdng d i n giai

Tuf khoi lu'dng cua CO2, H2O ta suy ra so mol. Neu goi cong tMc cua X 1^
C,Hy tiJf phU'dng trinh phan tfng do't chay X se tim diTdc moi quan he x vdi y
va tiJf Mx suy ra gia tri bang so' cua chiing.
Cdch 1: nco2 = 22 : 44 = 0,5 mol;
= 13,5 : 18 = 0,75 mol
CMy
+ (x + y/4)02 > X C O 2 + y/2H20
1 mol
mol
0,5y mol

0,5 mol
0,75 mol
Ta c6: x : 0,5 = 0,5y : 0,75 => 0,75x = 0,25y => 3x = y
(1)
Mx = 12x + y = 30
Giai phming trinh (1), (2) ta diTdc x = 2 va y = 6
Cong thu-c phan tuf cua X la C2H6.
Cong thtfc cau tao CH3 - CH3.
^ ^^ ''
Cdch 2 : nc = n^oj = 0-5 mol;
2nH20 =
2 = 1,5 mol
U H J O

X

U H =

CxHy => X : y = Uc : HH = 0,5 : 1,5 = 1 : 3.

Cong thiJc dc(n gian nhat cua X la CH3. Cong thtfc phan tuf la
(CH3)„ = M = 30 => 15n = 30 ^

n = 2.

Cong thiJc phan tuf cua X (CH3)2 hay CsHfi,
Cong thufc cau tao CH3-CH3.
'''
5. Dot chay 3 gam mot chat hffu cd A thu dxidc 6,6 gam C O 2 va 3,6 gam H2O.
a) Xac dinh cong thtfc cua A, bie't khoi lu'dng phan tuT ciia A la 60 dvC. '

b) Viet cong thiJc cau tao c6 the c6 cua A.
Hrfdng d i n giai

'

va m^^Q tim du'dc mc va mn; dxia vao djnh luat bao toan khoi Ixidng sc
suy ra trong A c6 nguyen to' oxi hay khong (vi dot A > CO2 + H2O ncn
trong A CO the c6 oxi).
mco2

Cdch

1:

nc

=

TICQ^ = 6,6

: 44

=

0,15

mol

mc = 0,15. 12= 1,8 g
2 n H 2 0 = 2.3,6 : 18 = 0 , 4 m o l = > m H = 0,4 g

Theo dinh luat bao toan kho'i lU'cJng ta c6:
m A = mc + m H + rno => mo = 3 - 1,8 - 0,4 = 0,8 g
206
UH

=

thtfc phan tuf cua A : C x H y O ,
12x
y
16z
60
mH
mc
mo
3
12x . y _ 16z
60
1,8
0,4
0,8
3
||ai c&c phuWng trinh tren ta diTCfc x = 3, y = 8, z = 1.
-'
: 6 n g thi?c phan tuf cua A la C^U^O.
Zung CO the tim cong thtfc dcJn gian nhS't cua A roi suy ra cong thi?c phan tuf:
m o _ 1,8 0,4 0,8
z = 12
1
16

12
1 16
= 0,15 : 0,4 : 0,05= 3 : 8 : 1
:6ng thtfc ddn gian nhat cua A la CsHjO ; Cong thufc phan tuf la (C3H80)„ = 60
(12. 3 + 8 + 16)„ = 60
n = 1
f^y cong thiJc ddn gian nhat cua A cung la cong thtfc phan tuf.
ich 2:
v6i 3 gam A tao ra 6,6 gam C O 2 va 3,6 gam H 2 O
Neu do't chay (1 mol) 60 g A tao ra x g CO2 va y g H2O
60.6,6
. 0 ^
,
= 132 g
3
;^6ng

X

=

y=

60.3,6

'

-

• nco2 = 132

= 72g

nH = 2 n H 2 0 = 4. 2 = 8 mol

'H2O

: 44

=3

mol

=

nc.

= 7 2 : 18 = 4 mol

Vay trong 1 mol A c6 3 mol C vk 8 mol H, cong thtfc phan tur cua A 1^:
CsHsO, = 60
•.rami
3. 12 + 8 + 16z = 60=> z = 1
Cong thiJc phan tuf ciia A la CiW.
C H 3 - C H 2 - C H 2 O H (1) CH3-CH(OH)-CH3 (2)
C H 3 - O - C H 2 - C H 3 (3)
Hdp chat \\\ixx cd A c6 thanh phan khoi lifdng cac nguyen to' nhiT sau:
C = 53,33%, H = 15,55%, N = 31,12%.
,
Xdc dinh cong thtfc phan tiJf cua A, bie't rkng A chi c6 mot nguySn tijf N trong

phan tuf.
Vie't cong thufc cau tao rut gpn ciia A, bie't N c6 hod tri III.
H\i6ng d i n g i a i
Cdch I :
a.a...-. KiNeu gpi cong thtfc phan tuf cua A la C x H y N , . Tuf thanh phan phan tram kho'i
IWng nguyen to ta tim dufdc ti 16 so' nguyen tiJf cac nguyen to x : y : z. ,
Sau d6 thay z = 1 ta tim diTdc cong thufc phan tuf.
207


X

^

XT

%C
—

XT

a) C H y N , : x : y : z =

%H
: —

Mc

:

MH

%N
—

•

^
=

12

'

4,444

1

^

14

CxH,

„:

: 15.55 : 2.222 =

2 : 7 : 1

^

b) Cong thiJc cau tao rut gpn cua A la: C H 3 - C H 2 - N H 2 hoSc la C H 3 - N H - C H 3

"CO2

(DCjH,

'^^^

12x
53.33
y

31,12.12x= 14.53,33

14
31,12
_

14

15,55 ~ 31,12

(2) C 2 H ,

(3) C2H4

C2H2

C2H2

C6ng thtfc cau tao:

Tu'dng tiT trong phan tuf CjHyN;, c6 xC cacbon; y H g hidro va N g nitd

53.33g

yg
15.55g

(1)

14g

CH3-CH3

3 1 . 1 2 y = 14.15,55

b) Xet c a p ( l ) :

=>x«2

C2Hfi + 7/2O2

=>y«7

C2H4

7. D o t chay 1,12 l i t hon hdp hai hidrocacbon (the k h i ) c6 ciing so nguyen [\i

cacbon, dan san pham phan uTng ch^y Ian liTcJt qua binh (1) diTng P 2 O 5 , binh

+

41

+302-

>2C02

Xet c a p (2)

c moi

a) Xac dinh cong thtfc phan tijT cua hai hidrocacbon trong h o n hcJp. V i e t cong

> 2 C O 2 + H2O
c moi

a + c = 0,05

thiJc cau tao cua chiing.

" H 2 0

J: N e u g p i cong thiJc cua hai hidrocacbon trong hon hpp la CxHy

->3a + c = 0,l

a = 0,025 va c = 0,025

M o i k h i c h i c m 5 0 % the tich hon hdp
X e t c a p (3). DiTa vao phan tfug (2) va (3)
'hh

(HjO

'H20

hap t h u , C O 2 b i K O H hap t h u . dp tang k h o i liTdng cua cac binh ciing

c + b = 0,05
->c + 2b = 0 , l

b = 0,05 ::> c = 0 v6 l i .

chinh la k h o i liTdng cua san pham phan iJng chay) ta t i m dirpc x (so nguyc"

Let luan: H o n hdp gom hai k h i C2H6 va C2H2.

tijr cacbon)...

pong Ihu-c cau tao: C H 3 - C H 3 va C H ^ C H

,.

^

a) nhh= 1,12:22,4 = 0,05 m o i

208

if':'

.1

v6 l i

•3a + 2b = 0 , l

(2) tang t h e m 4.4 gam.

Hifdng d i n giai

m

2b m o i

C2H2 + 5/2O2

b) Tinh thanh phan phan tram the tich moi khi trong hon hpp. The tich k h i do d dkic

(1)

+ 2H2O

a + b-0,05
"HJO

3H2O

3a m o i

(2) diTng K O H . Sau t h i nghiem k h o i IiTdng binh (1) tang t h e m 1.8 gam, binh

P2O5

->2C02

b moi

Cong thurc cau tao (rut gpn) la: C H 3 - C H 2 - N H 2 hoSc la C H 3 - N H - C H 3 .

C x H , , Tir phircJng trinh phan tfng dot chdy va k h o i lurpng san pham

CH=CH

CH^CH

a myl
>

(3) C H 2 = C H 2

(2) C H 3 - C H 3

CH2=CH2

31,12g

Cong thtfc phan tijr cua A ^ C 2 H 7 N .

Cdch

x = 2

= x a + xb = 0 , l

C2H4

Cvt lOOg A t h i C O 53.33 g C ; 15,55g H va 31.12g N

.

xb m o i

Hon hpp hai hidrocacbon c6 the la :

Cdch 2: Cong thtfc phan tur cua A la C^HyN.

12xg

XCO2 + Z / 2 H 2 O

(x + z/4)02

= a + b = 0,05

'hh

tur N (z = 1) nen cong thtfc D G N cung la cong thiJc phan tuf C 2 H 7 N .

-m|afeli§«0

+

b moi

• , ..„,

Cong Ihtfc ddn gian nhat ( D G N ) cua A \h C 2 H 7 N . V i phan tiJf chi c6 1 nguyf>

'>^-' -

xa moi

a moi

MN

53,33 . 15.55 . 31,12
-

-> XCO2 + y / 2 H 2 0

+ ( X + y/4)02

C,Hy

I 6 i k h i chiem 50% the tich hon hdp.

*

; ] ,5, ; r

mH20= ^0 tang k h o i IiTpng cua binh (1) = l , 8 g => nH^Q = 1,8 : 18 = 0,1 mol

tach

m c o 2 = ^9 tang k h o i liTpng cua binh (2) = 4,4g =0 Ucoj = 4,4 : 44 = 0,1

rong hon hdp, cong thiJc chung cua chiing la C x H -

2: N e u g p i y la so nguyen tuT hidro trung binh cua hai hidrocacbon
, .

...
209


C,H-

+

(x+

y/4)02

>

1 mol

XCO2
mol

0,5 y m o l

'

0,1 m o l

0,1 m o l

'*

X

0,05 m o l

Tir etan:

C H 3 - C H 3 + CI2

y =4

(mot hidrocacbon phai c6 so' nguyen lu'

CaC03

,

> CH3 - CH2 - OH + NaCl

Xac dinh cong thiJc phan tuf va viet cong thiJc ca'u tao rut gon cua X.

>

CaO + C02

p

;

'""^
(c)

>

v
^.rh

Ca(0H)2 + C2H2

>

3C2H2

k h o i liTdng la 44 : 27. Phan tiJ k h o i cua X la 30 dvC.

C6H5

'

'

,v•,,^.M
• ^

2. H d p chat hffu cd A c6 thanh phan k h o i lifdng cac nguyen to nhU' sau: C =

Bai 3.
1. Hay sap xep cac hidrocacbon cho du'di day theo thuT tu" tang dan nhiet do soi:

54,5%; H = 9 , 1 % ; O = 36,4%. Xac dinh cong thiJc phan tuT cua A biet rang
0,88 gam hdi chat A chie'm the tich 224 c m \

C4Hi(), CH4, C3HX, C2H6 va Ci{)H22,
2. Cho biet d 20"C benzen c6 k h o i lifdng rieng la 0,879 g/ml. N e u hoa long 7,8
kg hdi benzen xuong 20"C thi thu duTdc bao nhieu l i t benzen.

THI

B a i giai

V i ^ ' t c 6 n g thufc c a u t a o ,
V i e t phu'orng t r i n h p h a n ufng, Diiu

1. So nguyen tur cacbon cang nhieu, nhiet do soi cang tang, do do thu* M tang

che

nhiet do soi la C H 4 < C2H6 < CjHx < C4HU, < C,oH22.

B a i 1. Ngifdi ta c6 the dieu che metan tiJf C va H2 (co mat N i , t"), tif n h o m
cacbua (AI3C4) tac dung v d i nU'dc hoSc nung nong natri axetat v d i N a O H khi
CO mat CaO xiic tac, bie't rang trong tru'dng hdp nay ngoai metan chi c6 mgi
san pham m u o i v6 cd. V i e t ta't ca cac PTPL/ xay ra.

2. Hoa long 7,8kg hdi benzen xuong 20"C thu diWc:
I
•r

.

7 ' 8 x ^ 0 ' g = 8873 m l hay 8,873 l i t .
0,879g/ml

Bai 4.

B a i giai
c +

> C H 3 - C H 2 - CI + H C l

^"^'""^ > C a C 2 + CO

C a C 2 + 2H2O

1. Do't chay mot lu'cfng hidrocacbon X, ngU'di ta thu diTdc C O 2 va H2O theo li if.

a)

'"

CaO + 3 C

I I . Bai tap \\f g l i i

Chii de 1.

'"^''"^

2 TH than da, da v o i d i e u che' axetilen, benzen

nho hdn 4 la C2H2 va mot hidrocacbon c6 so' nguyen tuf H Idn hdn 4 la C2Hf,)

C. BAI T A P L U Y E N

CH2 = CH2

CH3 - CH2CI + NaOH

(1 : 0,05) = ( x : 0 , 1 ) = ^ x = 2
(1 : 0,05) = (0,5 y : 0,1)

C H ^ C H + H2

TCr axetilen:

y/2H20

+

<

\

..liifh^l

.

:^/.;Hf«:W-r-..l

1. Cho hon hdp cac k h i C H 4 , C2H4, C2H2, SO2, C O 2 di qua nu-dc brom. V i e t c^c
phu'dng trinh phan tfng xay ra.

2 H 2 - ^ C H 4

2. Cho 1 l i t benzen (d = 0,879 g/ml) tac dung v d i 112 l i t CI2 (ct dktc) k h i c6 mSt
b)

AI4C3 + I2H2O

c)

CH3 - COONa + NaOH

> 4A1(0H)3
'"

+ 3CH4

> CH4

xiic tac la bot s^t thu dirdc 450g clobenzen. T i n h hieu sua't phan iJng.
B a i giai

+ Na2C03

1- K h i cho h o n hdp k h i qua nU'dc brom chi CO cac phan iJng sau:

B a i 2.
1. Co the dieu che etilen tiJf ru'du etyliq, tiJf axetilen va tijr etan. V i e t cac phv[iM
trinh phan iJng xay ra.
2. TiJf than d a , da v o i , viet cac phu'dng trinh phan tfng d i e u che' axetilen, benzen
(c6 ghi dieu k i e n phan tfng).
B a i giai

'
S

1. Cic phuWng trinh dieu che etilen:
a) Tur riTdu elylic:

, ,'

C2H5 - O H

> C2H4

C2H4 + Br2

> C2H4Br2

C2H2 + 2Br2

> C2H2Br4

..ui l«
SO2 + 2H2O + Br2
> H2SO4 + 2 H B r
2- Phan iJng giiJa benzen va clo k h i CO bot s^t xuc tac. '
CfiHe + CI2

—
'*
-

'

'

'

HCl

^, ^
1000x0,879
,
112

.
.
Tinh n c , H 6 =
= 11.27raol; nc,^ - —
- 5mol

• ^

+ H2O
211

210


" c 6 H 5 a = Y Y ^ = 4mol

C O 2 + Ca(OH)2

> CaCOj i

+ H2O

T^ch l o a i tap chat:
V a y h i e u sua't p h a n rfng p h a i t i n h t h e o c l o : h % = ^^^^^

= go%

'

\f ach C O 2 k h o i C2H2: C h o k h i qua dung dich k i e m duT, ( t h i du N a O H )

*

B a i 5.

