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

Trong phong thi nghiem c6 7 binh thuy tinh khong mau bi mat nhan, m"J binh diTng mot chat long hoac mot chat khi sau day: etan, etilen, benzen, ^t^' cacbonic, k h i sunfurd, riTdu etyli

mjd sau phan ling = r r i j j C a ( O H ) 2 + " ^ 0 0 2 ~ " ^ C a T O ^

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

1 6 2 M 5 Nong do m u o i CaCHCOj) trong dung dich la: ^^^^^^ 100 = 0,4%

y = 0,025; x = 0,025

Cdch 2: CO2 + C a ( 0 H ) 2 > CaCOjto + H2O v , „ _ , , ,

2CO2 + Ca(OH)2 > Ca(HC03)2

2y mol y mol y mol _ , ^

nco2 = x + 2y = 0,075 1

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

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

Cdch 3: CO2 + Ca(OH)2 > CaCOjio + H2O

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

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

II Bal tap ti/ g i i i

1 Dot chay 12,6 gam hon hdp ba khi CH4, C2H4 va C2H2, ta thu difdc 39,6 gam

CO2 Neu cho 12,6 gam hon hdp tren qua binh diTng dung dich brom thi c6 80

gam brom tham gia phan tfng Xac dinh thanh phan phan tram khoi Wdng

m o i hidrocacbon trong hon hdp

2 Dot chay 11,2 lit (do d dktc) hon hdp metan va etilen roi 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 tram theo the tich m o i khi trong

hon hdp

Dang 4: X ^ c djnh th^nh ph^n h6n hpp U p c 6 n g thQc phan ttf

I Bdl tap CO Idfl giSl

1 Biet 0,01 mol hidrocacbon A lam mat mau vCfa dii 100 m l dung dich brom

0,1M V a y A la hidrocacbon nao trong so cac chat sau :

n

n B , 2 = 0 , l = 0 , 0 1 mol; : ng^^ = 0,01 : 0,01 = 1 : 1

Vay A l a C2H4 , o u - •

2, Dot chay 3 gam chat hffu cd, thu duTdc 8,8 gam k h i CO2 va 5,4 gam H2O

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

b) Biet phan tuf k h o i cua A nho hdn 40 T i m cong thiJc phan tuT cija A ' c) Chat A CO lam mat mau dung dich brom khong ? " -'^

d) Viet phu'dng trinh hoa hoc cua A vdi clo khi c6 anh sang

b) Dat cong thiJc A la C,Hy ta c6: x : y = mc mH_ 2,4 0,6

12 • 1 ~ 12 • T"

Cong thu-c A CO d a n g (CH3)„

Ta CO ( 1 2 + 3 ) n < 40 n < 2,67

Nghiem t h i c h h d p n = 2 ' ' ' Cong thtfc p h a n tur A : C2H6 < j i -

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

d) CH3-CH3 + CI2 """"''"^ ) CH3-CH2-CI +HC1

3 K h i do't c h a y h o a n t o ^ n m o t the tich h i d r o c a c b o n A c a n 6 the t i c h o x i , sinh ra

4 the 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 cac the t i c h do d

Tii phu'dng t r i n h p h a n vtng do't c h a y , vdi 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

Cong thtfc cau tao ciia A c6 the la:

CH3-CH2-CH=CH2 hoac la C H 3 -CH=CH-CH3

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

H i f d n g d i n g i a i

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; U H J O = 13,5 : 18 = 0,75 mol

CMy + (x + y/4)02 > XCO2 + y/2H20

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; U H = 2nH20 = 2 = 1,5 mol

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,

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 ' m c o 2 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)

||ai c&c phuWng trinh tren ta diTCfc x = 3, y = 8, z = 1 - '

:6ng 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:

v6i 3 gam A tao ra 6,6 gam C O 2 va 3,6 gam H2O

Neu do't chay (1 mol) 60 g A tao ra x g CO2 va y g H2O 60.6,6 0 ^ ,

207

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

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

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

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

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

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

''>^-' - -m|afeli§«0 12xg yg 14g

53.33g 15.55g 31,12g '^^^ 12x 14 31,12.12x= 14.53,33 = > x « 2

> 3 1 1 2 y = 14.15,55 = > y « 7

53.33 31,12

y _ 14

15,55 ~ 31,12

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

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

7 D o t chay 1,12 lit hon hdp hai hidrocacbon (the khi) c6 ciing so nguyen [\i

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

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

(2) tang them 4.4 gam

a) Xac dinh cong thtfc phan tijT cua hai hidrocacbon trong hon hcJp V i e t cong

thiJc cau tao cua chiing

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

H i f d n g d i n giai

Cdch J: N e u gpi cong thiJc cua hai hidrocacbon trong hon hpp la CxHy

C x H , , Tir phircJng trinh phan tfng dot chdy va khoi lurpng san pham (HjO

P 2 O 5 hap thu, C O 2 bi K O H hap thu dp tang k h o i liTdng cua cac binh ciing

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

tijr cacbon) , ^

a) nhh= 1,12:22,4 = 0,05 moi

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

mco2 = ^9 tang k h o i liTpng cua binh (2) = 4,4g =0 Ucoj = 4,4 : 44 = 0,1

( D C j H , (2) C 2 H , (3) C2H4 C2H4 C2H2

C6ng thtfc cau tao:

(1) C H 3 - C H 3

C H 2 = C H 2 b) Xet c a p ( l ) :

M o i k h i c h i c m 5 0 % the tich hon hdp

Xet ca p (3) DiTa vao phan tfug (2) va (3)

' h h ' H 2 0

c + b = 0,05 ->c + 2b = 0,l b = 0,05 ::> c = 0 v6 l i

Let luan: H o n hdp gom hai k h i C2H6 va C2H2 * pong Ihu-c cau tao: C H 3 - C H 3 va C H ^ C H

