Chuyen de Hidrocacbon Vip

7/31/2019 Chuyen de Hidrocacbon Vip

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Khoa Ho Hc Chuyn HidrocacbonTrng H ng Thp GVHD: TS Dng Huy Cn

CHUYN HIDROCACBON

C nhiu cch phn loi bi tp hidrocacbon nh: da vo bt bo ho c

hiddrocacbon no, khng no, thm hoc Thit lp CTPT,thnh phn % hn hp trongchuyn ny chng ti phn loi dc vo tnh cht ho hc in hnh caHidrocacbon.

I. Gii bi ton v phn ng t chy hirocacbonI.1 Kin thc cn nm:Tm CTPT khng cn qua CT n gin nht ( khi bit M, m, hoc bit mt trong

cc ch s nguyn t).

12

12% % 100

C H

x y M

m m m

x y MC H

= =

= =

x, y.

CxHy + (x+4

y)O2 x CO2 +

2

yH2O

CnH2n+2-2k + OHknnCOOkn to

222 )1(2

13++

+

Nu 22 COOH nn > A l ankan v ACOOH nnn = 22Nu

22

COOH nn = A l anken (hay xicloankan)

Nu 22 COOH nn < A l ankin, ankadien hoc aren. Nu A c C l mch h A l ankin hay ankadien, lc AOHCO nnn = 22

t chy mt hidrocacbon hay hn hp hidrocacbon bt k, ta lun c:

2 2 2

1

2puO CO H On n n= +

Lp cng thc n gin nht t lp CTPT,Tnh mC, mH, hoc %C, %H

Tnh t l x:y =% %

:12 1 12 1

C Hm m C H= =

Cng thc nguyn :(CxHy)n dng M hoc m bin lun tm CTPT

MT S LU : Nu hp cht hu c t bi CuO, sau phn ng, khi lng bnh dng

CuO gim ag th chnh l khi lng Oxi phn ng. Nu hp cht hu c t thu c CO2, H2O, Na2CO3 th thnh phn nht

thit phi c C, H, Na ngoi ra c th c Oxi.mc(trong hp cht hu c) = 2 2 3( )CO C Na COm m+

Nu hp cht hu c t thu c CO2, H2O, HCl th thnh phn nht

thit phi C, H, Cl, ngoi ra c th c Oxi.mc(trong hp cht hu c) = 2( ) ( )H H O H HClm m+

Khi cho hn hp CO2, H2O vo bnh Ca(OH)2 hoc Ba(OH)2 th :

Nhm SVTH: Nhm 3 1 Lp ho 07B, nm hc 2009-2010


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m bnh tng = 2 2CO H Om m+

m dung dch tng = 2 2CO H Om m m+

m dung dch gim = 2 2( )CO H Om m m +

I.2 Bi tp p dng:

1. t chy hon ton hn hp M gm mt ankan X v mt ankin Y, thuc s mol CO2 bng s mol nc. Tnh thnh phn phn trm v s mol ca Xv Y trong hn hp M

gii OHnnCOHC Onn 2222 )1(2 +++

a na (n+1)a

OHmmCOHC O

mm 2222

)1(2 +b mb (m-1)b

Theo :%50%

)()()(

)1()1(22

===++=+

++=+=

ankinankan

OHCO

nnba

bambnambna

bmanmbnann

2. t chy han ton mt th tch kh thin nhin gm metan, etan,propan bng oxi khng kh ( trong khng kh, Oxi chim 20% th tch), thu c7,84 lt kh CO2 (ktc), v 9,9g nc.Th tch khng kh nh nht cn dng t chy hon tan lng kh thin nhin trn l bao nhiu?

giiKhi t chy 1hidrocacbon bt k ta lun c:

lV

molnnn

kk

OHCOpuOt

7020

100).4,22.625,0(

625,055,0.2

135,05,0

22

==

=+=+=

3. t chy mt ankin A thu c 5,4g nc v cho sn phm chy quadung dch Ca(OH)2 d, dung dch c khi lng gim so vi dung dch nc vi

trong ban u l 19,8g. Xc nh CTPT ca A.gii:

Ta c: moln OH 3,018

5,52

==

gi amolnnamoln COCaCOCO === 232khi lng dung dch gim:

45,0

8,19)4,544(100

8,19)(223

==+

=+

a

aa

mmm OHCOCaCO

OHnnCOHC Onn 2222 )1(2 +



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0,45 0,3

0,3n = 0,45(n-1) 433 HCn =

Bi tp t rn luyn :

1. Hn hp gm hirocacbon X v oxi c t l s mol tng ng l 1:10. t chyhon ton hn hp trn thu c hn hp kh Y. Cho Y qua dung dch H 2SO4 c, thuc hn hp kh Z c t khi i vi hiro bng 19. Cng thc phn t ca X l (cho H= 1, C = 12, O = 16)

2. Cho 4,48 lt hn hp X ( ktc) gm 2 hirocacbon mch h li t t qua bnhcha 1,4 lt dung dch Br2 0,5M. Sau khi phn ng han ton, s mol Br2 gim i mtna v khi lng bnh tng thm 6,7 gam. Xc nh Cng thc phn t ca 2hirocacbon l (cho H = 1, C = 12)

3. Ba hirocacbon X, Y, Z k tip nhau trong dy ng ng, trong khi lngphn t Z gp i khi lng phn t X. t chy 0,1 mol cht Y, sn phm kh hp thhon ton vo dung dch Ca(OH)2 (d), thu c s gam kt ta l (cho H = 1, C = 12,O = 16, Ca = 40)

4. ot chay hoan toan 1,12 lt mot anken X (ktc) thuc 5,60 lt kh CO2 (ktc). CTPT X la:

A. C3H6 B.C4H8 C. C4H10 D. C5H10

5. ot chay hoan toan mot the tch hon hp gom

anken X va hidrocacbon Y thu c 5,56 lt kh CO2 ( ktc) va 5,40

g nc. Y thuoc loai hirocacbon co cong thc phan t

dang.

A.CnH2n B.CnH2n-2 C. CnH2n+2 D. CnH2n

6. ot chay hoan toan 1,3g ankin X thu c 2,24 lt kh

CO2 (ktc) . Cong thc phan t cua X la :

A. C2H2 B. C3H4 C.C4H6 D. C5H8

7. ot chay hoan toan 5,4g ankaien X thu c 8,96 lt kh

CO2 ( ktc) . Cong thc phan t cua X la

A. C4H4 B. C4H8 C.C4H6 D. C4H10

8. ot chay hoan toan hon hp kh gom ankin X va

hidrocacbon Y mach h co cung so nguyen t C, thu c

san pham chay co the tch hi nc bang the tch kh CO2

(cac the tch o cung ieu kien). Y thuoc loai:



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A.ankin B. anken C. xicloankan D. ankan

9. ot chay 1 so mol nh nhau cua 3 hidrocacbon A, B, C

thu c lng CO2 nh nhau, con t le so mol CO2 va H2O

oi vi A, B, C lan lt la 0,5: 1:1,5. CTPT cua A, B, C

A. CH4, C2H6, C3H8 B. C2H4, C3H6, C4H8

C. C2H2, C2H4, C2H6 D. C2H6, C2H4, C2H2

10. ot chay ht 2,24 lt ankan X (ktc), dan ton b san

pham chay vao dd nc voi trong d thay co 40g. CTPT X

A. C2H6 B. C4H10 C. C3H6 D. C3H8

11. ot chay hoan toan 0,15 mol hon hp 2 ankan thu

c 9,45g H2O. Cho san pham chay qua bnh ng Ca(OH)2 dth khoi lng ket tua thu c la bao nhieu ?

A.37,5g B. 52,5g C. 15g D.42,5g

12. ot chay hoan toan 0,15 mol hon hp 2 ankan thu

c 9,45g H2O. Cho san pham chay qua bnh ng Ca(OH)2 d

th khoi lng ket tua thu c la bao nhieu ?

A.37,5g B. 52,5g C. 15g

D.42

II. Bi tp v phn ng th halogen ca Hidrocacbon

II.1. C s l thuyt:

{ 2 22

asn nC H

X+

2 1

2

........

n n

n n

C H X

C H X

+

2

2 2n n

HX

X

C H +

Xt phng trnh phn ng:

Nhn xt:- Nu phn ng xy ra theo t l 1:1 (x=1) trong phn t ankan c bao nhiu

v tr cacbon khc nhau cn hiro s cho ta by nhiu dn xut monohalogen. Khi sn

phm chnh u tin th vo cacbon bc cao.- Nu cho kh Y qua dd NaOH d m vn c kh thot ra trong hh Y c

ankan cn d.


