pp giẢi toÁn ĐỐt chÁy h.c

C H2 2 2 2

x

y 2

2

2

1

PHNG PHP GII NHANH BI TON T CHY HIROCACBONI. Mt s lu v phng php Hirocacbon CxHy hoc CnH2n +2 -2k ( n 1; k 0) Vi k l bt bo ha (tng s lin kt CxHy + + (x y + 4 3n + 1 - k O2 2 nCO2 ) O2 v vng no) y 2 H2 O

xCO2

Hay CnH2n + 2 - 2k +

+ (n + 1 k)H2O

Da vo sn phm ca phn ng t chy hirocacbon +n > n H2O + nHO CO 2

suy ra cht l ankan CnH2n +2 v n

= nankan H2O

nCO2

= n CO

suy ra cht l anken (hoc xicloankan) CnH2n2

+ n mC H (phn ng)

= mC + mH = 12 nCO2 + 2 n H2 OH2 O

n

1 = n CO2 + O2 (p) n 2

Mt s cng thc cn nh + Khi lng mol trung bnh: M = + S nguyn t C = n CO nx y

m hh n hh

+ S nguyn t C =

n CO2 n hh

1

Copyright V Ngc Bnh, Dy v hc Ha hc

3

3

2 2 2

?

2

3

2

2

3

2 n1 .a + n 2 a+b

+ n= .b

trong , n , n1

l s nguyn t C ca cht 1, cht 2.2

a, b l s mol ca cht 1, cht 2. Khi s nguyn t C trung bnh bng trung bnh cng ca 2 s nguyn t C th 2 cht c s mol bng nhau. Tc l, n = n2 n1 + 2 Thng cho sn phm chy thu c dn qua bnh (1) ng cht hp th H2O: P2O5, H2SO4 c, CaCl2 khan, bnh (2) ng cht hp th CO2 nh: NaOH, KOH, Ca(OH)2, Ba(OH)2, Khi , khi lng bnh (1) tng = mH2O , khi lng bnh (2) tng = m CO2 Nu cho ton b sn phm chy qua dung dch Ca(OH)2, Ba(OH)2 th khi lng bnh tng = mCO2 + m H2 O. Khi , khi lng dung dch tng hoc gim so vi khi lng dung dch ban u. + Khi lng dung dch tng: m + Khi lng dung dch gim: m = (mdd CO 2

=> a = b

+mH 2O

) m +m )H2O

dd

= m (m

CO 2

+ Lc b kt ta, un nng li c kt ta => trong dung dch c mui hirocacbonatt M(HCO ) MCO + CO 2 + H 2 O0

II. Mt s bi ton Bi ton 1: Bi ton t chy cho tng loi hirocacbon V d 1: t chy hon ton 0,15 mol hn hp 2 ankan thu c 9,45 gam H2O. Cho sn phm chy vo dung dch Ca(OH)2 d th khi lng kt ta thu c l A. 37,5 Su y lun: n ankan = n H n CO2 =O

B. 52,5 n CO = nH

C. 15 n ankan

D. 42,5

n CO

O

9, 45 18

0,15 = 0, 375 mol

n CaCO = n CO = 0, 375 => m CaCO = 0, 375.100 = 37, 5 gam p n A. mol

2


2

22

2

2

3 V d 2: t chy hon ton hn hp 2 hirrocacbon lin tip trong dy ng ng thu c 22,4 lt CO2 (ktc) v 25,2 gam H2O. Hai hirocacbon l A. C2H6 v C3H8 C. C4H10 v C5H12 Su y lun: nH2O

B. C3H8 v C4H10 D. C5H12 v C6H14

=

25, 2 18

= 1, 4 mol

CO2

= 1 mol

;n

nH O > n CO CHn 2n + 2

=> 2 cht thuc dy ankan. Gi n l s nguyn t C trung bnh.

+

3n + 1 2

O2

nCO + (n + 1)H O2 2

Ta c:

1 = => n = 2, n+1 4 5

n

C3H 6 v C3 H8

Hoc n ankan = n 2 O n CO2 = 1, 4 1 = 0, 4 mol H n= n CO 2 n hh p n A. 1 = = 2, 5 0, 4 C H v C H3 6 3 8

V d 3: t chy hon ton V lt (ktc) mt ankin th tch kh thu c CO2 v H2O c tng khi lng 25,2 gam. Nu cho sn phm chy i qua dd Ca(OH)2 d thu c 4,5 gam kt ta. a) V c gi tr l: A. 6,72 lt Su y lun: nCO2

B. 2,24 lt

C. 4,48 lt 25, 2 0, 45.44 18

D. 3,36 lt

= n CaCO3

= 45 100

= 0, 45 mol

n H2 =O

= 0, 3 mol

n ankin = n CO n H

O

= 0, 45 0, 3 = 0,15

mol Vy Vankin = 0,15.22,4 = 3,36 lt p n D. b) Cng thc phn t ankin l A. C2H2 Su y lun: S nguy n t C = n CO B. C3H4 C. C4H6 D. C5H8

n ankin =

0,

45 = 3 0,15

C3H 4 p n B.