C 0 2 + 2NaOH

> N a j C O j + H2O

,

^

^

f d c h C2H4 k h o i C O 2 : C h o k h i d i qua nifdc brom duf:

1. V i e t cac phUcfng t r i n h p h a n iJng d o t c h a y cac cha't c h o d i / d i d a y t h a n h C O j ^

C2H4 + Br2

H2O: CfiHfi, C„H2n ^ 2, C , H y O „ N H 2 - C H 2 - C O O H ( b i e t n i t d b i c h a y t h a n h N^),''

> C2H4Br2

Loai C 2 H 5 O H k h o i C H 3 - C O O H : Cho hon hdp tac dung v d i dung dich k . e m ,

2. D o t c h a y r i f d u CnH2„ + 1 O H b l i n g C u O , b i e t r ^ n g s a n p h a m c h a y g o m C o

jioSc muo'i cacbonat, luc do chi c6 axit phan tfng tao thanh muo'i, thi d u :

H2O v a C u . V i e t phiTdng t r i n h p h a n iJng.

2 C H 3 - C O O H + CaO

B a i giai

> CaCCHj - C O O ) 2 + H 2 O

Pun dudi rifdu bay h d i . S a u 66 cho axit sunfuric tac dung v d i C a ( C H 3 -

1. C d c p h a n uTng d o t c h a y :
CfiHfi + 7 , 5 0 2

> 6CO2 + 3H2O

C„H2„.2+^^02

> nC02 + ( n + 1 ) H 2 0

C H y O , + (X + ^ - - ) 0 2

> X C O 2 + ^ H2O

C 0 0 ) 2 (dun nong) de axit bay hdi va l a m ngiTng tu thu di/dc C H 3 - C O O H .
C a ( C H 3 - C O O ) 2 + H2SO4 —

^

2CH3 - C O O H T

+ CaS04 i it

Bai 2.

1. Metan bi Ian mot I t tap cha't la C O 2 , C2H4, C2H2. T n n h bay phifdng phap hoa
sj,.

4

2

2

2NH2 - C H 2 - C O O H + 4,502

> 4 C O 2 + 5 H 2 O + N2

C„H2„ + , O H + 3 n C u O

2.

^0^jj,

hoc de loai het tap chat k h o i metan.
2. Benzen b i Ian mot it nu'dc va riTdu, l a m the nao de c 6 benzen tinh khiS't.
B a i giai

> n C O j + (n + 1 )H20 + 3nCu

-'^'i

^- '

1. Cho hon hdp k h i Ian Itfdt d i qua binh nu'dc brom dtf, luc do l o a i het C2H2 va
C2H4nhdphanu'ng:

Chu de 2.

N h i n b i e t - T a c h h o n h
Bail.

C2H4 + Br2

> C2H4Br2

C2H2 + 2Br2

> C2H2Br4

_

.0 4 a

>'

1. T n n h b a y phiTdng p h a p hoa h o c n h a n b i e t cac b i n h k h i : C H 4 , H2, C2H4, C O : .

Sau do cho k h i con l a i qua binh diTng dung dich k i e m d\i ( N a O H , C a ( 0 H ) 2 ,

2. T r i n h b a y phiTdng p h d p hoa h o c d e l a m sach t a p cha't:

v.v...), luc do C O 2 bi hap thu het do phan lirng:

a ) L o a i C O 2 k h o i C2H2

*

b ) L o a i C2H4 k h o i C O 2

1. PhiTdng p h a p hoa h o c n h a n b i e t b i n h k h i C H 4 , H2, C2H4, C O 2 .
C o n h i e u e a c h n h a n bie't. D i f d i d a y la m o t e a c h d d n g i a n :

T

C h o k h i d i q u a b i n h nxidc b r o m , b i n h n a o k h i l a m ma't m a u nufdc b r o m , do la
C H . = CH2 + B r j

k h i CO2: CO2 + Ca(0H)2

> CaCO., i

+ HjO

CH4 + 2O2
212

) 2H2O

— C O 2

+ 2H2O

> NaOH + 0,5H21

C 2 H , 0 H + Na

> CjHsONa + O.SHjT

C 6 the phan biet m u d i an, d i T d n g kinh bang each dot chay hay khong?
^' K h i dot chay k h i A thu d i T d c C O 2 va H 2 O , k h i dot chay k h i B thu difdc C O 2
SO2 con k h i dot chay k h i C thu diTdc C O 2 , H2O va N2. H o i cac k h i A , B , C

5,

D o t c h a y 2 b i n h c o n l a i v a c h o san p h a m c h a y q u a v o i , n d i n a o b i d u e la W"'^

'"

H2O + Na
Bii3.

> CH2Br - CH2Br

C h o k h i i r o n g ba b i n h c o n l a i q u a nu'dc v o i t r o n g , n d i n a o nu'dc v o i b i due

k h i C H 4 : 2H2 + O2

^

rifdu tac dung v d i Na tao ra cAc san pham khong tan trong benzen:

B a i giai

etilen:

> NazCOj + H2O

2- Cach ddn gian la cho benzen bi Ian tap cha't tac dung v d i Na diT, luc do niTdc

c) * L o a i C 2 H 5 O H k h o i C H 3 C O O H .

-

C 0 2 + 2NaOH

K h i con l a i la C H 4 nguyen cha't.

CO phai l a hdp chat hiJu c d hay khong?
J

B a t giai

Co the phan biet m u d i an va d i T d n g k i n h bkng each dot chay, v i m u d i an
•^hong chay, van la chat r ^ n mau tr^ng, trong k h i do d i T d n g bi chay het thanh
va H , 0 :

213

C,2H220|, + 1202

2. K h i dot chay k h i A thu diTdc CO2 va H2O. D i e u do chtfng to chat A chi?a
nguyen to C, H va c6 the c6 hoac khong c6 oxi. V a y A la hdp cha't hffu cd.
Thidu:

C2H6 +3,502

)• 2CO2 + 3H2O

C 2 H 6 O + 3O2

> 2CO2 + 3H2O

V i e t cac phu'dng trinh phan \ing (ne'u c6)
§

^

CS2 + 3O2

l^j^;

(NH4)2C03

+ 1,502

CO2 + 4H2O + N2

— ^

2NH2 - CH2 - C O O H + 4,502 — ^

'«j

Ca(0H)2 + CO2

— >

De phan biet CO2 va SO2 c6 the dung

brom con

S02 + 2H20 + Br2

Chat

C+

> SO2

Chat

D + O2

Chat
Chat

F + O2

CM

SO2

lam

ma't

mau niTdc
1

.

,

3.

nUdc brom,

cha't

lam mat mau

la etilen, con

)• C2H4Br2
L a p c 6 n g thiifc p h S n tuT

Bai 1. D o t chay 2,24 l i t hidrocacbon X 0 dktc) va cho san p h a m chay Ian li/dt
di qua binh diTng P2O5 va binh 2 diTng K O H r ^ n . Sau k h i k e t thuc t h i nghiem

B a i giai

tha'y k h o i liTdng binh 1 tang 9 gam va binh 2 tang 17,6 gam. T i m CTPT, viet

Co the difa v^o san pham dot chay de suy luan hdp chat dem do't chay la h(}r
chat hij"u cd hay v6

metan.

C2H4 + Br2

> CO2 + Na2C03
> CO2 + H2O + CaO

E + O2

niTdc brom:

the sue k h i CO2 vao 2 ket tua)

khong la

> CO2 + H2O + CI2

+ H2O

> 2 H B r + H2S04

Cho 2 k h i con l a i tac dung v d i

> CO2

' • - ^ ''-'^

i,! :

Chat
Chat

B + O2

'

CO2 thi khong:

(CO

> CO2 + H2O

' ^

>CaS03i+H20

cho d\idi day la hdp cha't hffu cd dtfdc khong:
A + O2

' '

v6i.

CaCOji

Ca(OH)2 + S02

4CO2 + 5H2O + N2

B a i 4. Co the diTa vao san pham dot chay de suy luan cac chat dem dot chay

O2

> Ca(CH3COO)2 + H2O + CO21

2 chat tao ket tua v d i niidc v o i trong la CO2 va SO2;

'

COOH).

niTdc la benzen C6H6.

2. Nhan bie't cac chat k h i .
.

cd ((NH4)2C03) ho|c hiiru cd (NH2 - CH, -

-

2 chat tan trong nvldc la C2H5OH va C H 3 C O O H , dung da v o i de nhan bie't

Con C2H5OH khong tac dung v d i da

' V a y chat C c6 the la hdp chat v6

C T C T cua hidrocacbon X .

cd.

-

Chat A la hdp cha't hSu cd chiJa C, H hoSc C, H , O

-

Chat B chi c6 the la C hoSc CO (v6

cd)

-

Chat C chi c6 the la liTu huynh S (v6

cd)

-

Chat D la hdp chat hihi cd chuTa C, H, C I hoSc C, H , C I , O

-

Chat E la hdp chat huTu cd chiJa C, O, Na.

-

^

2CH3COOH + CaCOj

nguyen to C, H, N va c6 the c6 hoac khong co o x i .

•

CH3COOH.

hUucd.

K h i dot chay chat C tao ra CO2, H2O va N j , dieu do chuTng to cha't C chiJa cac

Thi du:

B a i giai

Chat khong tan trong nu'dc, n o i tren mat
,

> CO2 + 2SO2

V a y B la hdp cha't

^

J Nhan biet cac chat long:

K h i dot chay k h i B tao ra CO2 va SO2, dieu do chtfng to chat B phai chuTa C, s
Thi du:

Iren.

niidc v o i trong, nu'dc brom, da v o i , hay cho bie't each nhan biet 7 chat

) I2CQ2+ IIH2O

'"

Baigi^i
jPhan ti-ng dot chay: CJiy + (x + ^ ) 0 2

••J^(it^mnm
> xCOj + ^ H j O

(1)

• '

'*''

Chat F la hdp chat hffu cd chtfa C, H , O, Na.

B a i 5. Trong phong thi nghiem c6 7 binh thuy tinh khong mau b i mat nhan, m"J
binh diTng mot chat long hoac mot chat k h i sau day: etan, etilen, benzen, ^t^'

cacbonic, k h i sunfurd, riTdu etylic va axit axetic. Chi dU'dc diing them ni/'^'^'

k h o i lu'dng binh 1 tang do H2O bi hap thu b d i P2O5
P2O5 + 3H2O
I D o do: ny^^Q=— 9

, .

> 2H3PO4

f'j

" ^ , 5 '

= 0,5 m o l

1
K h o i lu'dng binh 2 tang do CO2 bi hap thu b d i K O H :
CO2 + 2 K 0 H — > K2CO3 + H2O
'

' • '

'
215


17,6
44 = 0,4 mol
2,24 = 0,1 mol nen theo phan uTng (1) ta c6 he phufcfng trinh;

Vi so mol X =
22,4

Dodo n C02

'0,lx = 0 , 4 = > x = 4
0,l.^ = 0,5=>y = 10
Vay cong thtfc phan tur cua X la C4Hi(). Cac cong thufc cau tao c6 the c6
CH3 - CH2 - CH2 - CH3 hoac CH3 - CH - CH3
CH3

Bai 2. De dot chay 1 the tich khi hidrocacbon Y (d dktc, so nguyen tuf C nho hdn
5) can dung 6,5 the tich O2 (d dktc). Tim CTPT cua Y.
Bai giai
Phan u-ng dot chay Y: C,H„ + (x + ^ )02
> XCO2 + - H2O
(2)
4
2
cm 1 the tich Y can 6,5 the tich O2, tuTc cur 1 mol Y can 6,5 mol O2, nen theo
phan uTng (2) ta c6: x + - = 6,5 hay 4x + y = 26
4
1 4
2
3
X
-'fv
22
18
14

10
y
LapKetbang:
C4Hi()
luan
loai
loai
loai
Bai 3. Dot chay hoan toan 6 gam cha't A chiJa cac nguyen to' C, H, O ta thu diTdc
4,48 lit CO2 (d dktc) va 3,6 gam niTdc. Biet 1 lit hdi chat A (tinh theo dktc)
nang 2,679 gam. Tinh CTPT cua chat A.
Bai giai
Goi cong thtfc cua A la CxHyO,
,
Phan u-ng dot chay A: C.HyO, + (x + ^ - - )02
> XCO2 + - H 2 O (1)
4 2
'2/
Tim qua cong thiJc dcfn gian nhat (ttfc ti 16 ddn gian nha't cua so' nguyen ti'
cac nguyen to'). TriTdc het can tinh khoi lifcJng cua cac nguyen to
4,48x l 2,^ = ^2,4g;
,
„0,4g;
.
mc= —
- 3,6x2
=
mo = 6 - 2,4 - 0,4 = 3,2g
r^- .
2,4 0,4 3,2 , - ,

Ta CO X : y : z =
: :— =1:2:1
12 1 16
Cong thu-c ddn gian nhat la (CH20)„ (trong do n la boi so' cua CTDGN de
216cong thiJc phan tur).

Tinh khoi lifdng mol cua A = 2,679 x 22,4 - 60g
Vay cong thiJc phan tur cua A la C2H4O2
p^j 4. De dot chay hoan toan 4,6 gam chat B chiJa cac nguyen to' C, H, O can
dung 6,72 lit O2, thu diTdc CO2 va H2O theo ti le the tich Vco^ : VH20= 2 : 3.
Tim cong thtfc phan tur, vie't cong thdrc cau tao cua B.
Biet 1 gam chat B d dktc chiem the tich 0,487 lit.
"" ' " ' *
Bai giai
Gpi cong thtfc phan tur cua B la C,HyO,:
''
Phan u-ng do't chay:

+ (x + - - - )02 - -> XCO2+ -2H 2 O (1)
4 2
(a)
X 2
Theo dieu kien cho ta c6 ti le: — = — => y =^ 3x x + y x22,4
4
Theo phiTdng trinh (1) ta c6' ti le: 12x + y + 16z ^
(b)
6,72
4,6
The (a) vao (b), rut ra: x = 2z
Vay CO ti le X : y : z = 2 : 6 : 1

Cong thi?c ddn gian nha't cua B la (C2H60)„.
Tinh KL mol PT cua B b^ng 22,4 «46
0,487
Tim CTPT cua B: (CzHfiOn = 46n = 46 do do n = 1
.,
Vay CTPT cua B la CjHfiO
B&i 5. Tim cong thtfc phan tijr cua mot chat hiJu cd A chiJa 25% hidro va 75%
cacbon. 1 lit cha't A (d dktc) nhe hdn 2 Ian so vdi 1 lit O2 (c( dktc). Bai giai
Vi hidro chic'm 25%) va cacbon chiem 75*% chuTng to A la mot hidrocacbon
C,Hy. Vi 1 lit A nhc hdn 2 Ian 1 lit O2 chu'ng to khoi li/dng mol phan tur cua A
32
nho hdn khoi
lifdng mol
tur
cija O2 ,2 Ian: MA =
16x75
, phan
16x25
=4
Vay: x = —rr = l; y =
100x12
100x1
Vay CTPT cua A la CH4
6. Hidrocacbon B chita 20% hidro. 1 lit khi B (d dktc) nSng 1,34 gam.
Tim cong thiJc phan tiJ cua B.
Ne'u khong biet khoi liTdng cua 1 lit khi B co tim difdc cong thiJc phan tur hay
khong?
217
rr.,

1

^

,

CxHyO,

/ . N


Bai giai

9. De do't chay m gam chat A chtfa cac nguyen to C, H , O can 0,3 m o l O2

a) K h o i lifting m o l phan lit cua B b^ng: 1,34 x 22,4 = 30 g/mol

thu diTdc 0,2 mol C O 2 va 0,3 mol H2O.