Hn h p A Dn xut Hh kh Yd

d

xHClC H

n 2 n + 2 +xCl 2 C Hn 2 n + 2 +Cl- x

a s


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- Nu dd trong NaOH sau phn ng c tnh oxi ha trong hh Y c kh clod.

V:2NaOH + Cl2 NaClO + NaCl + H2O

v NaClO sinh ra c tnh oxi ha mnh.- Sn phm th (dn xut) thng dng lng v ktc. Trong trng hp to ra

hai sn phm th monohalogen v ihalogen th nn vit hai phng trnh c lp xutpht t ankan v Cl2 ly s mol mi sn phm lm n s v lp hai phng trnh ton

hc.- Ngoi ankan, cn c phng trnh phn ng:

CH2=CH-CH3 + Cl2 CH2=CH-CH2Cl + HClV d:

Trn 6g C2H6 v 14,2lit (ktc)Cl2 c chiu sng thu c sn phm th mono v

iclo. Cho hn hp kh i qua dd NaOH d th cn li 22,4 lit kh duy nht thot ra (ktc). Dung dch trong NaOH c kh nng oxi ha 200ml dd FeSO4 0,5M. Phn trm smol mi sn th l bao nhiu? Bit cc p xy ra hon ton.Gii:

( )2 6

60, 2

30C H bd

n m= =

Gi a, b ln lc l s mol ca 2 sp th mono v iclo:

C2H6 + Cl2 C2H5Cl HCl+

a s

a a a a

C2H6 + 2Cl2 C2H4Cl2 2HCl+

b 2b b 2b

V sau khi qua dd NaOH d thu c mt kh duy nht l C2H6 (d)=0,1mol

ddNaOH

2( ) 2 6

2 6( )

( 2 )

(0, 2 2 ) (0,1 )

(0, 2 )

du

du

HCl a b

Cl a b C H mol

C H a b

+


5000C

( )2

14,20, 2

22,4Cl bd

n m= =

Hn hp kh sau p:

HCl NaOH NaCl H2O

Cl2 2NaOH NaCl NaClO H2O

+ +

+ + +

2 60, 2 0,1 0,1C Hn a b a b= = + =

(1)d


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Trong 2 mui th ch c NaClO c tnh oxi ha mnh, v oxi ha Fe2+ thnh Fe3+

t (1) v (2) suy ra

Phn trm s mol sn phm th l: C2H5Cl = C2H4Cl2 = 50%

II.2 Mt s bi tp:

Bi 1: HC mch h X ch cha lk v c hai nguyn t C bc ba trong mt phn t.t chy hon ton mt th tch X sinh ra 6 th tch CO2 (cng iu kin nhit , psut). Khi cho X tc dng vi Clo (t l 1:1), s dn xut monoclo ti a sinh ra l:

A. 3B. 4C. 2D. 5

(tuyn sinh H khi B-08)

Hng dn gii:HC mch h X ch cha lk X l ankan

222

2 COHCO

nn +

V 6V146:6 HCXn =

c hai nguyn t C bc ba trong mt phn t CTCT ca X l:CH3CH(CH3)CH(CH3)CH3da vo CTCT X X tc dng vi Clo (t l 1:1) s dn xut monoclo l 2 n n C

Bi 2: Khi brom ha mt ankan ch thu c mt dn xut monobrom duy nht c tkhi hi i vi hidro l 75,5. Tn ankan l:

A. 3,3-imetylhexanB. 2,2-imetylpropanC. IsopentanD. 2,2,,3-trimetylpentan

(tuyn sinh H khi B-07)


FeSO4 NaOH Fe(OH)2 Na2SO4

2Fe(OH)2 NaClO 2Fe(OH)3 NaCl

+ +

+ ++ H2O

2 2 4( )

10, 05 0, 2 2

2Cl NaClO Fe OH FeSOn n n n a b= = = = =

(2)

{0,050,05a mol b mol

=

=


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Khoa Ho Hc Chuyn HidrocacbonTrng H ng Thp GVHD: TS Dng Huy CnHng dn gii:

HBrBrHCBrHC nnnn ++ ++ 12222

Ta c: 1512.5,7512

==+

BrHC nnM

1255

1518114

HCn

n

=

=+

2

1 25

B rHC mt dn xut monobrom duy nht

CTCT ca ankan l: C(CH3)4 :2,2-imetylpropan n n B.