I 6 i k h i chiem 50% the tich hon hdp ; ] ,5, ; r

tach 2: Neu gpi y la so nguyen tuT hidro trung binh cua hai hidrocacbon

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

209

C , H - + ( x + y / 4) 0 2 > X C O 2 + y / 2 H 2 0

1 m o l X mol 0,5 y m o l '

0,05 m o l 0,1 m o l 0,1 m o l '*

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

(1 : 0,05) = (0,5 y : 0,1) y = 4 (mot hidrocacbon phai c6 so' nguyen lu'

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

II Bai tap \\f g l i i

1 Do't chay mot lu'cfng hidrocacbon X, ngU'di ta thu diTdc CO2 va H2O theo li if

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

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

2 Hdp chat hffu cd A c6 thanh phan khoi lifdng cac nguyen to nhU' sau: C =

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 \

C B A I T A P L U Y E N T H I

Chii de 1 Vi^'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 c h e

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 vdi nU'dc hoSc nung nong natri axetat vdi NaOH 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

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 voi, viet cac phu'dng trinh phan tfng dieu che' axetilen, benzen

(c6 ghi dieu k i e n phan tfng)

Bai giai '

a) Tur riTdu elylic: C2H5 - O H > C2H4 + H2O

2 Cho biet d 20"C benzen c6 k h o i lifdng rieng la 0,879 g/ml Neu hoa long 7,8

kg hdi benzen xuong 20"C thi thu duTdc bao nhieu l i t benzen

B a i giai

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

nhiet do soi la CH4 < C2H6 < CjHx < C4HU, < C,oH22

2 Hoa long 7,8kg hdi benzen xuong 20"C thu diWc: < \

I 7 ' 8 x ^ 0 ' g = 8873 m l hay 8,873 lit

-•r 0,879g/ml l i i f h ^ l

Bai 4 : ^ / ; H f « : W - r - l

1 Cho hon hdp cac k h i CH4, C2H4, C2H2, SO2, CO2 di qua nu-dc brom V i e t c^c

phu'dng trinh phan tfng xay ra

2 Cho 1 lit benzen (d = 0,879 g/ml) tac dung vdi 112 lit CI2 (ct dktc) k h i c6 mSt

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

B a i giai , , '

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

C2H4 + Br2 > C2H4Br2 ui l« </« < C2H2 + 2Br2 > C2H2Br4

-SO2 + 2H2O + Br2 > H2SO4 + 2HBr ' *

2- Phan iJng giiJa benzen va clo khi CO bot s^t xuc tac ' - ' ' '

CfiHe + CI2 —<f£U CftHsCl + H C l

^, ^ 1000x0,879 , 112 • ^ Tinh n c , H 6 = = 11.27raol; nc,^ - — - 5mol

211

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

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

C o n h i e u each n h a n bie't D i f d i d a y la m o t each d d n g i a n :

- C h o k h i d i qua b i n h nxidc b r o m , b i n h nao k h i l a m ma't m a u nufdc b r o m , do la

e t i l e n : C H = C H 2 + B r j > C H 2 B r - C H 2 B r

T C h o k h i i r o n g ba b i n h c o n l a i qua 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 O 2 : C O 2 + Ca(0H)2 > CaCO., i + HjO 5,

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

k h i CH4: 2H2 + O2 '" ) 2H2O

CH4 + 2O2 — C O 2 + 2H2O

2 1 2

C O 2 + C a ( O H ) 2 > C a C O j i + H 2 O

T^ch loai tap chat:

\ f ach C O 2 k h o i C2H2: C h o k h i qua dung dich k i e m duT, (thi du NaOH)

* C 0 2 + 2 N a O H > N a j C O j + H2O , ^ ^

f dch C2H4 k h o i C O 2 : C h o k h i di qua nifdc brom duf:

C2H4 + Br2 > C2H4Br2

Loai C2H5OH khoi C H 3 - C O O H : Cho hon hdp tac dung v d i dung dich k e m ,

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

2 C H 3 - C O O H + CaO > CaCCHj - C O O ) 2 + H2O

Pun dudi rifdu bay hdi Sau 66 cho axit sunfuric tac dung v d i Ca(CH3

-C 0 0 ) 2 (dun nong) de axit bay hdi va lam ngiTng tu thu di/dc -C H 3 - -COOH

C a ( C H 3 - C O O ) 2 + H2SO4 — ^ 2CH3 - C O O H T + C a S 0 4 i i t

Bai 2

1 Metan bi Ian mot It tap cha't la CO2, C2H4, C2H2 T n n h bay phifdng phap hoa

hoc de loai het tap chat khoi metan

2 Benzen b i Ian mot it nu'dc va riTdu, lam the nao de c6 benzen tinh khiS't

B a i giai - ' ^ ' i ^- '

1 Cho hon hdp k h i Ian Itfdt di qua binh nu'dc brom dtf, luc do loai het C2H2 va

C2H4nhdphanu'ng:

C2H4 + Br2 > C2H4Br2 _ 0 4 a >' C2H2 + 2Br2 > C2H2Br4

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

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

* C 0 2 + 2 N a O H > NazCOj + H2O ^

K h i con lai la C H 4 nguyen cha't

2- Cach ddn gian la cho benzen bi Ian tap cha't tac dung v d i Na diT, luc do niTdc rifdu tac dung vdi Na tao ra cAc san pham khong tan trong benzen:

H2O + Na > N a O H + 0 , 5 H 2 1

C 2 H , 0 H + Na > CjHsONa + O.SHjT

B i i 3

C6 the phan biet mudi an, d i T d n g kinh bang each dot chay hay khong?