So nguyen lu* H = ^^^^'^ = 6 ; So nguyen tu" C = ^ ^ ^ ^ ^ = 2
100x1
^ '
100x12 <

'

g) T i n h kho'i liTOng m .

.

J,) T i m C T P T cua A , vie't cong thiJc ca'u tao cua A .

V a y CTPT cua B la C2H6

Bai giai

b) N c u khong biet k h o i liTOng cua 1 lit B, nghla la khong biet K L P T , luc do ta

Phan u-ng dot chay A : A + O2 — »

chi mcti biet t i le so nguyen^uT C va H cua C^Hy.
x : y = ^ : ^ = l:3
^
12 1

.

V a y cong thufc dcfn gian nhat cua B la (CH3)n.

V i so nguyen tu" H phai la so chan nen n chi c6 cac gia t r i 2, 4, 6. K h i n = 2 la
CO cha't C:Hf„ diing hoa trj cua oacbon
K h i n = 4 ta c6 cong ihiJc C4H12, cong thiJc nay khong dung v i cacbon c6 h o a
tri l(^n h(Jn 4.

m A + nio2 = mco2 +
niA

rnuxo

= 0,2 x44 + 0,3 X 18 - 0,3 x 32 = 4,6 gam

b) Goi cong thiJc cua A la CxHyO,:
So'mol H = 2 Ian so m o l H2O = 2 x 0,3 = 0,6 m o l
S6' m o l O trong A = to'ng so' m o l O trong C O 2 va H2O triT s6' m o l O trong O2
= 2 x 0,2 + 0 , 3 - 2

X

0,3 = 0,1 m o l

V a y t i 16 x : y : z = 0,2 : 0,6 : 0,1 = 2 : 6 : 1

K h i n = 6, 8... thi cong thiJc thu diTdc cang sai.
C6 the vie't C „ H 3 „ r o i b i e n luan theo tifng loai hidrocacbon.
Bai 7. T i le khoi lu'cing cua cacbon va hidro trong hidrocacbon X la mc : m H = 12.
T i m cong thufc phan tu" cua X biet k h o i liTcfng phan tur cua X Idn gap 1,3 I a n
khoi liTdng phan tuT cua axit axctic.

Cong thiJc ddn gian nhat cua A la ( C j H f i O n . Ta thay n c h i c6 the bang 1 v i
neu n bang 2 t h i d\i hoa tri cua cacbon C4H12O2 (so nguyen tuf H t o i da bang
10, nghTa la C4H|,)02).
V a y cong thtfc phan tur cua A la C2H6O.

12.1
x :y= —: 12'l

bkng nhau thu diTdc 3,52 gam C O 2 va 1,62 gam H2O. T i m CTPT, vie't C T C T
1:1

cua cac hidrocacbon.

Bai giai

V a y cong thuTc ddn gian nhat cua X la (CH)„. K L P T ciJa C H 3 - C O O H bling

Trirdc het can tinh so m o l C O 2 va H2O

60. K L P T cua X = 60 x 1,3 = 78
V i (12 + l ) n = 78

«St* ii < ' •••

Bai 10. Do't chay hoan toan hon hcJp 2 hidrocacbon CxHy va CxH,, c6 so' m o l

B a i giai
Goi cong thiJc cua X la C^Hy ta c6 t i le:

C O 2 + H2O

2) Theo djnh luat bao toan khoi lifdng:

' ^^*^'

Co the t i m cong thuTc phan tuT b^ng each bien luan nhiT sau:

' '

n = 6. Do do C T P T cua X la CfiH^.

nco2 =

'

3 52
1 62
^ = 0,08mol; U H ^ O = ^ = 0,09mol

; :

Ji-i-'* 'Iv •

B a i 8. T i m t i le so nguyen tijT C , H , O trong hdp chat Y chiJa 6 , 6 7 % H , 18,67'7rN
va 42,67% O. B i e t rSng khi dot chay hoan toan 1 m o l Y thu duTdc 11,2 lit N:

V i so m o l H2O nhieu hcfn so m o l C O 2 nen phai c6 1 ankan (trU'dng hdp 2

(d dktc). T i m C T P T cua Y .

ankan can l o a i v i luc do y = z = 2x + 2 nghTa la chi c6 1 chat chtf khong phai

r

hon hdp).

B a i giai

ke't d o i - (c6 the n o i tdng so l i e n ke't n) va so' vong.
= 0,5 m o l N2, dieu

chtfng 16 trong m o i phan tuf c6 1 nguyen tu" N , do do cong thuTc phan tuT cua
m C2H5O2N.

218

•

trong do a to'ng so lien ke't d o i , l i e n ke't ba - 1 lien ke't ba tiTdng diTdng 2 l i e n

^ ^ ^ ,
=2:5:2:1

V i k h i dot chay 1 m o l Y thu diTdc 11,21 N2 ttjTc

••

Goi cong thiJc cua ankan la CxH2x+2 va cua hidrocacbon thi? hai la CxH2x+2-2a.

Gpi cong thuTc cua hdp chat la C x H y O , N , ta c6 t i le:
32 6,67 42,67 18,67
X : y : /.: t = — :
:
:— —
•
12
1
16
14

- M «'»

4

Cac phan iifng dot chay.
CxH2x.2 + ^ ^ 0 2 —

^

CxH2x.2-2a + ^ ^ " " ^ " ' 0 2

XCO2 + ( X + 1)H20
XCO2 + ( X + 1 -

.,07
a)H20
219


9ni
9
V
V i dot m gam A thu diTcJc — gam H2O nen ta c6: - ( 1 2 x + y) = M S = 9y

V i so' mol ciia 2 hidrocacbon b^ng nhau nen ta c6 ti le so' mol C O 2 va HjCj
nhiT sau:
^

"CO2
HHJO

2x
2x

Hay y = 2x, ttfc CTDGN Ih (C^UiA-

^ 0,08

+ 2-a

Rut ra X = 8 - 4a hay a =

2,3

nr.;-

0,09

8-x

X

KLPT cua A n^m trong khodng

29 = 66,7 va 2,5

x

29 = 72,5

Tu-c 66,7 < 14x < 72,5, gia tri x duy nhai t^ng 5. Vay CTPT cua A la CsH,,,.
pai 13. Hon hcfp X gom 2 hidrocacbon A, B thupc loai ankan (no), hoSc anken

4

(CO 1 lien ket doi) hoSc ankin (c6 1 lien ket ba). T i le KLPT cua chung la 22:

V i a la so nguyen nen chi co nghiem x = 4 va a = I duy nhat

13. Dot chay hoan toan 0,3 mol X v^ cho san pham chay hap thu vao binh

CTPT cua hai hidrocacbon la

dung dich Ba(OH)2 dvl thay khoi liTdng binh dung dich tang 46,5 gam va c6

C4H1,,

va

C4HX.

CHi - (pH - C H 3

CTCT: C H 3 - CH2 - CH2 - C H 3

147,75 gam ket tua.
1. Tim CTPT cua A, B.

CH2 = C H - CH2 - CH3
C H 3 - C = CH2

CH3 - CH = CH - CH3
CH2-CH2

CH3

2. Cho 0,3 mol X di tijf tvt qua 0,5 lit dung dich nMdc Br2 0,2M thay nifdc brom

CH2

CH2-CH2

mat mau hoan toan va c6 5,04 lit khi bay ra (dktc).
•',1

CH3-CH-CH3

1. Phan urng tao ket tua: Ba(0H)2 + CO2

5rt Viet CTCT cua tat ca cic dan xuat chiJa brom c6 the c6.
Goi X la dan xuat chiJa nhieu brom nhat c6 CTPT C^HyBr,, KLPT cija X bang:
x

202

ti?c z toi da Ih. 2. Do do c6 the viet lai CTPT cua X la

-

12x + y + 80x 2 = 202. Hay 12x + y = 42.
v.^:.

A

TriTcfng hdp 1: 2 anken
„
.
Taco:

80
X

y

1

2

3

30

18

6

C,H2,

14x

=
CyHjy

Lap bang:

(1)

Dodo so'molnirdc nH20 = ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ = 0,75moL ,
18
Vi 0^02 = " H20
CO 2 triTcfng hdp:

2 = 202 dvC

Tac6: 12x + y + SOz = 202.
Vay z <

> B a C O j i + H2O

S 6 ' m o l C 0 2 = nB,co3 = ^ ^ ^ - 0 , 7 5 m o l

Bai giai
101

t.'":

Bai giai

Bai 11. Cho hidrocacbon A tdc dung v6i BT2 thu diTcJc mot so dan xuaft chij-a
brom, trong do dan xua't chuTa brom nhieu nhat c6 ti khoi so vdi H2 bhng 101.

I iiJ- -U

Hoi thu diTcJc san pham gi, bao nhieu gam?

22
X
22 ^ . 44
= — t t f c - = — h c a c — v . v . . . loai
14y 13
y 13
• 26

Trirdng hdp 2: 1 anken va 1 ankin

•
CxH2x+2 +

O2

> XCO2 + (x + 1)H20

C,H2,.2 + ^

O2

> yC02 + (y - 1 )H20

(2)

Vay CTPT cua X la C3H6Br2.
Bai 12. A la mot h d p chat hHu ccJ chiJa 2 nguyen to. Dot chay hoan toan m gam
9m

•>
'
A thu di^rtc -y gam niTdc. T i khoi cua A so vdi khong khi nam trong khoang
2,3 den 2,5. Tim CTPT, viet CTCT cua tat ca cac dong phan cua A. ,
f

Bai giai

^*

(3)

5oi a, b la so mol cua ankan va ankin, ta c6 a + b = 0,3
nco2 =ax + by = 0,75

'

. |,^.

n^^^o = a(x +1) + b(y - 1 ) = ax + by + a - b = 0,75

,

Hdp chat A chiJa 2 nguyen to cacbon va hidro, c6 the bieu dien A Ik C^Hy,
CO phan i^ng chay:

Giai he phu'cJng trinh tim di/dc a = b =

•

CxHy + (X + ^ )02

> XCO2 + ^ H2O

= 0,15mol.

Xet ti le KLPT
221

220


a)

Mankan
Mankin

_

14x

+

14y

- 2

22

2

13

V i a = b = 0,15 nen x + y =

hay

s6 m o l H2O nhieu hdn so' m o l CO2 nen A la 1 ankan. Thee phan tirng do't

1 3x + 5 = 2 2 y

0.75
0,15

chiy A:
=5

C„H2„.2 +

Ke't hdp 2 phi/dng trmh cuoi cung ta c6 x = 3 va y = 2.

Tac6tyl$:

Vay CTPT cua cac hidrocacbon la C^Hx va C2H2.
b)

M ankin
ankan

14X-2

22

14y + 2

13

hay 22x = I 3 y - 5

> nC02 + (n + D H j O .

=

Rdt ra n = 6. V a y cong thuTc phan tuf cua A la: CeHu. Suy ra do't chiy

loai.

3 - 1,5 = 1,5 m o l CO2 va 2,5 - 1,75 = 0,75 m o l H2O.

C H = C H + Br2

CHBr = CHBr

(4)

C H B r = C H B r + Br2

CHBr2 - C H B r j

(5)

5,04
So m o l k h i d i ra k h o i bmh midc brom bang '-^^ = 0,225mol
22,4

vi X khong l a m mat m a u dung dich nifdc brom n e n B phai thupc loai
hidrocacbon thdm (aren), tiJc CnHnph^i thoa m a n :
n = 2n - 6, tuTc n = 6 va c6ng thtfc phan tuf cua B la CeHe (benzen).
2. G p i X , y la so m o l cua A, B . Theo phdn tfng chay:

Nhir vay so m o l C2H2 da tham gia phan uTng b^ng: 0,3 - 0,225 = 0,075 m o l

CfiHu + 9 , 5 0 2 — ^

G o i p, q la s o m o l C2H2 tham gia cac phan tfng (4, 5) ta c6:

CfiHfi + 7,502 — ^

,;

6 C O 2 + 3H2O

T a c 6 6 x = l , 5 v a 6 y = l , 5 , tij'cx = y

p + 2q = n g ^ = 0,5 x 0,2 = 0 , 1 p = 0,05; q = 0,025

M o i chat A , B c h i e m 5 0 % so m o l .
143
— - 3 , 0 = 0,25 m o l CO2 v
44

K h o i li/dng C H B r ^ - C H B r j b^ng 0,025 x 346 = 8,65 gam.
B a i 14. Dot chay hoan toan 41 gam hon h d p X gom 2 hidrocacbon A, B thu dUdc
132 gam CO2 va 45 gam H2O. Neu them vao X mot nuTa liTdng A c6 trong X roi

So nguyen tuf cacbon =
0,05

dem dot chay hoan toan thi thu dU'dc 165 gam CO2 va 60,75 gam H2O.
1. T i m C T P T cua A , B biet hon h d p X khong l a m mat m a u dung dich niTctc
brom. A, B thuoc cac loai hidrocacbon trong c h i T d n g trinh da hoc.
3. T h e m 0,05 m o l hidrocacbon D vao hon hdp X d tren r o i d e m dot chdy hoan
toan t h i thu d i T d c 143 g CO2 va 49,5 gam H2O. T i m CTPT, viet tat ca dong
^

I^TPT

49 5
a 18

2,5 = 0,25 m o l H2O.

= 5 ; So nguyen tuf hidro =
x 2 = 10
^ '
0,05

cua D : C5H10

B i i 15. H o n hdp k h i X (dktc) g o m 1 ankan (C„H2„ + 2)

1 anken ( C ^ H z J . Cho

Nhif vay do't chay A se thu dU'dc:
2 = 1,5 mol CO2 va

(3,375 - 2,5) X 2 = 1,75 m o l H2O

6,72 l i t hon hdp X nSng 13 gam.
1- T i m C T P T cua ankan va anken, biet so nguyen tijf cacbon trong m o i phan tuf
kh6ngqud4.
v^o dung dich N a O H diT, sau 66 them BaCl2 diT t h i thu diTdc bao nhieu gam

= 3 m o l CO2 va ^ = 2,5 m o l H2O
18

di/dc — = 3,75 m o l CO2 va
• 44

X

J

2- D o t chay hoan toan 3,36 l i t h o n hdp X va cho tat ca san p h a m chdy hap thu

B a i giai

(3,75 - 3)

^ ' :

3,36 l i t h o n hdp X qua binh nirdc brom dU' tha'y c6 8 g a m b o m phan tfng. B i e t

2. T i n h % so m o l cua A, B t r o n g X .

1. D o t chay X diTdc ^
44

; ,,

3. Dot chay hoan toan 0,05 m o l D diTdc:

K h o i l i T d n g C H B r = C H B r bang 0,05 x 186 = 9,3 gam

phan ciia D.

«

6 C O 2 + 7H2O

p + q = 0,075

Do't chay X+-k
^
2

B diTdc:

V i so m o l CO2 ga'p d o i so' m o l H2O nen cong thiJc ddn gian cua B la (CH)2 va

2. Cac phan ufng:

rtRn^i,,^a

O2

18

= 3,375 m o l H2O.

ke't tua.
B a i giai

\jUid:jj^,..j
1^.^.