Bi 3: Mt ankan phn ng vi clo theo t l 1:2 thu c sn phm cha 83,53% clo vkhi lng. CTPT ca ankan l:

A. mtanB. tan

C. propanD. butan

Hng dn gii:

1:2

2 2 2 2 22 2n n n nC H C l C H C l + +

Ta c: %53,83%100.71145,35.2

% =+

=n

Cl

1= n ankan X l mtan

n n ABi 4: Cho 19,5g benzen tc dng vi 48g brom lng, c bt Fe lm xc tc, thu dc27,475g brombenzen. Hiu sut ca phn ng trn l?

A. 40%B. 50%C. 60%D. 70%

Hng dn gii:

moln

moln

Br

HC

3,0

25,0

2

66

=

=

moln BrHC 175,0157475,27

66==

Da vo t l s mol ta tnh hiu sut theo benzen

rHBrHCBrHCtFe

B+ + 66,

266

0

0,25 0,3 0,175 (mol)

Theo PTP ta c:

%70%100.25,0

175,0% ==H

p n D



8/24

Khoa Ho Hc Chuyn HidrocacbonTrng H ng Thp GVHD: TS Dng Huy CnBi 5: Cht A c CTPT C7H8. Cho A tc dng vi Ag2O/NH3 thu c kt ta B. MB >MA l 214vC. S ng phn tha mn iu kin trn l:

A. 2B. 3C. 4

D. 5Hng dn gii:Gi x l s H b thay th bi Ag.

OxHA gHCOxA gHC xxN H

28728722 3 + +

+

Ta c: MB - MA = 214 (92-x+108x)-92=214 x=2

CTCT c th c l:

CHC-CH2-CH2-CH2-CCHCHC-C(CH3)2-CCHCHC-CH2-CH(CH3)-CCH

p n B

Bi tp tham kho

Bi 1: Nitro ha benzen thu dc 2 cht X, Y km nhau mt nhm NO2. t chy honton 19,4g hh X, Y thu c CO2, H2O v 2.24l kh N2 (kc). Tm CTCT ca X, Y?

Bi 2: A l mt HC mch h. t chy A thu c t s mol CO2 gp i s mol nc.Mt khc ly 0,05mol A phn ng va ht vi agNO3/NH3 thu c 7,5g kt ta. tmCTCT ca A

Bi 3: Cho m(g) HC A thuc dy ng ng ca mtan tc dng vi Clo c chiu sng,ch thu c mt dn xut clo duy nht B vi khi lng 8,52g. trung ha ht khHCl sinh ra, cn 80ml KOH 1M. Tm CTPT ca A, B. Vit CTCT ca cc ng phnA v gi tn chng.

III phn ng cngIII.1 C s l thuyt:

222222 ++ + nnknn HCkHHC

kknnknn BrHCkBrHC 22222222 ++ +

Thng thng cc bi ton v phn ng cng thng dn hn hp ca anken ng ng phnng vi Br2, H2 sau phn ng khi lng bnh tng ln agam, hoc % ca brom trong hp chtl x% xc nh cng thc ca hidrocacbon .

Vi nhng dng ny ta cn xc nh_

n ca hidrocacbon ri t suy ra CTPT ca HC

III.2 Bi tp p dngBi. 1,792 lt hn hp (X) gm 2 anken lin tip (o 0 0 C, 2.5atm) cho qua nc brom

d th khi lng bnh nc brom tng 7gam.



9/24

Khoa Ho Hc Chuyn HidrocacbonTrng H ng Thp GVHD: TS Dng Huy Cna. xc nh CTPT ca 2 anken

b. t chy hon ton hn hp X ri cho sn phm chy vo 500ml dung dch NaOH 1,8M ththu c bao nhiu gam mui khan.