^' K h i dot chay k h i A thu diTdc C O 2 va H2O, khi 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

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

J B a t giai

Co the phan biet mudi an va d i T d n g kinh bkng each dot chay, v i mudi an

•^hong chay, van la chat r ^ n mau tr^ng, trong khi do d i T d n g bi chay het thanh

v a H , 0 :

213

C,2H220|, + 1202 '" ) I2CQ2+ IIH2O

2 K h i dot chay khi A thu diTdc CO2 va H2O Dieu do chtfng to chat A chi?a

nguyen to C, H va c6 the c6 hoac khong c6 oxi Vay A la hdp cha't hffu cd

T h i d u : C2H6 +3,502 )• 2CO2 + 3H2O

C2H6O + 3O2 > 2CO2 + 3H2O

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

T h i d u : CS2 + 3O2 > CO2 + 2SO2

Vay B la hdp cha't hUucd

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

nguyen to C, H, N va c6 the c6 hoac khong co oxi '«j

T h i d u : (NH4)2C03 + 1,502 — ^ CO2 + 4H2O + N2 '

l^j^; 2NH2 - CH2 - C O O H + 4,502 — ^ 4CO2 + 5H2O + N2

' V a y chat C c6 the la hdp chat v6 cd ((NH4)2C03) ho|c hiiru cd (NH2 CH,

-COOH)

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

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

Chat A + O2 > CO2 + H2O

Chat B + O2 > CO2

Chat C + O2 > SO2

Chat D + O2 > CO2 + H2O + CI2

Chat E + O2 > CO2 + Na2C03

Chat F + O2 > CO2 + H2O + CaO

B a i giai

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

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

- 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 bi mat nhan, m"J

binh diTng mot chat long hoac mot chat khi sau day: etan, etilen, benzen, ^t^'

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

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

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

2 Nhan bie't cac chat k h i

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

Ca(0H)2 + CO2 — > C a C O j i + H2O Ca(OH)2 + S02 > C a S 0 3 i + H 2 0

De phan biet CO2 va SO2 c6 the dung niTdc brom: SO2 lam ma't mau niTdc

S02 + 2H20 + Br2 > 2 H B r + H2S04 , (CO the sue k h i CO2 vao 2 ket tua)

Cho 2 k h i con lai tac dung vdi nUdc brom, cha't lam mat mau la etilen, con khong la metan

C2H4 + Br2 )• C2H4Br2

CM 3 L a p c 6 n g thiifc p h S n tuT

Bai 1 Dot chay 2,24 lit hidrocacbon X 0 dktc) va cho san pham chay Ian li/dt

di qua binh diTng P2O5 va binh 2 diTng K O H r^n Sau k h i ket thuc thi nghiem

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

C T C T cua hidrocacbon X

jPhan ti-ng dot chay: CJiy + (x + ^)02 > xCOj + ^ H j O (1) • '

k h o i lu'dng binh 1 tang do H2O bi hap thu bdi P2O5 , f ' j

P2O5 + 3H2O > 2H3PO4 " ^ , 5 '

9

1

K h o i lu'dng binh 2 tang do CO2 bi hap thu bdi K O H :

CO2 + 2 K 0 H — > K2CO3 + H2O ' ' • ' '

215

I Do do: ny^^Q=— = 0,5 mol

Dodo n C02

Vi so mol X =

17,6

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

22,4 '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

Lap bang:

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

Phan u-ng dot chay A: C.HyO, + (x + ^ - - )02 > XCO2 + - H 2 O (1)

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 4,48 ,^ ^ , 3,6x2 „

cong thiJc phan tur)

216

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: CxHyO, + (x + - - - )02

-4 2

X 2 Theo dieu kien cho ta c6 ti le: — = — => y =^ 3x

-> XCO2+ - H 2 O (1)

2

(a)

rr., 1 ^ , / N ' 12x + y + 16z Theo phiTdng trinh (1) ta c6 ti le: ^

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)„

x + y -4 x22,4 6,72 (b)

Tinh KL mol PT cua B b^ng 22,4 0,487 «46 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

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

So nguyen lu* H = ^^^^'^ = 6 ; So nguyen tu" C = ^^^^^ = 2 '

100x1 ^ ' 100x12 <

V a y CTPT cua B la C2H6

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

chi mcti biet ti 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 ' ^^*^'

Co the tim cong thuTc phan tuT b^ng each bien luan nhiT sau:

V i so nguyen tu" H phai la so chan nen n chi c6 cac gia tri 2, 4, 6 K h i n = 2 la

CO cha't C:Hf„ diing hoa trj cua oacbon

Khi n = 4 ta c6 cong ihiJc C4H12, cong thiJc nay khong dung v i cacbon c6 hoa

tri l(^n h(Jn 4

Khi n = 6, 8 thi cong thiJc thu diTdc cang sai

C6 the vie't C„H3„ roi bien luan theo tifng loai hidrocacbon

Bai 7 T i le khoi lu'cing cua cacbon va hidro trong hidrocacbon X la mc : mH = 12

Tim cong thufc phan tu" cua X biet khoi liTcfng phan tur cua X Idn gap 1,3 Ian

khoi liTdng phan tuT cua axit axctic

Bai 8 T i m ti le so nguyen tijT C , H , O trong hdp chat Y chiJa 6,67%H, 18,67'7rN

va 42,67% O Biet rSng khi dot chay hoan toan 1 mol Y thu duTdc 11,2 lit N:

V i khi dot chay 1 mol Y thu diTdc 11,21 N2 ttjTc = 0,5 mol N2, dieu

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

2) Theo djnh luat bao toan khoi lifdng:

mA + nio2 = mco2 + rnuxo

niA = 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 mol H2O = 2 x 0,3 = 0,6 mol S6' mol O trong A = to'ng so' mol O trong CO2 va H2O triT s6' mol O trong O2

= 2 x 0,2 + 0 , 3 - 2 X 0,3 = 0,1 mol Vay ti 16 x : y : z = 0,2 : 0,6 : 0,1 = 2 : 6 : 1

Cong thiJc ddn gian nhat cua A la (CjHfiOn Ta thay n chi c6 the bang 1 v i

neu n bang 2 thi d\i hoa tri cua cacbon C4H12O2 (so nguyen tuf H toi da bang

10, nghTa la C4H|,)02)