Khi cho hon hdp k h i qua n\Xdc brom:
CnH2n + 2 + Br2
CniH2m + Br2

)• khoug phau tfng i 5|

'''

<'

'*

> CmH2mBr2

Gpi X va y la so' m o l cua ankan v i anken trong 3,36 l i t h o n hdp X , ta c6 he
Phudng t n n h :
^«
.^^
,UA,, :..
223


3.36

p a i 3. Do't chay m gam hidrocacbon A thu difdc 2,688 lit CO2 (dktc) va 4,32 g

,
= 0,15mol

X + y =

H2O. L a p cong thtfc phan tur cua A

22,4

HuTdngdln

8
=y = — - 0 , 0 5 m o l

nankcn=

^

Do do: x = 0 , 1 5 - 0 , 0 5 = 0,1 m o l

-n

M a t khac, theo k h o i lufdng h o n hcfp ta c6:

nc : ttH = 0,12 : 0,48 = 1 : 4 => Cong thtfc la CH4 (d day cong thuTc dcfn gian
cung la cong thtfc phan tuf). Hay dat C T P T la (CH4)„, ta thay n chi c6 the

.

bang 1.

'"^'^^

(14n + 2) X 0,1 + 14m x 0,05 = ll^iM^^6,5

pai 4-

'">

hidrocacbon A mach hd, the k h i . K h o i lu'dng V l i t k h i nay bang 2 Ian

khoi lifctng V l i t N2 d cung dieu k i e n nhiet do va ap sua't. L a p CTPT

6,72

Hidrocacbon do

HoSc (rut gon): 2n + m = 9

1

n

1

2

3

4

MA = 2 MN2 = 2.28 = 56 dvC.

m

7

5

3

1

Dat cong thtfc cua hidrocacbon la C^Hy ta c6: 12x + y = 56.

CTPT

CH4

C2H6

C3HX

C4Hi()

C,7H,4(cha'tl6ng)

C5H10

C3H6

CH2

loai

loai

C3H8; C3H6

Loai

K ^ t luan

K e t luan: C h i c6 cap CjHs va

> 3CO2 + 4H2O

(1)

C3H6 + 4 , 5 0 2

> 3CO2 + 3H2O

(2)

CO2 + 2 N a O H

> Na2C03 + H2O

(3)

BaCl2 + Na2C03

> BaCOji

(4)

Hi^ng din
lien ket v d i nhau. D o la C4H1,).
Cx + 1H3X. L a p Cong thtfc phan tu" cua A
Hifdng d i n
x = 2

khi va hdi dan qua binh diTng H2SO4 dac thi the tich g i a m hdn mot nuTa. X
thuoc day dong ding

la 1 : 2. L a p cong thiJc phan tuT cua X

nao?

^

_ , ^. _ ^ . ,

" ^ " ^ din
" C O 2 • " H 2 0 = 1 : 2 => He : nH = 1 : 4 => Cong thiJc la CH4

, ^ • ,^

Hi^dng d i n
San pham chay g o m CO2 va hdi nifdc di qua H2SO4 dac t h i h d niTdc b i giff
lai. The tich g i a m hdn mot nuTa ttfc la VH20 hm > Vco2

mm^ mmn oi.d ^rm »(ti u-^

"HJO > "CO2 •

X la ankan
B^i 8. M o t hon hdp 2 ankan ik dong d^ng ke tiep c6 k h o i lifdng la 24,8 g, the

B a i 2. Hidrocacbon A c6 7 5 % C ve k h o i luTdng. L a p C T P T cua A
r

tich tifdng u-ng cua hon hdp la 11,2 lit (dktc). Lap C T P T cac ankan
Hufdngdin

L a y 100 g A thi c6 75 g C va 25 g H

75 25
n c : n H = — : — = 6 , 2 5 : 2 5 = 1 :4
12 1
Cong thu-c dcfn g i a n cua A la CH4. D o cung la cong thtfc phan tit cua A
,

C3H6 (phij hdp); x = 3 => C4Hy (loai)

Bai 7. D o t chay hoan toan hidrocacbon X b^ng mot liTdng o x i vira d u . San pha'm

giii

,

,,

x > 4 deu loai v i hidrocacbon d the k h i .

^

B a i 1. D o t chay hidrocacbon X thu diTdc CO2 va H2O c6 t i le so m o l ti/dng """g

224

, , ^ ,

Bai 6. Hidrocacbon A la cha't k h i d dieu k i e n thadng, cong thtfc phan tuTcd dang

x = 1 => C2H3 (loai);

HrfdngdSn

' ^ ^

V i - C2H5 la goc hidrocacbon no hoa tri I nen phan tu' chi c6 the g o m 2 goc

K h o i lu-dng k e t tua BaC03 = 0,45 x 197 = 88,65 gam.

il. B^i t$p III

'

Bai 5. M o t chcft co cong thiJc ddn gian nha't la C2H5. L a p cong thtfc phan tur cua

Theo c a c phan tfng ( 1 , 2, 3, 4):
= 0,45mol

" i'

chat do.

CjHg + 5O2

"3,003 = n , , , c o 3 = n c o , = 3 n , = 3 x ^

'

Vay, cong thiJc cua hidrocacbon A la C4H«

'

+2NaCl

, .jd} gfihi,

Hidrocacbon the k h i c6 so'C < 4.Nghiem thich hdp x = 4, y = 8

dHf la phil hcfp v d i d i e u k i ^ n cho.

Cac phan iJng:

•M

.j^,,

Hir(Jngd§n

Xet:

;,'

M2a„Ra„==49,6

, .

'•
,?

C-H-^^ " ^ ^ n + 2 = 49,6 => n = 3 , 4

=> CjHs va C4H1,,
225


Bai 9. Crackinh hoan loan mot ankan X thu dtfcJc hon hdp Y c6 ti khoi hdi so
Hzbkng 18. L a p C T P T c u a X

Vfjj

0,1

_

n+1

M Y = 18.2 = 3 6

V i khi crackinh thi so' mol khi tang gap doi, ta c6:
Mx =

25
nc = n c o 2

Mx = 2.36 = 72 ^ CsH.z

= "cacoj =

C

Bai 10. Do't chdy hoan toan hon hdp 2 hidrocacbon mach hd, lien tiep trong day

8,96
tao H2O

; ano.

Lap CTPT 2 hidrocacbon
Hvtdng d§n

na„kan=

"H20

HH^Q

'

"CO,

= 1,4 (mol)

18

=>nH20 >

> CO2

0,25mol

-0,25 = 0,15 (mol)

O2

+

11.

> 2H2O
:•• ,t": }

nc:nH = 0,25:0,6 = 5 : 12 => CsH.z
=
~ 2,5
"ankan

C2H6,

O-^

CjHx.

Slim

0,3mol <-- 0,15mol

day ankan.

"coa

= 1 . 4 - 1 = 0,4 (mol) => n =

=> hai hidrocacbon la:

-

2H2

25,2

O2

0,25mol ->

!

= ^ ' ^ ^ ^^^^^

+

dong dang thu diTdc 22,4 lit C O 2 (dktc) va 25,2 g H 2 O .

"coz =

CsHg

Hxtdng d§n

_ m x ^ m x ^ 2mx ^ 2mY ^

mx = my, ny = 2nx ^

C2H6

p^i 13. Do't chay hoan toan mot li/dng hidrocacbon can c6 8,96 lit O 2 (dktc).
Cho san pham chay di vao dung dich Ca(OH)2 dM thu dtfdc 25 g k6't tua. Lap
CTPT cua hidrocacbon
j

Hvldng d i n

^

n = 2,5 => hai hidrocacbon la

0,14

Bai 14. Dot chay hoan to^n V lit hon hdp 2 hidrocacbon dong d i n g lien tid'p thu
di/dc 1,12 lit khi C O 2 (dktc) va 1,26 g H2O.
Lap CTPT ciia 2 hidrocacbon

Bai 11. Dot 10 cm^ mot hidrocacbon no b^ng 80 c m ' o x i (lay dxi). San pham thu

HUdng d i n

diTdc sau k h i cho hdi niTdc ngUYig tu con 65 c m ' trong do c6 25 c m ' la o x i (cac
the tich di/dc do d cung dieu kien). Lap CTPT cua hidrocacbon
Hvldng d i n
Vco2 = 65 - 25 = 40 c m ' . Vc02 g^'P 4 Ian V^^Hy • Do la C4H,o
C4H|() + O2

n'H20

„ , ^ J-i^=
18
=> n^^o >

0,07 (mol); nco2
"CO2

d6 la ankan.
-> n C 0 2 + ( n + 1)H20

-> 4 C O 2 + 5H2O

10cm'

40cm'

0,05

Bai 12. Dot chay hoan toan 2 hidrocacbon ke' tiep nhau trong day dong dang
San pham chay cho Ian liTdt qua binh 1 diTng H2SO4 dac va binh 2 diTng K O H
r^n thay k h o i liTdng binh 1 tang 2,52 g va binh 2 tang 4,4 g. Lap CTPT hai

n +1

n = 2,5

=:> C 2 H 6 , C3H8

0,07

^ i i 15. Dot chdy hon hdp 2 hidrocacbon ke tiep nhau trong day d6ng ding, thu
^ dirdc 48,4 g C O 2 va 28,8 g H2O. Lap CTPT cac hidrocacbon

hidrocacbon

Hrfdngdin
Hvtdng d§n

1H2O

22,4 = 0-05 (mol)

^ ' ^ ^ =0,14 (mol);
18
•
-

nco2

=

44

=0,1 (mol)

Ta c6: n^jo > "coa ^ Thupc day ankan
3n + l

O2

48,4

28,8

"2°

""Is" = 1.6 (mol);

n^o^

44

= ^'^ ^"^"^^

^

nH20

>nco2

= 1,1.12+ 1,6.2= 16,4. ^
'^ankan= ^ H J O ~ " C O 2

-

M = 1M=32,8

14n + 2 = 32,8 => n =2,2

^ ' ^ = 0,5

=> m2 ankan

0,5

> n C 0 2 + ( n + 1)H20

Hai hidrocacbon d6 la

C2H6

va CjHs.
227

226

Ik.


B a i 16. K h i dot chay hoan toan 2 hidrocacbon lien tiep trong day dong d i n g thu

Htf
diroc 16,8 l i t C O 2 (dktc) va 13,5 g H 2 O . Hai hidrocacbon do thupc day d^og

k u y luan: n,„kan =

d i n g nao?
HiMngdln

'

= "-75 ( m o l ) ; n^^o ~

=

0.75

(mol)

=

= 0 , 1 2 5 (mol)

Vnkan la: C„H2n+2. anken la: CJ\2m,

D o t hidrocacbon cho n ^ j o = " C O 2 thi dc) la anken.

a . , ,

ta c6 0,05n + 0,025m = 0,125

2n + m = 5 ( l ) .

.

|l) chi C O n g h i c m khi m = 3, n = 1 => C H 4 v a C 3 H 6

B a i 17. D o t chay hoan toan 2 hidrocacbon mach hd trong ciing day dong dang

.

[i 21. H o n hdp g o m mot ankan v a mot anken c6 ciing so' nguyen tuf C trong

thu dufdc 1,12 lit C O 2 (dktc) va 0,9 g HjO. Hai hidrocacbon do thupc day

phan tijf v a c6 ciing so' mol. L a y m gam hon hdp nay l a m mat mau viTa du

dong dang nao?

80g dung d i c h Br2 2 0 % trong dung m o i

Hifdng d i n
= 0,05 ( m o l ) ; U H ^ O = ^

nco2 = ^

loU

12 5
\nco2 = "caC03

nco2 - ^

= 0.05 (mol); n,„kc„ = " B O = 7 7 7 : = 0.025 (mol)

224UU

D o t hidrocacbon cho

nyi^Q = TIQQ^

= 0.05

80.20

thi do la anken.

-

C O 2 nhuf nhau, c 6 n t i le so mo l H2O va C O 2 cua chung tiTdng iJng la 1,5 : 1.

.

L a p C T P T cua chiing
Hiring d i n
= 1,5 ^

-

"CO2

C„H2„.2 +

O2

C2H6

i^i
^Ta

> n C 0 2 + (n + 1 ) H 2 0

2
•^^^ = - => n = 2
n
2

^ , ^

vay anken la

C2H4

day dong d i n g nao?

0,1 m o l

>0,lnmol

c6: 0 , l n = 0,3 => n = 3 => CjHx,

hp

Oi:
C3H6

Bai 22, Cho 14 g hon hdp g o m 2 anken la dong dang ke tiep di qua dung d i c h
brom l a m mat mau vifa du dung dich chuTa 64 g Br2.
Lap C T P T cua c a c anken

Hifdngdln
64
Hankcn " " " ^ 1 2 = ^ ^ =

0,4

Hai anken la:

" H 2 0 ~ " C O 2 ~ 0,5 (mol). Vay do la anken.

= 0,3 (mol).
> nC02 + nH20

Manken = —

Hifdng d i n

,^

CnH2„ + O 2

B a i 19. Dot chay hoan toan hon hdp 2 hidrocacbon mach hcl trong cting day
dong d i n g thu diTdc 11,2 lit C O 2 (dktc) va 9 g H2O. Hai hidrocacbon do thuot

> A =;

Ankan v a anken c6 ciing so nguyen tuT C va c6 cDng so m o l nen chay cho

,
do la ankan

;:

"-""="'^^=1^516^=^'^^'""^^

cung so'mol C O 2 v a bang

= 1 => do la anken;

D o t chay hoan toan m gam hon

HiMngdln

(mol)

B a i 18. Dot chay so mo l nhU' nhau cua 2 hidrocacbon mach hof thu diTPc so mol

"CO2

CCI4.

hdp do thu dufdc 0,6 m o l C O 2 . Lap C T P T ciia ankan v a anken

= 35

0.4

=>

(mol)

14H

V<

= 35

=>

n = 2,5

^r

,

:*>

C2H4, C3H6

Bai 23. M o t hon hdp g o m 1 ankan v a 1 anken c6 l i le so m o l 1 : 1. So nguyen tOr
C-H,-

+ —

O2

> nC02+

nH20

B a i 20. Hon hdp k h i X gom 1 ankan va 1 anken. Cho 1680 m l k h i X l o i chan'
qua dung djch Br2 thay lam mat mau viifa du dung djch chiJa 4 g Br2 va co"^
lai 1120 m l k h i .
M a t khac ncu dot chay hoiin toan 1680 m l X r o i cho san pham chay di va'^
binh dirng dung djch C a ( 0 H ) 2 d\i thu diTdc 12,5 g ket tiia. L a p CTPT ci'"^^

C cua ankan ga'p 2 Ian so nguyen tuf C cua anken. Lay a gam hon hdp t h i lam
^ma't mau viTa du dung d i c h chi?a 0,1 mol B r j . Dot chay hoan toan a gam hon
[idp thu di/dc 0,6 m o l C O j . Lap C T P T cua chiing
A

HiftJngdIn
|'''Br2

l"co2

-

" i n k c n = Hankan = 0,1

,

(mol)

ankan gap d o i cua anken; n^oj ^^ua anken =

= 0,2 m o l

hidrocacbon
anken c6 2 C, vay ankan c6

228

4 C. Do lii C 2 H 4 va

C4Hio.