Gii:

mol

RT

PVnhhX 2,0

273.082,0

792,1.5,2===

a. t CTPT trung bnh ca 2 anken l nnHC 2Ta c phn ng: 2222 BrHCBrHC nnnn +mhhX = tng khi lng bnh nc brom =7g

5,2352,0

714)(

2===== n

n

mnM

hhX

hhXHC

nn

2 anken l lin tip nhau suy ra CTPT ca 2 anken l: 6342 HvCHC

b. phng trnh phn ng chy: OHnCOnOn

HC tnn 22

022

)2

3( ++

0,2mol 0,2 n moln(CO2)=0,2 n = 0,2.2,5=0,5 moln(NaOH) = 0,5.1,8 = 0,9 mol

nhn thy t l: 1< 25,0

9,0

2


10/24


Vy CTPT ca hai anken l: C4H8 v C5H10

Bi: Dn hn hp X gm etilen v axetilen qua dung dch brom d khi lng bnh

brom tng 1,34g. Cn khi cho tc dng ht vi dung dch AgNO3/NH3 thu c 7,2g

kt ta. Thnh phn phn trm th tch ca etilen v axetilen ln lt l :

A. 50% ; 50% B. 35,5% ; 64,5% C. 40% ; 60% D. 25% ; 75%

Gii:

Phng trnh phn ng:

242242 BrHCBrHC +

x x

422222 2 BrHCBrHC ++

y y 3433 222 NONHCAgAgCNHAgNOCHHC +++

y y

Gi x, y (mol) ln lt l s mol ca etylen v axetylen

khi lng bnh brom tng 1,34g l khi lng ca hn hp X

28x + 26y = 1,34

moly 03,0

242.108

2,7=

+

=

=x 0,02mol

cng iu kin v hit v p sut th t l v s mol bng t l v th tch

%V(C2H4) = %40100.05,0

02,0=

%V(C2H2) = %60100.05,0

03,0=

Bi 4. A l 1 HC mch h , cht kh iu kin thng .4,48 lt kh A ktc tc dng

va vi 0,4 mol Brom to ra sn phm B cha 85,562% brom v khi lng. Cng

thc phn t ca A l:

a. C2H6 b. C3H4 c. C4H6 d. C6H6

Gii:

n molA 2,0=

t CTPT trung bnh ca HC l knnHC 222 +

pt phn ng: kknnknn BrHCkBrHC 22222222 ++ +



11/24


0,2 0,2k

2= k

Ta li c 4562,85100.2.1602.2214

2.160% ==

++= n

nBr

chn p n C

Bi 5. Cho 10,2g hn hp kh A gm CH4 v anken ng ng lin tip i qua dd nc

brom d, thy khi lng bnh tng 7g, ng thi th tch hn hp gim i mt

na.Cng thc phn t cc anken l:

A. C2H4, C3H6 B. C3H6, C4H10 C.C4H8,C5H10 D. C5H10, C6H12

Gii:

t CTPT trung bnh ca anken l: nnHC

2

gi x, y(mol) ln lt s mol ca CH4 v anken

Phng trnh phn ng:

2222 BrHCBrHC nnnn +

y y

ta c 16x + 14n y = 10,2 (1)

khi hn hp kh qua Br2 th th tch gim mt na yx = (2)

khi lng bnh brom tng 7g l khi lng ca hn hp anken

ny

M 147

== (3)

t (3) v (1) 5.22,0,5,0 ==== nyxyn

Vy chon p n A

Bi tp tham kho1. Cho 2,24 lt anken X (ktc) tac dung vi dd brom thu c

san pham co khoi lng ln hn khoi lng anken la

A. 0,8 g B. 10,0g C. 12,0 g D. 16,0g

2. Anken X tac dung vi nc (xuc tac axit) to ra hon hp 2

ancol ong phan cua nhau. d/N2 = 2,00. Ten ca X la

A. iso-penten B. but-1-en C. but-2-en

D. pent-1en



12/24

Khoa Ho Hc Chuyn HidrocacbonTrng H ng Thp GVHD: TS Dng Huy Cn3. Anken Y tac dung vi dd brom tao thanh dan suat ibrom

trong o % khoi lng C bang 17,82 %. CTPT Y la:

A. C3H6 B.C4H8 C. C4H10 D. C5H10

4, Cho 14g hn hp 2 anken l ng ng lin tip i qua dung dch nc Br2 thy lm

mt mu va dd cha 64g Br2.Cng thc phn t ca cc anken l:

A. C2H4, C3H6 B. C3H8, C4H10 C. C4H10, C5H12 D.

C5H10, C6H12

5. Cho 4,48 lt hn hp X (ktc) gm hai hidrocacbon mch h li t t qua bnh cha

1,4 lt dung dch Br2 0,5 M. sau phn ng xy ra hon ton, s mol Br2 gim i mt na

v khi lng bnh tng thm 6,7gam. Xc nh cng thc phn t ca hai hidrocacbon

.6. Mt hirocacbon cng hp vi HCl theo t l 1:1 to ra sn phm c thnh phn khi

lng Clo l 45,223%. Xc nh CTPT X

IV.s dng cc gi tr trung bnh gii nhanh cc bi tp v

Hidrocacbon

IV.1/ c s l thuyt:

Nu c mt hn hp nhiu cht cng tc dng vi mt cht khc (m cc PTP

cng loi, cng hiu sut, sn phm phn ng tng t nhau) ta c th thay hn hp ny

bng mt cht tng ng.

Gi s hn hp gm cc chtA, B, C,( cha C, H, O), c th thay bng cht tng

ng x y zC H O : Mvi:

-Khi lng phn t (nguyn t) trung bnh:

...

...

hh A A B B K K

hh A B K

m n M n M n M M

n n n n

+ + += =

+ + +

(nA, nB, nKc th l s mol, th tch hay % s mol,)

Lun c MA < M


13/24


1 2...

...A B K K

A B K

n y n y n yy

n n n

+ + +=

+ + + (vi y1


16/24


4,34,2

14,3

4,305,0

17,0


17/24


)3(12,02

)2(

)2(17,0)1(

2

2

molybay

n

molxbaxn

OH

CO

=++

=

=++=

T (2) ta c: (a + b)x + b = 0,17

b = 0,17-0,05xta c 0 < b < 0,05

0 < 0,17-0,05x < 0,05

2,4 < x < 3,4

3=x

03,0

02,0)3.05,0(17,0

===

a

b

Thay gi tr a, b vo (3) ta c y=4

Vy CTPT ca 2 cht l C3H4 v C4H6.

Bi 5 (bi 6.11 trang 43SBT HH 11)

HH kh A cha H2 v 2 anken k tip nhau trong dy ng ng. T khi ca A i vi

H2 l 8,26. un nh hh A c mt xc tc Ni th A bin thnh hh B khng lm mt mu

nc Brom v c t khi i vi H2 l 11,8.

Xc nh CTPT v % th tch ca tng cht trong hh A v hh B?

Hng dn gii:

t CTC ca 2 anken l: )(2 xmolHC nn v H2 (1-x)mol. ( trong 1 mol A)

52,162.26,8)1(214 ==+= xxnMA (1)

2222 ++ nnnn HCHHC

x x x (mol)

molx

xMB

3,0

6,232.8,111

52,16

=

==

=

Thay x vo (1) c 6,3=n

Vy CTPT 2 anken l: C3H6 (amol) v C4H8 (bmol)

Ta c h:

=+

+=+

6,343

03,0

ba

ba

ba



18/24


==

)(18,0

)(12,0

molb

mola

hh A: C3H6 : 12% ; C4H8 : 18% ; H2 : 70%

hh B: C3H8 : %17%100.7,0

12,0

=

C4H10 : %26%100.7,018,0

=

H2 : 57%

Bi 6(bi 7.27 trang 57 SBT HH 11)

HH M cha 2 HC k tip nhau trong mt dy ng ng. Khi t chy hon ton 13,2g

hh M thu c 20,72l CO2 (kc). Xc nh CTPT v % khi lng tng cht trong hh

M.

Hng dn gii:

molnCO 925,04,2272,20

2==

gmC 1,1112.925,0 ==

moln

gm

OH

H

05,118

1,21,21,112,13

2==

==

V 22 COOH nn > nn 2 cht trong hh M u l ankan

4,705,1

925,0

1

)1(2

2322222

==

+

+++

++

nn

n

OHnCOnOn

HC nn

Vy CTPT 2 ankan l: C7H16(xmol) v C8H18 (ymol)

Gii h:

=+==+=925,087

2,13114100

2yxn

yxm

CO

M

==

moly

molx

05,0

075,0

% v khi lng ca C7H16 : 56,8% ; C8H18 : 43,2%



19/24

Khoa Ho Hc Chuyn HidrocacbonTrng H ng Thp GVHD: TS Dng Huy Cn V.Bi tp v phn ng th nguyn t hidro linh ng bng ion kim loi

(Ag)

V.1 c s l thuyt

Vi cc hidrocacbon c ni ba u mch th nguyn t H gn vi C ni ba trnn linh ng d tham gia phn ng th:

R-C CH + AgNO3 + NH3 R-C CAg + NH4NO3

Dng tng qut:

CxHy + AgNO3 + NH3 CxH y Ag + NH4NO3

Vi ankin:

Nu =1 ank-1-in (R-C CH)

Nu =2 HC CH (C2H2)

Kt ta d ho tan trong axit mnh

R-C CAg + HCl R-C CH + AgCl

V.2. Bi tp p dng:

Bi tp 1: Cho 17,92 lt hn hp kh X gm 3 hidrocacbon l ankan, anken v ankin ly

theo t l th tch tng ng l 1:1:2 li qua bnh cha dung dch AgNO 3/NH3 ly d thu

c 96g kt ta v hn hp kh Y. t chy ho ton hn hp kh Y thu c 13,44 ltCO2. Th tch cc kh o ktc. Tm CTPT ca 3 hidrocacbon?

Gii:

nX = 4,2292,17

=0,8 mol

V t l th tch bng t l s mol nn s mol tng ng l: CnH2n+2 : 0,2 mol ; CmH2m:

0,2 mol; CpH2p-2 : 0,4 mol.

CpH2p-2 + AgNO3 + NH3 CpH 22p Ag + NH4NO3

0,4 mol 0,4 mol

Ta c n = nankin = 0,4 mol =107214

96

+p 5,6p + 42,8 = 96,8

+ Nu =1 p = 9,6 (loi)

+ Nu = 2 p = 2 (nhn) CTPT ca ankin l C2H2

Kh Y gm : CnH2n+2: 0,2 mol v CmH2m: 0,2 mol

Ta c phn ng chy:



20/24


CnH2n+2 +

+

2

13nO2 nCO2 + (n+1)CO2

0,2 mol 0,2n mol

CmH2m +

2

3m

O2 mCO2 + mH2O

0,2 mol 0,2m mol

0,2 (n+m) =4,22

44,13= 0,6 mol

n+m = 3 n=1 v m=2

CTPT ankan l CH4

CTPT anken l C2H4

Bi tp 2: t chy ankin A to ra 11g CO2. Mt khc, khi cho 3,4g A tc dng vi

lng d dung dch AgNO3 trong NH3 thy to thnh a gam kt ta. CTPT ca A v a

l:

A. C2H2 ; 8,5g B. C3H4 ; 8,5g C. C4H6 ; 8,75g D. C5H8 ; 8,7

Gii:

Do A tc dng c vi AgNO3 trong NH3 nn A l ank-1-in:

CnH2n-2 +

2

13nO2 nCO2 + (n-1)H2O

214

4,3

nmol 0,25 mol

n 2CO = 4411

= 0,25 mol

Ta c: 214

4,3

n = n

25,0

n = 5 CTPT A l C5H8

C5H8 + AgNO3 + NH3 C5H7Ag + NH4NO3

0,05 mol 0,05 mol

n 85HC = 8,64,3

=0,05 mol

a = 0,05.175= 8,75g

p n D



21/24


Bi tp 3: Dn 3,36 lt hn hp A gm propin v etilen i vo mt lng d dung dch

AgNO3 trong NH3 thy cn 0,84 lt kh thot ra v c m gam kt ta. Cc th tch kh o

ktc.

a) Tnh thnh phn phn trm theo th tch ca cc kh trong A.

b) Tnh m.

Gii:

a) Khi dn A vo dd AgNO3/NH3 ch c propin phn ng

VC2H4= 0,84 lt

Vpropin= 3,36-0,84= 2,52 lt

%Vetilen= 36,384,0

.100%= 25%

%Vpropin= 100%-25%=75%

b) npropin= 4,2252,2

= 0,1125 mol

CH C-CH3 + AgNO3 + NH3 AgC C-CH3 + NH4NO3

0,1125 mol 0,1125 mol

m= 0,1125. 147=16,5375 g

Bi tp 4: Dn 6,72 lt hn hp kh X gm propan, etilen v axetilen qua dung dch

brom d, thy cn 1,68 lt kh khng b hp th. Nu dn 6,72 lt kh X trn qua dung

dch bc nitrat trong amoniac thy c 24,24 gam kt ta. Cc th tch kh o ktc.

a) Vit cc phng trnh ho hc gii thch phng trnh TN trn.

b) Tnh thnh phn phn trm theo th tch v theo khi lng ca mi kh trong

hn hp.