Vay cong thtfc phan tur cua A la C2H6O «St* ii < ' •••

Bai 10 Do't chay hoan toan hon hcJp 2 hidrocacbon CxHy va CxH,, c6 so' mol bkng nhau thu diTdc 3,52 gam CO2 va 1,62 gam H2O T i m CTPT, vie't CTCT

cua cac hidrocacbon

Bai giai Trirdc het can tinh so mol CO2 va H2O '

4

nco2 = ^ = 0,08mol; UH^O = ^ = 0,09mol ; :

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

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

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

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

ke't doi - (c6 the noi tdng so lien ke't n) va so' vong

Cac phan iifng dot chay

CxH2x.2 + ^ ^ 0 2 — ^ XCO2 + ( X + 1)H20 ,07

CxH2x.2-2a + ^ ^ " " ^ " ' 0 2 XCO2 + ( X + 1 - a)H20

219

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

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

CTPT cua hai hidrocacbon la C 4 H 1 , , va C 4 H X

CTCT: C H 3 - CH2 - CH2 - C H 3 CHi - (pH - C H 3

CH2 = C H - CH2 - CH3 CH3 - CH = CH - CH3

C H 3 - C = CH2 C H 2 - C H 2 CH2

CH3 C H 2 - C H 2 C H 3 - C H - C H 3

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

5rt Viet CTCT cua tat ca cic dan xuat chiJa brom c6 the c6

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 T i m CTPT, viet CTCT cua tat ca cac dong phan cua A ,

f Bai giai

Hdp chat A chiJa 2 nguyen to cacbon va hidro, c6 the bieu dien A Ik C^Hy,

CO phan i^ng chay: •

CxHy + (X + ^ )02 > XCO2 + ^ H2O

220

V i dot m gam A thu diTcJc — gam H2O nen ta c6: -(12x + y) = M S = 9y

Hay y = 2x, ttfc CTDGN Ih (C^UiA- KLPT cua A n^m trong khodng

2,3 X 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 (CO 1 lien ket doi) hoSc ankin (c6 1 lien ket ba) T i le KLPT cua chung la 22:

13 Dot chay hoan toan 0,3 mol X v^ cho san pham chay hap thu vao binh dung dich Ba(OH)2 dvl thay khoi liTdng binh dung dich tang 46,5 gam va c6 147,75 gam ket tua

1 Tim CTPT cua A, B

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

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

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

Giai he phu'cJng trinh tim di/dc a = b = = 0,15mol

Xet ti le KLPT

221

Vay CTPT cua cac hidrocacbon la C^Hx va C2H2

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

p + q = 0,075

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

K h o i liTdng C H B r = C H B r bang 0,05 x 186 = 9,3 gam

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

dem dot chay hoan toan thi thu dU'dc 165 gam CO2 va 60,75 gam H2O

1 T i m CTPT cua A, B biet hon h d p X khong l a m mat mau dung dich niTctc

brom A, B thuoc cac loai hidrocacbon trong c h i T d n g trinh da hoc

2 Tinh % so m o l cua A, B t r o n g X

3 T h e m 0,05 mol hidrocacbon D vao hon hdp X d tren r o i dem dot chdy hoan

toan thi thu diTdc 143 g CO2 va 49,5 gam H2O T i m CTPT, viet tat ca dong

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

vi X khong l a m mat mau dung dich nifdc brom nen B phai thupc loai hidrocacbon thdm (aren), tiJc CnHnph^i thoa man:

n = 2n - 6, tuTc n = 6 va c6ng thtfc phan tuf cua B la CeHe (benzen)

2 Gpi X, y la so m o l cua A, B Theo phdn tfng chay: « ; ,,

CfiHu + 9,502 — ^ 6CO2 + 7H2O

CfiHfi + 7,502 — ^ 6CO2 + 3H2O ^ ' : J

6,72 lit hon hdp X nSng 13 gam

1- T i m CTPT cua ankan va anken, biet so nguyen tijf cacbon trong m o i phan tuf

kh6ngqud4

2- Dot chay hoan toan 3,36 lit hon hdp X va cho tat ca san pham chdy hap thu

v^o dung dich N a O H diT, sau 66 them BaCl2 diT thi thu diTdc bao nhieu gam ke't tua

B a i giai \jUid:jj^, j<y:u-:H 1^.^ Khi cho hon hdp k h i qua n\Xdc brom:

K ^ t luan loai loai C3H8; C3H6 L o a i

K e t luan: C h i c6 cap CjHs va dHf la phil hcfp v d i dieu k i ^ n cho

CjHg + 5O2 > 3CO2 + 4H2O (1)

•M K h o i lu-dng ket tua BaC03 = 0,45 x 197 = 88,65 gam

il B^i t$p III giii

B a i 1 Dot chay hidrocacbon X thu diTdc CO2 va H2O c6 t i le so mol ti/dng """g

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

mm^ mmn oi.d ^rm »(ti u-^ " ^ " ^ d i n

pai 4- hidrocacbon A mach hd, the k h i Khoi lu'dng V l i t k h i nay bang 2 Ian khoi lifctng V lit N2 d cung dieu kien nhiet do va ap sua't Lap CTPT Hidrocacbon do 1 ;,'

MA = 2 MN2 = 2.28 = 56 dvC , jd} gfihi, '

Dat cong thtfc cua hidrocacbon la C^Hy ta c6: 12x + y = 56 " i ' ' Hidrocacbon the k h i c6 so'C < 4.Nghiem thich hdp x = 4, y = 8

Vay, cong thiJc cua hidrocacbon A la C4H«

Bai 5 M o t chcft co cong thiJc ddn gian nha't la C2H5 Lap cong thtfc phan tur cua

chat do

H i ^ n g d i n ' ^ ^

V i - C2H5 la goc hidrocacbon no hoa tri I nen phan tu' chi c6 the gom 2 goc lien ket v d i nhau Do la C4H1,) , , ^ , Bai 6 Hidrocacbon A la cha't khi d dieu kien thadng, cong thtfc phan tuTcd dang