229


Bai 2 4 . Dot chdy hoan toan 8,96 lit (dktc) hSn hop hai anken Ik d6ng ding \•^.
tie'p thu dtfOc m g H 2 O
(m + 39) g C O 2 . Lap CTPT hai anken

;=> nankin =

HcQz " " H J O

= 0.6-

0,4 = 0 , 2

(mol)

Htfdng d§n
"ankin

Dot anken cho U H - Q = " c o i
^
27 + 39
+ O2

C-H^0,4

~ = "^^"^^ => m = 2 7
18
44
,^
8,96

pai

n CO2 + n H2O

>

>

0,4n

Ta c6 0,4n = 1,5 => ii = 3,75 ^

Hai anken la C4HX va C3H6

Dot chdy hoan to^n V lit (dktc) mot ankin de khi thu dufOc CO2 v^ H2O
CO t^ng khoi Itfdng la 25,2 g. Neu cho san pham chay di qua binh difng ni/dc
v6i trong diTthu dUdc 45 g ket tua. Lap CTPT cua ankin
,(.!-,••,
HrfdingdSn
45
'^ ' '

"•
ncacoj = "coz = "c =
= 0.45 (mol) ^ m c o 2 = 44.0,45 = 19,8 (g)
100
28.

fflH^o = 25,2 - 19,8 = 5,4 (g)

Bai 25. Dot chay hoan toan 0,4 mol hon hcJp gom 2 anken dong ding lien tie'p
thu diTdc lu-dng CO2 nhieu hdn luTcJng H2O la 39 g. Lap CTPT cua cic anken
Hrfdngdin

Suy luan: Dot chay anken cho

C-H2-

+O2

0,4

>

=^

= 1 ('"o');

"H20 =

x = 1,5

^

^3ii-r
2
^

n-1

0,7

dong

Tinh thco phi^dlng trinh phan lifng,

nco2

a) nci2

= nc2H4

230.

, •;

c)

O2

= 0,7 (mol)

> nCOj + (n - 1)H20

< nc2H4 hay a > 1 ta thu diTcJc 1 mol C2H4CI2 va c6n (a - 1) mol C2H4

Bai 2.
1. Dot chay ho^n toan 0,56 lit butan (C4H1,)) ct dieu kien tieu chuan va cho ta't
ca san pham chay hap thu vao 750 ml dung dich Ba(0H)2 0,2M. Hoi thu
diTcJc bao nhieu gam ket tiia.
2- Dot chay ho^n to^n V lit butan (d dktc) va cho tat ci san pham chay hap thu
vio 500ml dung dich Ba(0H)2 0,2M thay tao thanh 15,76 gam ket tua.

f

inh the tich V.

Cac phan vlng:

nHjo

= ^ = 0,4 (mol)
18

:

C4H10 + 6,502

> 4C02 + 5H2O
> BaCOj i + W

CO2 + H2O + BaCOj

Tinh so'mol cac cha't:

nr,o,,=-^^
C4H1O

5 . ; - . f f r

Bai giai

CO2 + Ba(0H)2
Ne'udirC02thi

= 33,6 )g)
= 0-6 (mol);

= 1 = a ta chi thu difdc C2H4CI2 (1 mol)

^

n = 3,3 => hai ankin Ih. C3H4 va C4H6

"^~r]r^
44

''

Hif u suSt phan ufng, ndng dp dung dich
Bai 1. Cho 1 mol CI2 tac dung vdi a mol C2H4 thu diTcJc hon hdp X. Hieu suat
phan tfng 100%. Hoi trong hon hdp X c6 nhffng cha't gl, bao nhieu mol.

Bai giai
1. Phan u-ng gii^a CI2 va C2H4: C2H4 + CI2
> C2H4CI2
, :
j
.
Xet 3 truing hdp:
, v -

Hrfdng d i n
mH20

i

b) nci2 > nc2H4 hay a < 1 ta thu dufdc a mol C2H4CI2 va c6n (1 - a)mol CI2

Bai 27. Dot c h d y hoan toan V lit (dktc) mot ankin thu di/dc 7,2 g H2O. Neu cho
tat ca san pham c h a y ha'p thu het vao binh diTng niTdc voi trong diT thi khoi
lufdng binh tang 33,6 g. Lap CTPT ankin
mco2 +

• (u. ^

Hi^dng din

=^ f^ai chat thuoc day ankin.

=> nco2 > "H20

n c : n H = 0,45 :0,6 = 3 :4 => C3H4

CAH

Ta c6 0,4 n = 1,5 => ii = 3,75 => Hai anken la C4H8 va CjHg ^
Bai 26. Dot chay hoan toan 2 hidrocacbon mach hd lien tie'p trong day
ding thu di/dc 44 g CO2 va 12,6 g H2O. Lap CTPT hai hidrocacbon

nco2

= ^ ' ^ ('""D =>"H = 0,6 (mol)

rod

> KCO2+ nH20
0,4 ii

nH20 =7^

,„

n^^Q = TIQQ^ .

Goi X mol CO2 do cung la so mol H2O ta c6: 44x - 18x = 39

^,2

22,4

(D
(2)

> Ba(HC03)2

(3)

=0,025 mol;

'
,,

. . . i . , : ;

£
u.i

i?^:

231


n^jj^

=0,025

X

4 = 0,1 mol;

nBi,(on)2 > nco2

nB„(OH)2

=0,75 x 0,2 = 0,15 mol

4. Gia sur xang la hon hdp cung so' mol cua 2 hidrocacbon CsH|2 va CfiHu.

khong xay ra phan tfng (3) va so mol ket tu;,

J Hoi 1 lit hdi xang (d dktc) nang bao nhieu gam?
2. Can bao nhieu lit khong khi (1/5 the tich la O2) d dieu kien tieu chuan dc dot

BaCOj bang so mol CO2 = 0,1 mol.
i

Vay khoi liTdng ket tua bang: 0,1 x 197 = 19,7g

chay hoan toan 1 gam xang.
Bai giai

2. Cac phan ufng nhtfcau 1:
Tinh: nB„oH)2 =0-5 x 0,2 = 0,1 mol; Us^co^

= 0,08 mol

'

0,5

86 = 79g
79

Vay 1 lit hdi xang (d dktc) nang
= 3,527g.

Co hai triTcing hdp xay ra:

Trirdng hdp 1: CO: ihieu, luc do khong xay ra phan uTng (3)
Ta

CO

n^-o, = nBaco3 = 0,08 mol. Dodo

HQ^IO

X

72 + 0,5

=—^—• = 0,02 mol

-.0 aSv

X

'

Trifdng hdp 2: CO2 dvt, luc do xay ra ca phan uTng (2) va (3)
Tong so' mol CO2 tham gia ca phan iJng (2) va (3) bkng:

• ^ 22,4

" '

C5H,2 + 802

> 5 C 0 2 + 6H20

(1)

Cf,H,4 + 9,502

) . 6 C 0 2 + 7H20

(2)

0,5

pu (3)

Dodo nc^Hio = — =0.03. Ttfc V C ^ H , , , =0,03 x 22,4 = 0,672 lit

X

8 + 0,5

X

'
,, \

}; >

9,5 = 8,75 mol O2

Hay 8,75 x 5 = 43,75 mol khong khi.
Vay de dot chay 1 gam xang can:

x 22,4 = 0,554 x 22,4 - 12,4 1 khong khi.

Bai 5. Dot chay 56 lit khi l\i nhien (cf dktc) chii-a (% the tich) 89,6% CH4, 2,24%

Bai 3.
1. Dien phan 500 ml dung dich NaCl 4 M (dien ciTc trd c6 mang ngan xop). Sau
khi 80% NaCl bi dien phan, neu lay lu'dng clo thoat ra c6 the dieu che to'i da

CjHfi, 4%' CO2 va 4,16%) N2. Cho tat ca san pham chay hap thu vao 5 kg dung
djch NaOH 8%. Tinh nong do % cua cac chat trong dung djch sau phan tfng.

bao nhieu kg thuo'c trijf sau 666.

Bai giai

2. Tai sao ngay nay ngU'di la cam dijng thuoc truf sau 666, DDT.

Cac phan urng chiiy: CH4 + 2O2

Baigiai
1. Phan u-ng dien phan: 2NaCl + 2H2O

H 2 T + C U T + 2NaOH (1)

Tdng so mol NaCl = 0,5 x 4 = 2 mol.
2
'
2
Theo phan urng dieu che thuoc trif sau C 6 H 6 C 1 6 :
C,H, + 3CI2
Theo phan lirng (2)

n666

= 0,8 mol.

100

> C,H,CU
1
08
= -nci2 = -r-mol

(1)

> 2CO2 + 3H2O

(2)

-r- u
,
56x89,6
,

56x2,24
^ ..,
,
Tmhsomol: npu. =
^ = 2,24mol. nr^^^ =
=r.0,056mol.
*^2H6

22,4x100

Theo cac phan u'ng (1,2) tong so' mol CO2 bang:
2,24 X 1 + 0,056 X 2 = 2,352 mol

(2)

ScT mol NaOH b i n g l ^ l i i ^ = 10mol
. 'iH >
100x40
Vi so mol NaOH Idn hdn 2 Ian so mol CO2 nen chi tao ra muoi trung hoa
theo phan u'ng:

0 8
Vay khoi liTdng thuo'c trir sau 666 toi da bkng:

CjHft + 3,50:

> CO2 + 2H2O

22,4x100

So mol CU thoat ra = - so' mol NaCl bi dien phan = — x 2 x

x 291 = 77,6g.

2. Ngay nay ngu'di ta ca'm diing thuo'c trilf sau 666 vi thuo'c nay chuTa clo gay dp'-'

CO2 + 2NaOH

> NajCOj + H.O

(3)

Theo cac phan u'ng (1,2) tdng so mol nifdc bang:

Idn cho ngUdi, hdn niJa thuoc nay ben, ton dong lai trong da't, ni/dc rat la^"'

n^j^o = 2 x ncH4 + 3 x n^^Hf, = 2 x 2,24 + 3 x 0,056 = 4,648 mol

tham chi thuo'c nay co the gay benh ung thu".

Khoi lu'dng dung dich sau phan u'ng b^ng:

232

5

,w -

Theo cac phan tfng (1,2) dc do't chay 1 mol xang can:

0,1 +(0,1-0,08) =0,12 mol

'

.ife ^•;

2. Cac phan iJng dot chay xang:

Ttfc Vc^H,,) = 0.02 X 22,4 = 0,448 lit

pu (2)

J, V i gia sur xang la hon hdp dong so' mol C 5 H 1 2 va C 6 H 1 4 , nen 1 mol xang chtfa
0,5 mol C 5 H 1 2 va 0,5 mol C 6 H 1 4 , liJc 1 mol xang nang:

233


5000 + 2,352 X 44 + 4,648 x 18 = 5187,152g
Vay nong do cua NaaCOs va cua NaOH diTbang:
C% cua Na.C03 = 2352 x 106 x 100 ^ ^ ^ ^ ^ ^ ^
5187,152
C% cda NaOH =

Mziil^:M>^li2ll22 ,

• i. t((>.
4.08%

5187,152

••

tl

Bai 6.
1. TO 10 tan da't den chuTa 96% CaCj c6 the dieu che diTdc bao nhieu
axetilen (d dktc).
2. Lay 1/2 lifdng axetilen cho tac dung vdi H2 (c6 mat t", xiic tdc Pd) de dieu
che etilen va sau do tfung hdp thanh polietilen. Tinh khoi liTdng polietilen
thu du'dc, biet hieu suat moi phan ilrng la 60%.
3. Lay 1/2 liTdng axetilen cho tac dung vdi HCl (c6 mat t", xuc tac HgClz) dc
dieu che vinyl clorua, sau do trilng hdp thanh PVC. Tinh kh6'i liTdng PVC thu
du'dc, bic't hieu suat moi phan iJng la 75%.
Bai giai
1. Phan iJng dieu che axetilen: CaCj + 2H2O

X ^ c dinh th£knh ph^n h § n h^rp

ChudeS.

> Ca(0H)2 + C2H21 (1)

(nhd rang 1 tan = lO'^g = lO^kg; Im^ = 10^ dm^ = 10^ lit)
TU
u' >
10x10^96 , ^ , ^ 5 ,
Theo phan Mng (1): Uc
= "cac, =

" 1.5.10^mol.
^ ^
^
100x64
Vay the tich C2H2 b&ng: 1,5 x lO' x 22,4 = 3,36 x lO^'l = 3,36 x lO'm^

hi 1.
1 Hon hdp khi X chtfa % the tich khi nhif sau: 22,4% CO2, 44,8% CO, 32,8%
CH4. Tinh % kho'i lifdng cua moi khi trong hon hdp X. •
...
2. Hon hdp khi Y chtfa % khoi lifdng nhif sau: 22% CO2, 14% CO va 64% Hj.
Tinh % the tich cua moi khi trong h5n hdp Y.
Bai giai

1. Gia suf hon hdp X la 100 lit thl CO2 chiem 22,4 lit ttfc 1 mol, CO c6 44,8 lit
tiJc 2 mol va CH4 c6 32,8 lit ttfc ^
-1,464 mol.
22,4
Do do tdng khoi lifdng hon hdp b^ng

Vay % kho'i lifdng CO2 =

=35,65%; % C 0 = ^ ^ ^ ^ ? ^ = 45,37%
123,424

123,424

Va cua CH4 = 100 - 35,65 - 45,37 = 18,98%
22
2. Gia siJ hon hdp Y nang lOOg, trong d6 CO2 c6 22g ttfc — = 0,5 mol, CO c6

44
14g tiirc — = 0,5 mol va H2 c6 64g ttfc — = 32 mol.
28
2
V i doi vdi chaft khi ti 1$ the tich dung b^ng ti le so mol, nen ta c6 % the tich
cac khi nhif sau:
% V cua CO2 = % V cua CO =

(2)

nCH2 = CH2

(3)

,

mco2 +'Tico +i"cH4 = 1 x 4 4 + 2x28 +1,464x16 = 123,424g

2. Cac phan lirng cong hdp va trilng hdp:
CH = CH + H2 - ~ > CH2 = CH2

'

0,5x100

• = 1,515%
0.5 + 0,5 + 32

Va % V cua H2 = 100 - 2 X 1,515 = 96,97%
( - CH2 - CH2 -)„

Theo (2, 3) so mol m^t xich bkng ^

so mol C2H4, c6n so mol C2H4 bing

Cac phan iJng dot chay:

- x l , 5 x l 0 ^ x — x — x28 = 756xl0^g = 756kg.
2
100 100
. *
^
3. c a c phan uTng cpng hdp va trilng hdp:

nCH2 = CH —

^

CH2 = CH - CI

(-CH2 - CH2 -)„

hoan toan 2,24 lit A thu difdc 1,568 lit CO2 (5 dktc) va 2,34 gam H2O.
Tinh % the tich va % khoi lUdng moi khi trong hon hdp A.
Bai giai

— so mol C2H2. Do do khoi liTdng PE blng:
100

CH s CH + HCl — ^

Bai 2. Hon hdp khi A gom H2, CO va CH4 d dieu kien tieu chuan. Dot chay

H 2 + 0,502

> H2O

(1)

CO + 0,502

> CO2

(2)

> CO2 + 2H2O

(3)

CH4 + 2O2

(4)
(5)

'

I Goi X, y, z la so mol cua H2, CO, CH4, ta c6 he phifdng trinh:
I Tong so' mol hon hdp: x + y + z =

22,4

= O.lmol

1,568
Khoi liTdng PVC = - x 1,5 x 10^ x — x — x 62,5 = 2637 x 10^ g = 2637kg.
2
100 100
234

I To'ng so' mol CO2: y + z =

= 0,07mol

(a)
(b)

22,4
235


••v~

-

T o n g so m o l H2O: x + 2z =

= 0,13niol

(c)

^ •'

\ihn hdp D thi thu diTdc hon hdp k h i nang b^ng ctan (C2H6). Biet cac the tich

18

jihi deu do d dieu k i e n tieu chuan. T i m cong thuTc phan tOr cua X.