Gii:

a) Dn X qua dd Br2 d ch c etilen v axetilen phn ng

CH2=CH2 + Br2 BrCH2-CH2Br

CH CH + Br2 BrCH=CHBr

BrCH=CHBr + Br2

Br2CH-CHBr2Dn X qua dung dch AgNO3/NH3 ch c axetilen phn ng



22/24


CH CH + 2AgNO3 + 2NH3 AgC CAg + 2NH4NO3

b)

Khi dn qua dung dch brom d c 1,68 lt kh khng b hp th l propan

Vpropan= 1,68 lt

Vetilen+axetilen= 6,72 - 1,68 = 5,04 lt


0,101 mol 0,101 mol

n =240

24,24= 0,101 mol

Vaxetilen= 0,101. 22,4= 2,2624 lt Vetilen = 5,04 2,2624= 2,7776 lt

%Vpropan= 72,668,1

.100%= 25%

%Vaxetilen= 72,62624,2

.100%= 33,67%

%Vetilen= 41,33%

Bi tp 5: t chy hon ton 2,24 lt hirocacbon X thu c 6,72 lt CO2 ( ktc). X

tc dng vi dd AgNO3 trong NH3 sinh ra kt ta Y. Cng thc cu to ca X l:

a) CH3-CH=CH2

b) CH CH

c) CH3-C CH

d) CH2=CH-CH CH

Gii:X tc dng c vi dd AgNO3/NH3 X l ank-1-in

CnH2n-2 +

2

13nO2 nCO2 + (n-1)H2O

0,1 0,1.n mol

nX=0,1 mol

n2

CO = 0,3 mol

Ta c: 0,1n = 0,3 n=3 CTPT X l C3H4

CTCT X l: CH3-C CH



23/24

Khoa Ho Hc Chuyn HidrocacbonTrng H ng Thp GVHD: TS Dng Huy Cn p n d

Bi tp 6: Mt bnh kn ng hn hp hidro vi axetilen v mt t bt Niken. Nung

nng bnh mt thi gian sau a v nhit ban u. Nu cho mt na kh trong

bnh sau khi nung nng i qua dd AgNO3/NH3 th c 1,2g kt ta mu vng nht. Nu

cho na cn li qua bnh ng nc brom d thy khi lng bnh tng 0,41g.

a) Vit PTHH v gii thch cc hin tng TN.

b) Tnh khi lng axetilen cha phn ng, khi lng etilen to ra sau phn ng.

Gii:

a) Khi nung nng hn hp hidro v axetilen c xc tc NikenC2H2 + H2

0,tNi C2H4

C2H4 + H2 0,tNi C2H6

Khi dn kh thu c sau phn ng qua dd AgNO3/NH3 to kt ta mu vng nht

chng t hn hp kh thu c c C2H2 cn d.


Khi dn qua bnh ng nc brom d thy khi lng bnh tng 0,41g l khi

lng ca C2H2 v C2H4

CH2=CH2 + Br2 BrCH2-CH2Br

CH CH + 2Br2 Br2CH-CHBr2

b) Khi lng axetilen chua phn ng:


0,005 mol 0,005 mol

n = 2402,1

= 0,005 mol

maxetilen cha p= 0,005.26= 0,13 g

metilen to ra= 0,41- 0,13= 0,28 g

Bi tp 7: Khi t chy mt ankin A thu c mt khi lng nc ng bng khi

lng ankin em t. Tm cng thc phn t, vit cng thc cu to ca A. Gi tn caA, bit A to c kt ta vi AgNO3/NH3. Vit PTHH ca phn ng.



24/24

Khoa Ho Hc Chuyn HidrocacbonTrng H ng Thp GVHD: TS Dng Huy CnGii:

Gi m l khi lng ankin v cng l khi lng ca nc to thnh.

CnH2n-2 +

2

13nO2 nCO2 + (n-1)H2O

214 nm

mol (n-1)214 n

mmol

nankin=214 n

mmol

n OH2 = 18m

mol

Ta c:

(n-1)214 n

m =18m 18n -18 = 14n -2 4n = 16 n = 4

CTPT A l C4H6

V A c p vi AgNO3/NH3 nn CTCT ca A l:

CH C-CH2-CH3 : But-1-in

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Chuyen de Hidrocacbon Vip