Cx + 1H3X Lap Cong thtfc phan tu" cua A , ,

Hifdng d i n

x = 1 => C2H3 (loai); x = 2 C3H6 (phij hdp); x = 3 => C4Hy (loai)

x > 4 deu loai v i hidrocacbon d the khi

Bai 7 Dot chay hoan toan hidrocacbon X b^ng mot liTdng oxi vira du San pha'm khi va hdi dan qua binh diTng H2SO4 dac thi the tich giam hdn mot nuTa X thuoc day dong ding nao? _ , ^ _ ^ , , ^ • ,^

^ Hi^dng d i n San pham chay gom CO2 va hdi nifdc di qua H2SO4 dac thi h d niTdc bi giff lai The tich giam hdn mot nuTa ttfc la VH20 hm > Vco2 " H J O > " C O 2 •

X la ankan B^i 8 M o t hon hdp 2 ankan ik dong d^ng ke tiep c6 khoi lifdng la 24,8 g, the tich tifdng u-ng cua hon hdp la 11,2 lit (dktc) Lap CTPT cac ankan ' •

H u f d n g d i n , , ?

M 2 a „ R a „ = = 4 9 , 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 Vfjj _

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

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

=> hai hidrocacbon la: C 2 H 6 , CjHx

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

diTdc sau khi cho hdi niTdc ngUYig tu con 65 c m ' trong do c6 25 c m ' la oxi (cac

the tich di/dc do d cung dieu kien) Lap CTPT cua hidrocacbon

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 KOH

r^n thay khoi liTdng binh 1 tang 2,52 g va binh 2 tang 4,4 g Lap CTPT hai

p^i 13 Do't chay hoan toan mot li/dng hidrocacbon can c6 8,96 lit O2 (dktc)

Cho san pham chay di vao dung dich Ca(OH)2 dM thu dtfdc 25 g k6't tua Lap

! Slim

tao H2O 2H2 +

di/dc 1,12 lit khi CO2 (dktc) va 1,26 g H2O

Lap CTPT ciia 2 hidrocacbon

^ 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

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

diroc 16,8 lit C O 2 (dktc) va 13,5 g H 2 O Hai hidrocacbon do thupc day d^og

d i n g nao?

H i M n g d l n '

n c o 2 - ^ = "-75 (mol); n^^o ~ = 0.75 (mol)

Dot hidrocacbon cho n ^ j o = "CO2 thi dc) la anken

B a i 17 Dot chay hoan toan 2 hidrocacbon mach hd trong ciing day dong dang

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

dong dang nao?

Hifdng d i n

n c o 2 = ^ = 0,05 (mol); U H ^ O = ^ = 0.05 (mol)

Dot hidrocacbon cho nyi^Q = TIQQ^ thi do la anken

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

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

Lap CTPT cua chiing

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

day dong d i n g nao?

Hifdng d i n

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

C - H , - + — O2 > n C 0 2 + n H 2 0

B a i 20 Hon hdp khi X gom 1 ankan va 1 anken Cho 1680 m l khi X loi 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 roi cho san pham chay di va'^

binh dirng dung djch Ca(0H)2 d\i thu diTdc 12,5 g ket tiia Lap CTPT ci'"^^

Ankan va anken c6 ciing so nguyen tuT C va c6 cDng so mol nen chay cho

cung so'mol C O 2 va bang = 0,3 (mol)

Bai 23 M o t hon hdp gom 1 ankan va 1 anken c6 l i le so mol 1 : 1 So nguyen tOr

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 Brj Dot chay hoan toan a gam hon [idp thu di/dc 0,6 mol C O j Lap CTPT cua chiing

HiftJngdIn A ,

| ' ' ' B r 2 - " i n k c n = Hankan = 0,1 (mol)

l " c o 2 ankan gap doi cua anken; n^oj ^^ua anken = = 0,2 mol

anken c6 2 C, vay ankan c6 4 C Do lii C 2 H 4 va C 4 H i o

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

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 n^^Q = TIQQ^

Goi X mol CO2 do cung la so mol H2O ta c6: 44x - 18x = 39 x = 1,5

C-H2- +O2 > K C O 2 + n H 2 0

Ta c6 0,4 n = 1,5 => i i = 3,75 => Hai anken la C4H8 va CjHg ^

Bai 26 Dot chay hoan toan 2 hidrocacbon mach hd lien tie'p trong day dong

ding thu di/dc 44 g CO2 va 12,6 g H2O Lap CTPT hai hidrocacbon

Hi^dng din

nco2 = ^ = 1 ('"o'); " H 2 0 = ^ = 0,7 (mol)

=> nco2 > "H20 =^ f^ai chat thuoc day ankin

Bai 27 Dot chdy 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

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) ^ mco2 = 44.0,45 = 19,8 (g)

100

fflH^o = 25,2 - 19,8 = 5,4 (g) nH20 =7^ = ^ ' ^ ('""D =>"H = 0,6 (mol)

n c: n H = 0,45 :0,6 = 3 :4 => C3H4 • ( u ^

r o d i

C A H Tinh thco phi^dlng trinh phan lifng, ' '

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 a) nci2 = nc2H4 = 1 = a ta chi thu difdc C2H4CI2 (1 mol) , •;

-b) nci2 > nc2H4 hay a < 1 ta thu dufdc a mol C2H4CI2 va c6n (1 - a)mol CI2 c) < 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: C4H10 + 6,502 > 4C02 + 5H2O (D

CO2 + Ba(0H)2 > BaCOj i + W (2)

Ne'udirC02thi CO2 + H2O + BaCOj > Ba(HC03)2 (3) £

Tinh so'mol cac cha't: n r , o , , = - ^ ^ =0,025 mol; ' u i

C4H1O 22,4 , , i , : ; i?^:

231

n ^ j j ^ =0,025 X 4 = 0,1 mol; n B „ ( O H ) 2 =0,75 x 0,2 = 0,15 mol

nBi,(on)2 > nco2 khong xay ra phan tfng (3) va so mol ket tu;,

BaCOj bang so mol CO2 = 0,1 mol

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

2 Cac phan ufng nhtfcau 1:

Tinh: nB„oH)2 =0-5 x 0,2 = 0,1 mol; Us^co^ = 0,08 mol '

Co hai triTcing hdp xay ra:

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

Ta C O n^-o, = nBaco3 = 0,08 mol Dodo H Q ^ I O =—^—• = 0,02 mol

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

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:

0,1 + ( 0 , 1 - 0 , 0 8 ) =0,12 mol

pu (2) pu (3)

D o d o nc^Hio = — =0.03 Ttfc V C ^ H , , , =0,03 x 22,4 = 0,672 lit

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

bao nhieu kg thuo'c trijf sau 666

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

Baigiai

1 Phan u-ng dien phan: 2NaCl + 2H2O H2T + C U T + 2NaOH (1)

Tdng so mol NaCl = 0,5 x 4 = 2 mol

So mol CU thoat ra = - so' mol NaCl bi dien phan = — x 2 x = 0,8 mol

Theo phan urng dieu che thuoc trif sau C 6 H 6 C 1 6 :

C , H , + 3CI2 > C,H,CU (2)

1 0 8 Theo phan lirng (2) n666 = - n c i 2 = - r - m o l

0 8

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

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

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

232

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

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 chay hoan toan 1 gam xang

Bai giai

J, V i gia sur xang la hon hdp dong so' mol C5H12 va C 6 H 1 4 , nen 1 mol xang chtfa

0,5 mol C 5 H 1 2 va 0,5 mol C6H14, liJc 1 mol xang nang:

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%

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

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 dúdc, 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

dúdc, bic't hieu suat moi phan iJng la 75%

Bai giai

1 Phan iJng dieu che axetilen: CaCj + 2H2O > Că0H)2 + C2H21 (1)

(nhd rang 1 tan = lÓ^g = lÔkg; Im^ = 10^ dm^ = 10^ lit)

T U u ' > 1 0 x 1 0 ^ 9 6 , ^ , ^ 5 ,

Theo phan Mng (1): Uc = "cac, = " 1.5.10^mol

^ ^ ^ 100x64 Vay the tich C2H2 b&ng: 1,5 x lÓ x 22,4 = 3,36 x lỐl = 3,36 x lÓm^

2 Cac phan lirng cong hdp va trilng hdp:

nCH2 = CH2 ( - CH2 - CH2 -)„ (3)

Theo (2, 3) so mol m^t xich bkng ^ so mol C2H4, c6n so mol C2H4 bing

— so mol C2H2 Do do khoi liTdng PE blng:

1 Hon hdp khi X chtfa % the tich khi nhif sau: 22,4% CO2, 44,8% CO, 32,8%

CH4 Tinh % khói 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 Ỵ

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 ,

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

Vay % khói lifdng CO2 = =35,65%; % C 0 = ^ ^ ^ ^ ? ^ = 45,37%

CH4 + 2O2 ' > CO2 + 2H2O (3)

I Goi X, y, z la so mol cua H2, CO, CH4, ta c6 he phifdng trinh:

I Tong só mol hon hdp: x + y + z = = Ọlmol

22,4 1,568

I Tóng só mol CO2: y + z = = 0,07mol

22,4

(a)

(b)

235

••v~

- Tong so mol H2O: x + 2z = = 0,13niol (c) ^ •'

18 Giai he phifOng irinh (a, b, c) ta tint diTdc:

1 Tinh % khoi luTdng m o i nguyen to trong aminoaxit N H 2 - C H 2 - C O O H (glyxin)

2 Ne'u mot day hidrocacbon di/Oc bieu dien bdi cong thtfc chung CnH2n + 2 thi

thanh phan % cua hidro bien doi nhiT the nao khi gid tri n thay d d i

n + 1 7

V a y thanh phan % H bien thien trong khoang: 14,29% < % H < 25%

B a i 4

1 Hon hop k h i A gom 5 lit H2 va 15 lit C2H6, hon hop k h i B gom 10 lit CH4 va

10 lit C2H4 Cac the tich khi deu d dieu kien tieu chuan H o i hon hdp khi A

hay hon hcfp k h i B nSng hdn?

H5n hdp khi D gom 5 lit H2, 5 lit CH4 Neu them 15 lit hidrocacbon k h i X v^o

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

jihi deu do d dieu kien tieu chuan T i m cong thuTc phan tOr cua X

Bai giai

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

1 lit hon hdp hoSc 1 mol hon hdp, nhiTng trong bai nay ta chi can tinh khoi liTdng cua 20 lit hon hdp A cung nhiT B:

KhS"! lu'dng cua 20 lit A bang: — ^ x 2 + - ^ ^

Khoi liTdng cua 20 lit B bKng:

Khi noi hon hdp khi D nang bang etan (C2H6) c6 nghla la 1 lit D nang bang 1 lit etan hoac khoi Itfdng cua 1 mol D bang khoi lu'dng cua 1 m o l etan C2H6

b^ng 30g Do do ta c6 bieu thtfc ve khoi liTdng 1 mol D : f* u.^mt^ ;<tff« >

Goi cong thiJc cua X la C^Hy Ta c6 12x + y = 44 , , j ri,- 5

Bi^n luan nhU cac bai tren ta de dang tim di/dc CsHx » 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 Tinh % 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

ml dung dich Brj 0,04M Hoi % the tich cua CH4 bien doi trong khoang nao

^Vay % the tich cua C2H4 =

160 0,025x100 0,25 = 10%

I Va % the tich cua CH4 = 100 - 10 = 90%

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

C2H2 + 2Br2 > C2H2Br4 (2)

11'!,! > :

231

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

22,4

S6' mol brom = 0,5 x 0,04 = 0,02 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