G i a i he phifOng irinh (a, b, c) ta tint diTdc:
X

H5n hdp k h i D g o m 5 l i t H2, 5 l i t CH4. N e u them 15 lit hidrocacbon k h i X v^o

B a i giai

= 0,03 m o l ; y = 0,02 m o l va z = 0,05 m o l

V a y : I ) Phan tram the tich:

CuaCO=M^.20%

C a a H 2 = M 3 ^ . 3 0 % ;
0,1

0,1

^rl^hl'^^

Cua C H 4 = 100 - 30 - 20 = 50%
' "

KhS"! lu'dng cua 20 l i t A bang: — ^ x 2 + - ^ ^ x 3 0 - — =
22,4
22,4
22,4

'

0,03x2x100

r-> u
CuaH2=

= 4,23%
0,03

X

2 + 0,02

X

C a a CO

Khoi liTdng cua 20 l i t B bKng:

28 + 0,05 x 16
= 39,44%

1,42

'

Cua C H 4 = 100 - 4,23 - 39,44 = 56,33%
1. T i n h % khoi luTdng m o i nguyen to trong aminoaxit N H 2 - C H 2 - C O O H ( g l y x i n )
2. Ne'u mot day hidrocacbon di/Oc bieu dien b d i cong thtfc chung CnH2n + 2 thi
thanh phan % cua hidro bien d o i nhiT the nao k h i gid t r i n thay d d i .

» N = l l i l i ^ = 18.67%
75

%

%o=

^ 2 ^
75

'

= 42,67%

100

14n + 2

14n + 2

100

2n + 2
K h i n = 1 thi % H =

X

=—

2. K h i noi hon hdp k h i D nang bang etan (C2H6) c6 nghla la 1 l i t D nang bang 1
b^ng 30g. D o do ta c6 bieu thtfc ve khoi liTdng 1 m o l D :
15
-x2 + -xl6 +
22,4
V22,4
22,4

n +1

^_

u.^mt^

;<tff« >

22,4^5x2 + 5xl6 + 15xMx
25

~

25
, , jri,-.5

»v

Bai 5.
1. Cho 5,6 lit ( d dktc) hon hdp khi CH4 va C2H4 di qua nUdc brom dU' thay c6 4
gam brom tham gia phan tfng. T i n h % the tich moi khi trong hon hdp.
2. Cho biet 2,8 lit (c( dktc) hon hdp khi C H , , C2H4 va C2H2 tac dung vifa du vdi 500
B a i giai

C2H4 + Br2

n+1
= 25% (do la triTcJng hdp CH4)

coi bang khong, t h i : % H = —
7

V a y thanh phan % H bien thien trong khoang:

f*

|v K h i qua niTdc brom chi c6 C2H4 phan uTng:

1 +1
K h i n v6 ciing Idn thi

= 19,64gam
^

Ket luan: H o n hdp k h i A nSng hdn hon hdp k h i B.

2. Thanh phan % cua hidro trong ankan CnH2n+2 diTdc tinh theo bieu thiJc:
(2n + 2 ) x l 0 0

22,4

ml dung dich B r j 0,04M. H o i % the tich cua CH4 bien doi trong khoang nao.

Hoac ( % 0 = 100 - 32 - 18,67 - 6,67 = 42,67%)

%H =

22,4

440

x28 =

B i ^ n luan nhU cac bai tren ta de dang tim di/dc CsHx.

' ' ^ C = ' - i ^ . n %
75
= 6,67%

22,4

10

Goi cong thiJc cua X la C^Hy. Ta c6 12x + y = 44

NH2 - C H 2 - C O O H ( M = 75)

^
75

xl6 +

Rut ra: M x = 44.

B a i giai
1. T i n h % k h o i Itfdng m o i nguyen to' trong g l y x i n

=

10

20,53gam |,

lit etan hoac k h o i Itfdng cua 1 m o l D bang k h o i lu'dng cua 1 m o l etan C2H6

Bai3.

H

1 lit hon hdp hoSc 1 m o l hon hdp, nhiTng trong bai nay ta chi can tinh khoi
liTdng cua 20 l i t hon hdp A cung nhiT B:

^

2) Phan tram k h o i liTOng:

Ve nguyen tac de biet hon hdp k h i nao nang hdn, ta can tinh khoi lu'dng cua

= 42,29%

14,29% < % H < 25%

B a i 4.
1. H o n hop k h i A g o m 5 l i t H2 va 15 l i t C2H6, hon hop k h i B g o m 10 l i t CH4 va

(1)

> C2H4Br2

I T d n g so m o l h o n hdp b^ng —
= 0,25mol.
22,4
4
Theo phan u r n g ( l ) nc2H4 = "Br2 =
= 0,025mol.
160
^Vay % the tich cua C2H4 =

0,025x100
0,25

11'!,! > .:

= 10%

I V a % the tich cua CH4 = 100 - 10 = 9 0 %

K h i cho hon hdp khi qua niTdc brom c6 cac phan iJng: -

10 l i t C2H4. Cac the tich k h i deu d dieu k i e n t i e u chuan. H o i hon hdp k h i A

C2H4 + Br2

hay hon hcfp k h i B nSng hdn?

C2H2 + 2Br2

> C2H4Br2
> C2H2Br4

(1)
(2)

231


CnH^2n + 2+

2g
So'mol cua honhcfp bang —-— = 0,125moL
22,4
S6' mol

brom = 0,5

x 0,04

= 0,02

nhau tham gia

phan iJng.

1. Tinhtile V, : V2
2. Neu cho ciing the tich, tuTc V, = V2, thi ti 1$ khoi li/dng brom tham gia phan
tfng doi vdi hai triTdng hcfp nhu* the' nao?
Bai giai
1. Cac

phan iJng xay

ra vdi

brom:

Hon hdp A:
C2H4 + Br2
>• C2H4Br2
(1)
Hon hdp B:
C2H2 + 2Br2
> CiHiBti
(2)
Vi li/dng brom phan tfng b^ng nhau nen so mol C2H4 phai gap doi
C2H2, do do the

tich V, phai gap

doi

the

tich V2, ttfc V,

so

mol

: V2 = 2 : 1.

2. Theo cac phan liTng (1,2) thi so' mol Br2 d phan tfng 2 ga'p doi so mol Br2 cl
phan iJng 1, nen ti 16 khoi liTcfng Br2

la 1 : 2.

Bai 7. Hon hdp X gom CO2 va hidrocacbon A (C„H2n + 2). Tron 6,72 lit X vdi mpt
li/dng diT oxi roi dem dot chay hoan toan X. Cho san pham chay Ian lu'dt qua
binh 1 dirng P2O5 va binh 2 diTng liTdng diT dung dich Ba(0H)2 tha'y kho'i
liTdng binh 1 tang 7,2 gam va trong binh 2 c6 98,5 gam ke't tiia. Tim cong
thtfc phan tuf cua hidrocacbon A; tinh % the tich va % khoi Irfdng ciia A trong
hon hdp. Cac the tich khi do d dilu kien tieu chuan.
Bai giai
238Phan iJng dot chay A:

nC02 + (n+

(1)

(2)
(3)

1)H20

Binhl: H2O + P2O5
> 2HPO3
Binh 2: CO2 + Ba(0H)2
> BaCOj i + H2O
Gpi a va b la so mol A va CO2, ta c6 cic phiTPng trinh
(4)
72
Theo phan
phan iJng
urng (1),
(1, 3),
so molnH20
CO2:= ncoj
= an= + b == 0,4mol
= 0,5mol. (5)
Theo
so tong
mol niTdc:
a(n +1)
6 72
(6)
22,4 = 0,3mol.
Theo dieu kien cho: a + b = -r—
Lay bilu thtfc (5) triT (4) ta c6: b - a = 0,1.
K6't hdp vdi (6) ta de dang tim dufdc a = 0,1 mol va b = 0,2 mol.

Thaygid tria = 0,l vao(4)tac6n = 3.
,
Vay CTPT cua A la C3H8.
, .
Tinh tha nh pha n pha n tra m:

mol

Gpi X , y, z la so' mol cua C2H4, C2H2, CH4. Ta c6 he phifdng trinh:
Tong so mol hon hcJp: x + y + z = 0,125 (a)
Tdng so mol Brj: x + 2y = 0.02 (b)
Gia sur X = 0; the y tuf (b) vao (a), ta c6 z = 0,115 mol.
Gia sur y = 0; the x ti{ (b) vao (a), ta c6 z = 0,105 mol.
* f
Nhir vay % the tich ciia CH4 n^m trong khoang:
0,115x100
0,105x100
0,125
^
0,125
92% > %CH4 > 84%
Bai 6. Hon hdp khi A chiJa nhi^ng the tich bkng nhau CH4 va C2H4. Hon hcJp khi
B chtfa nhifng the tich b^ng nhau CH4 va C2H2. Cac hon hdp khi deu ct dieu
kien tieu chuan.
Cho V, lit A va V2 lit B qua ni/dc brom (diT) deu tha'y mot li/dng brom nh\i

3n + l- Op,2

0

1x100

- Phin tram thi tich cua A = ^ ^ ^ - ^ = 33,33%
0.1x44x100 = 33,33%
- Phan tram kh6i liTdng cua A = 0,1x44
+ 0,2x44
S6 dl % the tich bing % khoi liTdng vi hai khi c6 ciing khoi liTdng phan tiJf.
Bai 8. H5n hdp khi A gom H2, CO va C4H,o (butan). De dot chay 17,92 lit hSn
hdp A can 76,16 lit oxi, thu duTdc 49,28 lit CO2 v^ a gam niTdc.
1. Tinh % the tich cua C4H10 trong hon hdp A.
6!) h '
2. Tinh kh6'i liTdng niTdc a.
Cho cic the tich khi do d dktc.
Bai giai
Cac phan i?ng chiy:
H2O
(1)
H2 + 0,502
CO2
(2)
CO + 0,502

I

-> 4CO2 + 5H2O

C4H,o +6,502
|Goi

X , y, z la so mol

cua

|Theo bai toan x + y + z =

H2, CO,

17 92

(3)

C4H10

.

= 0,8mol

^
76

[leo cac phan tfng (1, 2, 3), Itfdng oxi b^ng: | +1 + 6,5z =

16

239
= 3,4mol


GiSi he phi/dng trinh ta c6 z = 0,5 mol. Ttfc ^ V C ^ H J O = Q g = 62,5%
49 28

2. Tong so mol CO2 =
= 2.2mol
22,4
Theo cdc phan iJng (2, 3):
So mol CO = so mol C O 2 = 2,2 - 4 x 0,5 = 0,2 mol.
Dodo
=0,8-0,5-0,2 = 0,lmol.
Theo phan uTng (1): n^jo ^
O.lmol
VSy tong so mol H2O = 0,1 + 5 x 0,5 = 2,6 mol
Kh6'i liTdng niTdc a = 2,6 x 18 = 46,8 gam.
Bai 9. D6't chdy ho^n toan 27,4 lit hon hdp khi A g6m CH4, CjHg va CO, thu
dirdc51,41itC02.
1. Tinh % the tich cua CaHg (propan) trong hon hcfp khi A.
2. H6i 1 lit hon hdp A nSng hay nhe hdn 1 lit N2
Biet cdc the tich do d dktc.
B^i giai
1. Cdc phan tfng d^t chdy:
C H 4 + 2O2

> C 0 2 + 2H20

(1)

CO

> CO2

(2)

> 3CO2 + 4H2O

(3)

+

0,502

C 3 H 8 + 5O2

Gpi V i , V2, V3 m the tich cua CH4, CO, CjHg, ta c6 cic phuTdng trinh
V| + V 2 + V 3 = 27,4 lit
V , + V 2 + 3 V 3 = 51,4 lit

Tilf d6 ta CO 2V3 = 51,4 - 27,4 = 24 => riit ra V3 = 12 lit
Vay%Vc3H«=^^=43,8%
2. Tong phan tr5m the tich cua CH4 CO bkng
100% - 43,8% = 56.2% (hoSc viet thanh 0,562)
Kh6i Itfdng cua 1 mol hon hdp A b^ng
M A = 0,438 x 44 + 28x + 16(0,562 - x)
(trong d6 x la % the tich cua CO va 0,562 - x la % the tich cua CH4).
Gia sur ngoai
phan c6n lai la C H 4 (khi nhe hdn CO) thi:
M A = 0,438 x 44 + 16 x 0,562 = 28,264
VI M A > M N 2 (28) do d 6 1 lit A n&ng hctn 1 lit N2.
(C6 the ddn thuan diTa vko bieu thuTc tinh MA c» tren v^flmM A khi x n^m 5 2 chn240
C3HS

Khi X = 0; M A = 28,264. Khi x = 0,562 ; M A = 35,008
NhiT vay 28,26 < M A < 35,008).

0ai 10. Cho biet 2 lit hon hdp khi gom hidro, metan v^ cacbon monoxit d dktc
nSng 1,715 gam. De dot chiy hokn toan 4 the tich hon hdp khi do can 19 the
tich khong khi (1/5 the tich la O2). Tinh % the tich moi khi trong hon hdp.
Bai giai
1; ,t'-s
'Cdc phan uTng dot ch^y:
2H2O

2H2 + O2

—

2 C O + O2

— -> 2 C O 2

C H 4 + 2O2

—

C O 2 + 2H2O

ilCh^ li/dng mol cua hon hdp b^ng: hJl^^^
= 19^2g/mol^5/'
C^ch 1: Goi V,, V2, V3 la the tich cua H2, CO va CH4 trong 4 lit hon hdp ta c6:
V,+ V 2 + V 3 = 4
(I)
(ID
°2 2 2 + 2V3=y = 3,8
r (I, II) rut ra V 3 = 1,2 lit. Vay % V cua C H 4 b^ng: 19x100

'
= 30%
4
Mat khdc theo khoi Itfdng mol cua hon hdp ta c6:
,
M = 19.2 = 16 X 0,3 + 28x + 2(1 - 0.3 - x)
(III) ,^,|
Trong do x la % the tich cua CO
Gidi phiTdng trinh (III) ta c6 x = 0.5 tuTc CO chiem 50% va H2 chiem:
100 - 3 0 - 50 = 20% thI tich.
;
Cdch 2: Gpi V la the tich CH4 trong 4 lit hon hdp ta c6 phufdng trinh:
Vo2=2V + l ( 4 - V ) = 3,8

...v.'V

Rut ra V = 1,2 lit ttfc CH4 chiem h ^ l ! ^ = 30%. phan tiep l^m theo each 1.
4
B^i 11. Cho 43.2 gam hon hdp X gom Ca va C a C 2 tic dung vira het vdi nifdc
thu drfdc hon hdp khi A (kho). Cho A vao binh kin (c6 mSt bot Ni xuc tac),
dun nong mot thdi gian diTdc hon hdp khi B. Chia B thanh 2 phan b^ng nhau.
Cho phan 1 di rat cham qua binh diTng dung dich niTdc brom diT. thay c6 4,48
lit (d dktc) hon hdp khi C di ra khoi binh v^ khoi liTdng binh tang them 2,7
gam. Biet 1 lit khi C nang 0,4018 gam.
. , ^ ,^ , , , ^ , 1- Tinh % the tich moi khi trong ttfng hon hdp A, B, C.
241
2. Dot chay hoan to^n phan 2. Tinh khoi liTdng CO2 vk H2O tao thanh.


Cty TNHH

B ^ i giai
C d c PTPLf:

MTV DWH

Khang

Viet

100

Ca + 2 H 2 O

Ca(OH)2 + H 2 t

CaCz + 2 H 2 O

-> Ca(OH)2 + C2H2

%H2

(1) Hon hcJp A gom
(2)

C2H2 v a H2

C2H2 + H2

C2H4

(3)

Hon hdp B gom

C2H4 + H2

C2H6

(4)

C2H4 va C2H6

C2H4 + Br2

-> C2H4Br2

=50%
0,6

•

| V i kho'i ItfcJng cacbon va hidro trong A cung nhiT trong B n6n c6 the tinh theo
ic chat trong A.

(5) Hon h d p B gom C2H4
(6) vh H2 CO trong 1/2 B

C2H2BU

C2H2 + 2Br2

= 0,3x

C2H2 + 2 , 5 0 2

2CO2 + H2O

(7)

H2 + 0 , 5 0 2

H2O

(8)

Choi li/dng H2O = 1 x i x (0,3 + 0,6) x 18 = 8,Igam

1. Kh6'i IiTdng cua hon hdp khi C bkng: 0,4018 x 4,48 = 1,8 gam.

Ipioi lurdng C O 2

Goi so' mol cua H2, C2H6 trong C l a a, b ta c6:

2a + 30b = 1,8

^ X

0,6

X

44 = 13,2gam

B^i 12. Dot chay ho^n toan V lit metan (dktc) va cho ta't ca san pham hSp thu
hoan to^n vao binh duTng 500 ml dung dich Ba(OH)2 0,2M thay tao thanh
15,76 gam ket tua.

Giai ra ta difdc a = 0,15 va b = 0,05
: . Vay % the tich cua H2 bang "'^^""^^^ = 75%
^

=

1. Tinh the tich V.

0,2

Va % the tich cua C2H6 bkng 100 - 75 = 25%

2. Hoi kho'i lu'dng bmh diTng dung dich Ba(0H)2 tang hay giam bao nhieu gam?

C6 the giai theo KLPTTB cua C

3. Hoi kho'i ItfcJng dung dich trong binh tang hay giam bao nhieu gam?
'

I

Mc =0,4018x22,4 = 9 = -30b + 2 ( 0 , 2 - b )

0,2

Cac phan iJng c6 the c6:

Khoi liTdng 1/2 B b^ng tong k h o i lifdng C v a k h o i liTdng binh tang (khoi
liTdng C2H4 v a C2H2) = 1,8 + 2,7 = 4,5 g a m .

Khoi liTdng A b^ng 2 Ian k h o i liTdng B = 4,5 x 2 = 9 g a m .
Theo c a c p h a n tfng (1,2) thi nca =
"caCa = " C 2 H 2 = y . n e n ta c6: •

= x

B a i giai

'

C H 4 + 2O2

> C O 2 + 2H2O

(1)

C02 + Ba(OH)2

>BaC03;+H20

(2)

Neu d i r C 0 2 : C O 2 + H2O + BaCOj

> Ba(HC03)2

= 0,08mol

'2x + 26y

=9

Trirdc het can tinh:

40x + 64y

=43,2

Trirdng hdp 1: C O 2 thieu, ttfc nco2 < 0.1 mol, luc d6 khong c6 p h a n tfng (3)

Giai ra ta difdc x = 0,6 va y = 0,3.

^) n c o 2

= "BaC03

^i^^(Q\i)2

=0,5xO,2 = 0,lmol;

(3)
15,76

Hgg^Q^ =•

= 0,08 mol = ncH4 . Vay Vcoj = 0,08 x 22,4 = 1,792 lit.

NhU' vay % the tich ciia cac khi trong A la:
%C2H2 =

0,3x100
0,3 + 0,6

2) Khoi liTdng binh dung dich tang (m).
= 33,33%; % H 2 = 1 0 0 - 33,3 = 66,77%

Trong B: so mol C2H6 = 0,05 x 2 = 0,1 => nH2 = 0,15 x 2 = 0,3

So mol C2H4 = so mol H2 d phan tfng (3)

m = mco2

+ I " H 2 0 = 0,08

x 44

+ 0,08

t ,1* ,(^>«'
x 2x

18 =

6,4g

IKhoi lirong dung djch trong binh giam (a) b^ng khoi liWng ket tua truT tong
ikhoi liwng C O 2 + H2O: a = 15,76 - 6,4 = 9,36 gam
pVirdng hdp 2: C O 2 du", tiJc nco2 > 0,1 mol, luc d6 xay ra p h a n tfng (3)

= nH2 b a n d a u - n H 2 d i r - nH2 tao ra C2H6 = 0,6 - 0,3 - 0,2 = 0,1

\ Tong so mol C O 2 = 0,1 + (0,1 - 0,08) = 0,12 mol

So mol C2H2 = nc2H2 ^an dau - n c 2 H 4 - " C z H f i = 0,3 - 0,1 - 0,1 = 0,1.

Vay % the tich cua % C2H6 = %C2H4 = %C2H2 = 0,1 x — = 16,67%
0,6
242

|

'

Vay Vco2 = 0,12 x 22,4 = 2,688 lit.

|)

Khoi lu'dng bmh dung dich tang (m')
m' =0,12 X 44 + 0,12 x 2 x 18 = 9,6 gam.
'
Khoi liTdng dung dich trong binh giam (a'): a' = 15,76 - 9,6 = 6,16 gam
243

I Cic phifcfng t r i n h p h a n iJng :

ChUOng V.

DIIV xuiT can mnocacsoN - POUME

A. L i T H U Y E T C d B A N V A N A N G C A O
I . RvKfu etylic C2H5OH
RiTdu etylic Ik mpt chat long, khdng m ^ u , tan v6 h a n trong n\idc.
+O2

Men nJOu

*

- t ^ ^ CH3-CH2-ONa

CH3-CH2OH

4-CH3COOH CH3-COO-CH2-CH3

+H2O

C H 3 - C H 2 O H + O2

Mengia'm

QH,206(dd,
(Dtfdng glucozcJ)

2CH3COOH +CuO

—-> (CH3COO)2Cu + H2O

2CH3COOH + Z n

—

(CH3COO)2Zn + H2T

2CH3COOH +CaC03

—

(CH3COO)2Ca + H2O + CO2T

CH3C00H + NaOH

— > cH3C00Na + H2O

2C02(k) + 3H20(„

> CzHsONad, + l/2H2(k)
) CH3COO-C2H5(,) + H2O

CH3COOH0, + C2H3OH0)
Do rifcJu =

^

V •

CH3COO-C2H5 + H2O

01. C h a t beo (RCOO)3C3H5 (ran ho§c long)
|Chat b e o la h o n hcJp nhieu este cua glixerol C3H5(OH)3 vdi cic
lCOOH(vdiRlaCnH35-; CnH33-;C,5H3,-...)

axit b^o

^
''

(RCOO)3C3H5 + 3H2O

> C3H5(OH)3 + 3 R C 0 0 H

IV. Glucozcr CfiHizOfi
GlucozcJ la cha't ran, mau tr^ng, ngpt de tan trong ntfdc.
Hgp chd't cua Ag
+H20, t"

*dungdichnr0u

Qujr tim
„ , ^,
—
• Hoi do
+CuO

(CHjCOOzCu
+Zn
(CH3COO)2Zn

• O , . xt, t'
C4H 10
CH3-COOH

, (CH3COO)2Ca
+NaOH
+C2H5OH

C3H5(OH)3 + 3RC00Na

(RCOO)3C3H5 + 3NaOH

^ n f ( 1 u n g u y e n c h a ' t ^ ^QQO

I I . Axit axetic CH3COOH
Axit axetic Ik mpt chS't long, khSng mku, vi chua, tan nhi^u trong nirdc.

C2H5OH
j
Men gifl'm

H2S04.cl»c.t»

mmwim .

?hin ufng xa phong hoa :

C2H50H(,)

C2H50H(,) + 302(k) — ^
CzHjOHd) +Na(rt

^ CH3COOH + H2O

?han iJng thuy phan:
> 2C2H50H(,) + 2C02(,)

C2H4(k) + HjOd)

2CH3COOH + H 2 0

|C6ng thiJc Chung : (RCOO)3C3H5

C^c phiTdng trinh phan tfng :

244

xt,t"

C H 3 C O O H + C2H5OH

C6H,206
C2H4

CO2

C4H,o + 2,502

(tinh bpt)

Men nr^u

CzHjOH

Cdc phiTdng trinh phSn uTng :
(-C6H,o05-)„ + nHaO
CfiH.zOft + Ag20
CfiHizOfi

axit, Io

-> nQHuOe

C6H,207 + 2Agw(g»Wngbac)
) 2C2H5OH + 2 C O 2

Saccaroztf C12H22O11
Saccarozd la chaft r^n, m^u tr^ng, c6 vi ngpt, d§ tan trong n\Xdc.

CHjCOONa
CH3COOC2H5

C6H,206

Axit gluconic + Ag

CI2H220II

>• Khong cho phan tfng trang giTcJng
+H2O, t", H *

Glucozd + Fructozd

, ,


Cty TNHH MTV DWH Khang Viet

Boi dudng hoc sink gidi Hod hoc 8, 9 - DAo HOu Vinh

CizHzzOn

axit.to

+H2O

GlucozcJ

Fructozd

VI. Tinh bOt va xenlulozcf (-CgHioOs -)„
* Tinh bot la chat r^n mau tr^ng, khong tan trong nifdc lanh, tan trong nirg
n6ng tao thanh ho tinh bot.
* Xenlulozd la cha't ran mau trang, khong tan trong niTdc lanh va ca niTdc nong
* Ca tinh bot va xenlulozcJ deu tham axit.t''
gia phan tfng thuy phan:

(-C6H,o05-)„ + nH20

* Deu duTdc tao ra trong cSy xanh nhcJ qua trinh quang hdp:

^^^^^ ) (-QH,o05-)„ + 6n02

6nC02 + 5 n H 2 0

* Tinh bot cho phan iJng vdi iot tao thanh san pham c6 mau xanh.
VII. Protein
* Protein c6 trong cac bo phan cua cd the sinh vat.
* Protein chiJa cac nguyen to C, H, O, N va c6 the c6 S, P, Fe...
* Protein diTcfc tao ra tilf cac aminoaxit (vi du: axit amino axetic H2N-CH2COOH)
* Protein tham gia phan tfng thuy phan:
Protein + nUdc axit hoacba/(f Hon hdp aminoaxit
* Mot s6 protein bi dong tu khi dun nong (vi du 16ng tr^ng trifng).
* Protein bi phan huy khi dun nong manh tao ra cha't bay hdi c6 mtii khet.
VIII. Polime
* Polime la nhffng chat c6 phan tur khoi rS't Idn do nhi^u m^t xich lien ket vdi
nhau tao nen.
* Polime diTdc chia lam 2 loai: polime thien nhien va polime tdng hdp.
* Polime CO cau tao mach thing hoSc mach nhanh hoSc mang liTdi khong gian.
* Polime thirdng la chat r^n, khong bay hdi.

B. BAI TAP THEO CHU o i
Dang

1: Nh$n bi^t -

Ikch

h6n hpp - Tinh ch^ c&c chdt

I. Bai tap c6 lari gi§i

Neu phifdng phap hod hoc de phan bi6t cac chS't sau: CH4, C2H4, C2H2, CO:Hurdng din giai
a) Nhan CO2 b^ng dung dich Ca(0H)2: CO2 lam van due nirdc voi trong.
Dilng dung dich Brj (vdi nhifng IvTcfng nhiT nhau) de phan biet CH,, C2H4
C2H2 (vdi nhffng the tich nhff nhau).
'246
1.

: Lam dung dich Br2 phai mau nhieu.
H4 : Lam dung dich Br2 phai mau it.
CH4 : Khong lam dung dich Br2 phai mau.
2 Trinh bay phffdng phdp hod hoc de nhan biet ba chat long : benzen, rffcJu
' etylic, axit axetic.
HtfdngdSngiai
• '
Difa vio tinh cha't cua axit axetic khac vdi tinh cha't cua rffdu etylic, khdc vdi
tinh chat cua benzen de nhan biet theo cac cdch sau :
^' "'
Cdch 1: Dung quy tim nhan ra axit axetic: quy tim hoa d6.
Cdch 2: Diing muoi Na2C03 hoac CaCOs nhan ra axit axetic: siii bot khi CO2.
2 C H 3 C O O H + CaCOj
> (CH3COO)2Ca + C O 2 1 + H2O
Cdch 3: Diing kim loai manh nhiT: Mg, Zn, Fe... nhan ra axit axetic; kim loai
W tan dan va cd khi H2 bay ra.
C2H2

f

2 C H 3 C O O H + Zn

> (CH3COO)2Zn + H 2 t

Sau khi nhan ra C H 3 C O O H ta phan biet riTdu etylic va benzen b^ng mot
||: trong hai each sau:
Cdch 1: Cho Ian lifdt tffng chat tac dung vdi Na, riTdu etylic c6 phdn iJng tao
khi H2 bay ra, benzen khong cd phan iJng.
^
2C2H5OH + 2Na
> 2C2H50Na + H 2 1
Cdch 2: Cho tiifng chat tdc dung vdi axit axetic cd them H2SO4 ddc, xuc tdc,
trffdng hdp cd miii thdm cua este la C2HSOH :
C H 3 C O O H + C2H5OH
>CH3COOC2H5 + H2O
•
(mill thdm)

II. B^i tap W giii

'>

1. Hay chi ra day chat n^o diTdi day khong the lam mat m^u dung dich niTdc
I brom (do phan uTng cpng )
'|0>H';>
" a) C H 3 - C H 3 ; CH2=CH2; CH3-CH2OH.

; b) CH2=CH2; C H s C H ; CH3-CH=CH2.

c) CH2=CH-CH2-OH ; CH^CH ; CH3-CH=CH2.
J d) CH=CH ; CH3-C=CH ; CH3-CH=CH-CH3

"^-^

•..J.Mmy

A, B la hai hdp chat hffu cd cung cd cong thtfc phan tur C2H6O. Cong thiJc cau
tao cua chiing nhiTsau :
1) C H 3 - O - C H 3 ;

2) CH3-CH2-OH.

Hay cho biet cong thtfc cau tao nao la cua chat A ? cua chat B ? Biet r^ng A
lal8 chat long tan v6 han trong nxTdc, tie dung vdi kali giai phdng hidro. B la
chat khi khong tan trong nffdc, khong tdc dung vdi kali.


Dang 2: V\4t c6ng thOc c^u tgo,
Ho^n th^nh phifOng trinh phdn Cfng - Oi^u ch^

1^) Etilen, axetilen, benzen deu khong lam mat mau dung dich brom.
g) Axetilen, etilen, deu lam ma't mau dung dich brom.
Hrfdng d§n giai

I. Bdi tSp CO lofi gidi
1. Nhifng cha't sau day c6 diem gi chung (thknh phan, ca'u tao, tinh chat) ?
a) Metan, etilen, axetilen, benzen.

Cau e.

5. Hay giai thich tai sao

va benzen lai k h o n g c6 phan uTng nay.

c) Protein, tinh bot, xenlulozd, polietilen.

Htfdng d i n giai

d) Etyl axetat, chat beo.