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 so mol

C2H2, do do the tich V, phai gap doi the tich V2, ttfc V, : 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

Phan iJng dot chay A:

238

CnH ^2n + 2 + - O 2 nC02 + (n+ 1)H20 3n + l p, 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

7 2 Theo phan iJng (1), so mol niTdc: nH20 = a(n +1) = = 0,4mol

(1)

(2) (3)

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 ,

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

2 Tinh kh6'i liTdng niTdc a 6!) h '

Cho cic the tich khi do d dktc

H2O CO2

(1) (2)

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

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 C 3 H S 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 d6 1 lit A n&ng hctn 1 lit N2

(C6 the ddn thuan diTa vko bieu thuTc tinh MA c» tren v^ flm MA khi x n^m 5 2

chn-240

Khi X = 0; MA = 28,264 Khi x = 0,562 ; MA = 35,008 NhiT vay 28,26 < MA < 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

C O 2 + 2H2O

1; ,t'-s

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:

Trong do x la % the tich cua CO

Gidi phiTdng trinh (III) ta c6 x = 0.5 tuTc CO chiem 50% va H2 chiem:

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

2 Dot chay hoan to^n phan 2 Tinh khoi liTdng CO2 vk H2O tao thanh 241

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

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

2a + 30b = 1,8 Giai ra ta difdc a = 0,15 va b = 0,05

: Vay % the tich cua H2 bang "'^^""^^^ = 75%

^ 0,2

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

C6 the giai theo KLPTTB cua C

30b + 2 ( 0 , 2 - b )

M c =0,4018x22,4 = 9 =

-0,2 Khoi liTdng 1/2 B b^ng tong k h o i lifdng C v a k h o i liTdng b i n h tang (khoi

liTdng C2H4 v a C2H2) = 1,8 + 2,7 = 4,5 gam

Khoi liTdng A b^ng 2 Ian k h o i liTdng B = 4,5 x 2 = 9 g a m

'2x + 26y = 9 40x + 64y =43,2

So mol C2H2 = nc2H2 ^an dau - n c 2 H 4 -"CzHfi = 0,3 - 0,1 - 0,1 = 0,1

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

2 C O 2 + H2O H2O

(7) (8)

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

Ipioi lurdng C O 2 = ^ 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

1 Tinh the tich V

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

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

= 0,08mol

t ,1* ,(^>«'

Trirdc het can tinh: ^i^^(Q\i)2 =0,5xO,2 = 0 , l m o l ; H g g ^ Q ^ = •

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

^) n c o 2 = "BaC03 = 0,08 mol = ncH4 Vay Vcoj = 0,08 x 22,4 = 1,792 lit

2) Khoi liTdng binh dung dich tang (m)

m = m c o 2 + I " H 2 0 = 0,08 x 44 + 0,08 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 n c o 2 > 0,1 mol, luc d6 xay ra p h a n tfng (3) |

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

Vay V c o 2 = 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

ChUOng V

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 han trong n\idc

C6H,206

C2H4

CO2 CH3-CH2OH

CzHjOHd) +Na(rt > CzHsONad, + l/2H2(k)

CH3COOH0, + C2H3OH0) ) CH3COO-C2H5(,) + H2O

Do rifcJu = ^nf(1unguyencha't ^ ^QQO

*dungdichnr0u

I I Axit axetic CH3COOH

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

•O, xt, t'

Qujr tim „ , ^,

— • Hoi do

+CuO C4H 10

, (CH3COO)2Ca +NaOH

I Cic phifcfng trinh p h a n iJng :

C4H,o + 2,502 CH3-CH2OH + O2 2CH3COOH +CuO —

2CH3COOH +CaC03 —

CH3C00H + NaOH — > cH3C00Na + H2O mmwim V • CH3COOH + C2H5OH H 2 S 0 4 c l » c t » ^ CH3COO-C2H5 + H2O

01 Chat beo (RCOO)3C3H5 (ran ho§c long)

|Chat beo la hon hcJp nhieu este cua glixerol C3H5(OH)3 vdi cic axit b^o

l C O O H ( v d i R l a C n H 3 5 - ; C n H 3 3 - ; C , 5 H 3 , - ) ^

|C6ng thiJc Chung : (RCOO)3C3H5

?han iJng thuy phan: ''

(RCOO)3C3H5 + 3H2O > C3H5(OH)3 + 3RC00H , ,

?hin ufng xa phong hoa :

Axit gluconic + Ag CzHjOH

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

CizHzzOn +H2O axit.t o

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 no

* Ca tinh bot va xenlulozcJ deu tham gia phan tfng thuy phan:

Cty TNHH MTV DWH Khang Viet

* 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-CH2-COOH)

* 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

1. 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

C2H2 : 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

2CH3COOH + 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

Sau khi nhan ra CH3COOH 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 + H21

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 :

(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

" a) CH3-CH3; CH2=CH2; CH3-CH2OH

; b) CH2=CH2; CHsCH ; CH3-CH=CH2 "^-^

J d) CH=CH ; CH3-C=CH ; CH3-CH=CH-CH3

A, B la hai hdp chat hffu cd cung cd cong thtfc phan tur C2H6O. Cong thiJc cau

tao cua chiing nhiTsau :