Drfa vao thanh phan v^ ca\ tao phan tur:

Hvtdng dSn giai

o

b) Deu la dan xuat cua hidrocacbon c6 3 nguyen to C, H , O.

kim loai kiem nhu* Na.

d) Deu la este, c6 nhom - C 0 0 - .

2C2H5-O-H + 2Na - — > 2C2H5-0-Na + Hzt

2. DiTa tren dSc diem nao, ngtfcfi ta xe'p cac chat sau vao ciing mot nhom :

2 C H 3 - C - O - H + 2Na

a) Dau mo, khi tiT nhien, than dd, go.

> 2CH3-C-0-Na + H j t

p.

O
O
Trong phan tijT CH4, C2H4 va Cf,U(, khong c6 nhom - O H , khong c6 nguyen tur
H Unh dong nen khong tac dung diTdc vdi kim loai kiem nhiT Na.

b) Glucozd, saccarozd, tinh bot, xenlulozd.
HuTdng d§n giai
a) Thuoc nhom nhien lieu (cha't dot).

6. Vie't cac phifdng trinh phan iJng thtfc hien cac bie'n hoa sau:

b) Thuoc nhom gluxit.

Etilen — r i T d u etylic

3. Vie't cac phiTdng trinh phan uTng thtfc hien cac bien hoa hoa hoc sau :
^'^ > Glucozd — R i T d u etylic — A x i t

Etyl a x e t a t — R i T d u etylic — E t i l e n

(-C6H,„05-) + nH20

)

CfiH.zOfi

Mennf^»u

C2H50H + 02

^""^^'^

C H 3 C O O H + C2H5OH
CH3COOC2H5 +

^

—

—Polietilen.

nCfiH.zOs

Natricacbonat—^^

natri axetat

(1) H2O (xuc tic axit); (2) C H 3 C O O H ; (3) oxi

16n men d 30 - 32"C ;

chtfc: nrdu, axit,

este.

b) Vi^t cAc phiTdng trinh phan tfng thifc hi^n sd do bien hod d6.
HiMng dSn giai

_ J l 2 S O l d ? c _ ^ C2H4 + H2O

c) Metan, etilen, benzen deu khong lam ma't mau dung dich brom.

i

7. a) Vj^t sd do hiiu diSn m^i quan h$ giCTa hidrocacbon \k h0p cha't c6 nhdm

H2O . "^^°^'"'"§=±CH3COOH + C2H50H

a) Metan, etilen, axetilen deu lam ma't mau dung djch brom.

• e t y l axetat

(4) dd NaOH; (5) Mg ; (6) C2HJOH (xuc tic H2SO4 dac).

. "2^°4'»g«=-'" ^ CH3COOC2H5 + H2O

b) Etilen, axetilen, benzen deu lam mat mau dung dich brom.

<—^-^—II

Cho tac dung vdi cic chaft:

?

) C H 3 C O O H + H20

4. Chpn nhi?ng cau dung trong cac cau sau :

* magie axetat

HritfngdSngiai

2C2H5OH + 2CO2

> (-CH2-CH2-)„

\
axit axetic

axetic —

Hvtdng d i n giai

248

'

Phan tuf CO nhom - O H nen nguyen tijT H linh dpng de dang bi thay the bdi

c) Deu la polime.

nCH2=CH2

i

Rtfdu etylic C 2 H 5 - O H va axit axetic C H 3 - C - O H ,

a) Deu la hdp cha't c6 2 nguyen to C va H. Khi dot chdy cho CO2 va H2O.

C2H5OH

etylic v^ axit axetic deu tac dung vdi kim loai

kiem, giai ph6ng khi hidro, trong khi do cic hidrocacbon nhU" metan, etilen,

b) Ri/du etylic, axit axetic, glucozd, protein.

Tinh bot

riTdu

^

a) etilen

^"2° > riTdu etylic
axit

b)

•

°2

> axit axetic

Men

> etylaxetat
H2SU4a

CH2=CH2 + HOH ^ 5 i i M M _ > CH3-CH2OH
C H 3 - C H 2 O H + O2
CH3COOH +CH3-CH2OH

> C H 3 C O O H + H2O
HgS04c6c
t"

>CH3COO-CH2-CH3+ H2O


Boi dwdng hoc sink gidi Hod hoc 8, 9 - Dao HOU Vinh

m H 2 0 = 1 8 . 0,1 = 1,8 g => V H 2 0 = 1,8

II. Bdi tSp tif g i i l
1. V i e t phiTdng trinh phan tfng (ghi ro dieu k i e n neu c6) di

thifc h i e n scf d6

chuyen hoa sau :
- ^ C z H s O H — ^

CH2 = CH2

CH3COOH—^

(CHjCOOzZn

e5r.du=
•

I)

CaCOa

+H2O

>CaO 2000"c
-> CaCz ^(3)

C2H2-^iilZd,^C2H4
•t"(4)

CHsCOONa

— ^

C2H5OH

(6)
CH3COOH

Dang 3: Tinh theo c6ng thOc phuong trinh phdn Qng,

hJQu sudt phdn Ong, ndng dQ dung djch

I. Bdl tSp CO IcTi g l i i

1. Dung dich A la hon hdp rtfcfu etyhc va nvldc. Cho 20,2 g A tac dung vdi Na
(lay dM) thay thodt ra 5,6 lit khi H2 (dktc).
a) Xdc dinh nong do nfcJu cua dung dich A ; bie't khoi li/cJng ri6ng ciia riTdu
etylic la 0,8 gam/ml, cua ntfdc la 1 gam/ml.
b) Neu diing rU'du 40" cho tac dung vdi Na, thi can bao nhieu gam riTdu nay dc
du'dc lifdng hidro noi tren.
Hufdng din giai
*.
Nirdc va ru'du etylic cung tac dung vdi Na giai phdng hidro, ne'u gpi x va y la
so m o l ru'du, niTdc trong hon hdp, se lap du'dc he phiTdng trinh ve so m o l khi,
va khoi lufdng hon hdp, tijT do tim ra x, y. Tilf so m o l m o i cha't ta suy ra khoi
li/dng va the tich cha't long va suy ra dp ru'du...
a) n H 2 = 5,6 : 22,4 = 0,25 mol
C2H5OH + Na
> C2H50Na+ O.5H2T
X

H2O

y ,

•

+ Na

•••••aI

> NaOH +

0,5x

0,5H2t

0,5y

nH =0,5x + 0,5y = 0,25"
2
^ => x = 0,4; y = 0,l
m^h-46x + 18y = 20,2 J
mczHsOH = 46.0,4 = 18,4g =^ V C ^ H J O H = 18-4 : 0,8 = 23 m l
250

,

H2O

b
nH2=

0.5a

+

0,5b

ml

,,,,40.

+ 1,8

•

+ Na

^CzHsONa +

0.5H2t

+ Na

>NaOH

0,5H2t

a

CH3COOC2H5

+C{2)

23

VC2H5OH+VH20

C2H5OH

(4)

—^CHjCOONa
2. Vie't phtfcfng trinh phan tfng c6 ghi dieu kien de thifc hien chuyen hoa trong
sd do :

, 2H00

VC2H.OH.IOO

•

: 1 = 1,8

=

0,25

i=>

a + b

=

+

•>

0,5a

0,5b

,^

0,5

(1)

= 46a =^ V C J H J O H =46a : 0,8 = 57,5a m l
raH20= 18b => V H 2 0 = 18b : 1 = 18b m l
57,5a. 100
Do rurdu = —
= 40
(2)
57,5a + 18b
Giai he phiTdng trinh (1), (2) ta dtfdc a = 0,086 ; b = 0,414
mczHjOH = 46a = 46.0,086 = 3,956 g
m H 2 0 = 18b = 18.0,414 = 7,452 g
Khoi liTdng riTdu 40" can dung : 3,956 + 7,452 = 11,408 g.
mczHjOH

2. Dun

,n,

-

nong hon hdp g o m 10 gam riTdu etylic va 3 gam axit axetic (c6

mat H2SO4

dSc l a m xuc tac). Hieu suat phan iJng dat 60%. Tinh liTdng este thu diTdc.
HuTdng din giai
DiTa vao phufdng trinh phan uTng, ta xet xem cha't nao he't, cha't nao con dvl

ne'u hieu suat la 100%

sau do tinh k h o i lifdng san pham (este) theo chat phan

tfng he't, suy ra k h o i li/dng thifc te thu diTdc theo hieu suat da
> CH3COOC2H5

C H 3 C O O H + C2H5OH

60g
3g

46g
x(g) < 3

88g

biet.

+ H2O

yg

Ttf cac ti le khoi lifdng cac chat phan lihg ta thay: x < 3g, theo dau bai mn,,^ = lOg.

I Vay riTdu etylic con diT, tinh k h o i Itfdng este theo k h o i luTdng C H 3 C O O H .
6 0 : 3 = 8 8 : y ^ y = ^ = 4,4g
60
Kho'i liTdng este thuTc te'thii diTdc: 4,4.
= 2,64 g
' ',
Cung CO the tinh theo each sau: ncH^cooH = 3 : 60 = 0,05 m o l ' '
" C 2 H 5 0 H = 10 : 46 = 0,27 m o l > n c H j C O O H do dd riTdu con dtf nhieu
'^CH3COOC2H5 - " C H 3 C 0 0 H = 0,05

mol

'

' - 2 5r^J -^'^1


60
K h o i ItfcJng este C H 3 C O O C 2 H 5 thvfc te thu difdc l a : 0,05. 88. —

= 2,64 g.

3. Cho 2 lit dung dich glucozd l e n men ruTcJu l ^ m thodt ra 17,92 l i t k h i cacbonic

I

Hifdng dSn giai
Tur the tich, suy ra nkhi = 5,6

: 22,4 = 0,25 m o l va 2,24 : 22,4 = 0,1 Tru^

Tit cdc phiTdng trinh p h a n i?ng ta l a p diTdc he phtfdng trinh l i f i n q u a n giQ.^ ^g-

(dktc). T i n h nong do m o l cua dung dich glucozd bie't h i e u suat cua qua trinh

m o l c h a t p h a n iJng (axit, ruTdu) va n^hi

16n men chi dat 4 0 % .

T r o n g 1/3 hon hdp c 6 x/3 m o l axit v i y/3 m o l rtfdu.

CH3COOH

a)

Htfdng d i n giai
difdc so m o l , k h o i liTcJng glucozd l e n men, tiif h i ? u suat qua trinh, ta t i m dirdc

C2H5OH

so m o l glucozcf C O t r o n g 2 l i t dung dich...

y/3 m o l

17,92

: 22,4

+0,5H2t

.,
+

Na

C2H50Na

>

x/6 m o l

-

"^e"30-32"c ^

2C2H50H + 2 C 0 2 t

0,4 m o l

C2H5OH

0,8 m o l

100
So m o l glucozd c6 t r o n g 2 l i t dung dich: 0,4. —
N o n g do m o l cua d u n g dich glucozd 1^ 1 : 2 =

=

> 2CH3COONa + CO2 T + H 2 O

1 mol

vhy

trong xa phong c6 6 0 % k h o i liTdng CnHssCOONa.
TH phifdng trinh p h ^ n tfng xa phong hod v^ so m o l (suy tCf kh6'i liTcfng) chat

x/3

J
= 8900 : 890 = 10 mol

SCnHssCOONa + C 3 H 5 ( 0 H ) 3

i

x/3

x/3

=

= 0,2 m o l (axit) \h cung c6 0,2 m o l C2H5OH tham gia phan tfng

= o,3 > 0,2 => nfdu dir. T i n h este theo

axit nhu" tren Ih phh hdp. K h o i lifdng este thu diTdc neu h = 6 0 % la:

10 m o l

60
0,2. 88. —

306)

Bai t§p tir gi§i

= 15300g hay 15,3kg.

Cho phan mot t a c dung \di N a d\i, thu difdc 5,6 l i t k h i .

-

Cho lifdng dir N a ^ C O s vao phan hai thay thodt ra 2,24 l i t k h i .

=

10,56 g C H 3 C O O C 2 H 5 .

trinh len men ngurfJi ta

thu difdc m k g dung dich CH3COOH 2%. H i ? u suat qua trinh len men la 90'??'.
Xac djnh gia t r i b k n g so cua m .

phan deu nhau.
-

100

^' Pha 15 lit riTdu 2 0 " vao niTdc r o i len men giSm, sau qui

gam axit axetic v d i b gam rtfdu etylic r o i chia lam
>

^' Cho 10,6 g a m N a 2 C 0 3 vdo dung dich C H 3 C O O H 0 , 5 M . Phan iJng x a y ra
hoan toan, liTdng k h i thoat ra diTdc dan v a o binh duTng 1 l i t dung
Ca(0H)2 0,075M. T i n h :

a) T i n h a va b. B i e t tht? tich k h i do d dktc.
b) D u n nong phan ba v d i H 2 S O 4 dSc, xuc tac. T i n h k h o i lifdng este tao i h ^ " '
hieu suat phan tfng e s t e hod dat 6 0 % .

= 46. 0,9 = 41,4 g [

(neu h i e u sua't 100%).

lOO

5. NgiTcJi ta tron d e u a

^

"2SO4 ^^'^ > C H 3 C O O C 2 H 5 + H 2 O

C 2 H 5 O H trong 1/3 h o n hdp la ^ = M

K h o i liTdng g l i x e r i n s i n h r a : 10. 9 2 = 9 2 0 ( g ) hay 0,92 kg.

30.306.—^

oO

"'^^

0 6

X

IiTcJng xa phong.

252

CH3COOH + C2H5OH

t.

h6o ta t i m diTcJc so m o l va k h o i ItTdng g l i x e r i n , m u o i natri, va suy ra khoi

b) K h o i lifdng xa p h o n g thu diTdc l a : (CnHssCOONa => M =

= 6. 0,15 = 0,9 m o l C 2 H 5 O H

™CH3COOH = a = 60. 0,6 = 3 6 g ; mcjHjOH = b
b)

Hifdng d§n giai

30 m o l

y : 6 = 0,15 m o l

Kho'iliTdngmoichS'tdemtrQnlania:

b) T i n h k h o i liTcfng xa p h 6 n g bdnh thu diTdc neu phan iJng xay ra ho^n toan va

10 m o l

13

> khong c6 phdn tfng xay ra.

V a y trong t o ^ n bp hon hdp c6 x = 6. 0,1 = 0,6 m o l C H 3 C O O H

a) T i n h k h o i Irfdng g l i x e r i n sinh ra.

(CnHjsCOOjCsHj + 3NaOH — ^

+

= ( x : 6 ) + ( y : 6 ) = 0,25mol

4. D u n 8,9 kg (C,7H35COO)3C3H5 v d i mot li/dng dung dich N a O H viTa d u .

a)

x/6 m o l

" C O 2 = x : 6 = 0,l(mol)
'"
, , . .,

0,5M.

Chat beo: ( C n H j s C O O j C s H j => M = 890g => n^o

'

+ 0,5H2T

x/3 m o l
CeUuOf,

*

y/6 m o l

2 C H 3 C O O H + NazCOj

In •.

= 0,8 m o l

>CH3COONa

Na

x/3 m o l

Tiir phi/cJng trinh p h J n tfng l e n m e n va the tich ( r o i suy ra so mol) CO2, dm

nco2=

+

i

a) The tich dung d i c h C H 3 C O O H da dilng (vifa dii).

^) K h o i liTdng ke't tua sinh ra trong blnh dtfng C a ( O H ) 2 .

^|'
'

d}ch