1) CH3-O-CH3 ; 2) CH3-CH2-OH

Hay cho biet cong thtfc cau tao nao la cua chat A ? cua chat B ? Biet r^ng A

la 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

l8

Dang 2: V\4t c6ng thOc c^u tgo,

Ho^n th^nh phifOng trinh phdn Cfng - Oi^u ch^

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

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

c) Protein, tinh bot, xenlulozd, polietilen

d) Etyl axetat, chat beo

Hvtdng dSn giai

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

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

c) Deu la polime

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

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

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

b) Glucozd, saccarozd, tinh bot, xenlulozd

HuTdng d§n giai

a) Thuoc nhom nhien lieu (cha't dot)

b) Thuoc nhom gluxit

3. Vie't cac phiTdng trinh phan uTng thtfc hien cac bien hoa hoa hoc sau :

Tinh bot ^'^ > Glucozd — R i T d u etylic — A x i t axetic —

C H 3 C O O H + C2H5OH "2^°4'»g«=-'" ^ CH3COOC2H5 + H2O

CH3COOC2H5 + H2O "^^°^'"'"§=±CH3COOH + C2H50H

C2H5OH _Jl2SOld?c_^ C2H4 + H2O

nCH2=CH2 > (-CH2-CH2-)„ ^

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

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

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

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

248

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

Cau e

5. Hay giai thich tai sao riTdu etylic v^ axit axetic deu tac dung vdi kim loai

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

va benzen lai khong c6 phan uTng nay

Htfdng din giai

Drfa vao thanh phan v^ ca\ tao phan tur: i Rtfdu etylic C2H5-OH va axit axetic C H 3 - C - O H ,

o '

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

kim loai kiem nhu* Na

2C2H5-O-H + 2Na -—> 2C2H5-0-Na + Hzt

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

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

Etilen — r i T d u etylic — \ * magie axetat

axit axetic Natricacbonat—^^ natri axetat <—^-^—II • e t y l axetat i

HritfngdSngiai Cho tac dung vdi cic chaft:

(1) H2O (xuc tic axit); (2) CH3COOH ; (3) oxi 16n men d 30 - 32"C ; (4) dd NaOH; (5) Mg ; (6) C2HJOH (xuc tic H2SO4 dac)

7 a) Vj^t sd do hiiu diSn m^i quan h$ giCTa hidrocacbon \k h0p cha't c6 nhdm chtfc: nrdu, axit, este

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

HiMng dSn giai

a) etilen ^"2° > riTdu etylic °2 > axit axetic > etylaxetat

axit • Men H2SU4a

b) CH2=CH2 + HOH ^ 5 i i M M _ > CH3-CH2OH

CH3-CH2OH + O2 > CH3COOH + H2O CH3COOH +CH3-CH2OH HgS04c6c >CH3COO-CH2-CH3+ H2O

t"

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

II Bdi tSp tif giil

1 Viet phiTdng trinh phan tfng (ghi ro dieu kien neu c6) di thifc hien scf d6

chuyen hoa sau :

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

hJQu sudt phdn Ong, ndng dQ dung djch

I Bdl tSp CO IcTi glii

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

-Khoi liTdng riTdu 40" can dung : 3,956 + 7,452 = 11,408 g

2 Dun nong hon hdp gom 10 gam riTdu etylic va 3 gam axit axetic (c6 mat H2SO4

dSc lam 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 khoi lifdng san pham (este) theo chat phan tfng he't, suy ra khoi li/dng thifc te thu diTdc theo hieu suat da biet

C H 3 C O O H + C 2 H 5 O H > C H 3 C O O C 2 H 5 + H 2 O

60g 46g 88g 3g x (g) < 3 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 khoi Itfdng este theo khoi 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 - r^J -^'^ ' ' 2 5 1

60

K h o i ItfcJng este C H 3 C O O C 2 H 5 thvfc te thu difdc la: 0,05 88 — = 2,64 g

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

(dktc) T i n h nong do mol cua dung dich glucozd bie't hieu suat cua qua trinh

16n men chi dat 4 0 %

Htfdng d i n giai

Tiir phi/cJng trinh p h J n tfng len men va the tich (roi suy ra so mol) CO2, dm

difdc so mol, k h o i liTcJng glucozd len men, tiif hi?u suat qua trinh, ta t i m dirdc

so m o l glucozcf C O trong 2 l i t dung dich

n c o 2 = 17,92 : 22,4 = 0,8 m o l In •

CeUuOf, - "^e"30-32"c ^ 2C2H50H + 2 C 0 2 t

0,4 m o l 0,8 mol

100 ' "

So mol glucozd c6 t r o n g 2 l i t dung dich: 0,4 — = 1 m o l , , ,

Nong do mol cua d u n g dich glucozd 1^ 1 : 2 = 0,5M

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

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

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

trong xa phong c6 6 0 % k h o i liTdng CnHssCOONa

Hifdng d§n giai

TH phifdng trinh p h ^ n tfng xa phong hod v^ so m o l (suy tCf kh6'i liTcfng) chat

h6o ta ti m diTcJc so m o l va k h o i ItTdng glixerin, muo i natri, va suy ra khoi

5 NgiTcJi ta tron deu a gam axit axetic v d i b gam rtfdu etylic r o i chia lam

phan deu nhau

- Cho phan mot t a c dung \di Na d\i, thu difdc 5,6 lit k h i >

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

a) T i n h a va b Biet 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 khoi lifdng este tao i h ^ " '

hieu suat phan tfng e s t e hod dat 60%

252

I

y : 6 = 0,15 m o l

Hifdng dSn giai

Tur the tich, suy ra nkhi = 5,6 : 22,4 = 0,25 mol va 2,24 : 22,4 = 0,1 Tru^

Tit cdc phiTdng trinh p h a n i?ng ta lap diTdc he phtfdng trinh lifin quan giQ.^

^g-m o l chat p h a n iJng (axit, ruTdu) va n^hi

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

X 0 6

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

(neu hieu sua't 100%)

C 2 H 5 O H trong 1/3 hon hdp la ^ = M = 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:

60

0,2 88 — = 10,56 g C H 3 C O O C 2 H 5

100

Bai t§p tir gi§i

^' Pha 15 lit riTdu 20" vao niTdc r o i len men giSm, sau qui trinh len men ngurfJi ta

thu difdc m kg dung dich CH3COOH 2% Hi?u suat qua trinh len men la 90'??'

Xac djnh gia tri bkng so cua m

^' Cho 10,6 g a m N a 2 C 0 3 vdo dung dich CH3COOH 0,5M 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 d}ch

Ca(0H)2 0,075M T i n h :

a) The tich dung dich 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 '

i