3


2 2 2

O2 2 2 2

4 V d 4: t chy hon ton mt lng hirocacbon X. Hp th ton b sn phm chy vo dung dch Ba(OH)2 d to ra 29,55 gam kt ta, dung dch sau phn ng c khi lng gim 19,35 gam so vi dung dch Ba(OH)2 ban u. Cng thc phn t ca X l A. C3H4 B. C2H6 C. C3H6 D. C3H8 (Trch TTS vo cc trng i hc khi A, 2010) Su y lun: nCO2

= nBaCO3

=

29, 55 197

= 0,15 molO

Khi lng dung dch gim = m BaCO (m CO2 + m H3

) = 19, 35

=> m H2O = 19, 35 + 0,15.44 29, 55 = 3, 6 gam => n H2 O = 0, 2 mol nHO

> n CO => X l ankan v n X = n H n CO nX

O

n CO = 0, 2 0,15 = 0, 05 mol

=> S nguyn t C (X) = p n D.

3 = 3 => X l: C H8

V d 5: Hn hp hirocacbon X v oxi c t l s mol tng ng l 1 : 10. t chy hon ton hn hp trn thu c hn hp kh Y. Cho Y qua dung dch H2SO4 c, thu c hn hp kh Z c t khi so vi H2 bng 19. Cng thc phn t ca X l A. C3H8 B. C3H6 C. C4H8 D. C3H4 (Trch TTS vo cc trng i hc, Cao ng khi A, 2007) Su y lun: Chn s mol cc cht theo ng h s phn ng C xH y + (x + 1 mol (x + y y )O 2 xCO 2 + HO 4 2 2 y 4 ) mol x mol y 2 mol y 4 )] mol

Hn hp kh Z gm: CO2 (x mol) v O2 d [10 - (x + CO2 44 38 O2 32 6 6

n CO n

=

1

2

1

x = 10 - x y 4

8x + y = 40

4


5

Ch c gi tr x = 4, y = 8 l tha mn => Cng thc phn t ca X l C4H8. p n C. Bi ton 2: t chy hn hp hirocacbon V d 6: t chy hon ton mt th tch kh thin nhin gm metan, etan, propan bng oxi khng kh (trong khng kh oxi chim 20% v th tch), thu c 7,84 lt kh CO2 ( ktc) v 9,9 gam H2O. Th tch khng kh (ktc) nh nht cn dng t chy hon ton lng kh thin nhin trn l A. 70,0 lt Su y lun: nO2

B. 78,4 lt

C. 84,0 lt

D. 56,0 lt

(Trch TTS vo cc trng Cao ng, 2007) 1H2O

= nCO2

+

=

n 2

1 9, 9 + . = 0, 625 mol 22, 4 2 18

7, 84

=> Vkk

100 = 0, 625.22, 4. = 70 lt ( V oxi chim 20% th tch khng kh). 20

p n A. V d 7: t chy hon ton m gam hn hp X gm: C3H8, C4H6, C5H10 v C6 H6 thu c 7,92 gam CO2 v 2,7 gam H2O. m c gi tr l: A. 2,82 Su y lun: S phn ng: X { C3H8, C4H6, C5H10, C6H6} O2, 0 t 7,92g CO2 + 2,7g H2O B. 2,67 C. 2,46 D. 2,31

Theo bo ton nguyn t C v H (C v H trong X chuyn ht thnh C trong CO2 v H trong H2O) nn ta c: m X = p n C. V d 8: Hn hp X c t khi so vi H2 l 21,2 gm propan, propen v propin. Khi t chy hon ton 0,1 mol X, tng khi lng ca CO2 v H2O thu c l A. 20,40 gam Su y lun: Cch 1: S dng phng php trung bnh * Ch : Khi t chy hn hp cc cht hu c c cng s nguyn t H nhng khc s nguyn t C v ngc li. Ta t mt cng thc chung cho c hn hp cc cht , trong B. 18,60 gam C. 18,96 gam D. 16,80 gam (Trch TTS vo cc trng i hc, Cao ng khi A, 2008) mC + m H = 7, 92 2, 7 .12 + .2 = 2, 46 gam 44 18

5


3

6,4

?

2

2

6 gi tr trung bnh l s nguyn t ca nguyn t khc nhau gia cc cht trong hn hp (quy bi ton v 1 cht). Cng thc phn t chung ca propan, propen v propin l C3 H y M = 42, 4 => 36 + y = 42, 4 => y = 6, 4 CH 0,1 3CO + 3,2H O 0,3 0,32t0

m = 44.0,3 + 18.0,32 = 18,96 gam p n C. Cch 2: S dng phng php quy i Ta thy hn hp X ch gm 2 nguyn t C v H, ta quy 0,1mol hn hp X v 0,3 mol C ( c 3 cht u c 3 nguyn t C) v y mol H nH = 4,24 0,3.12 = 0,64 mol. 0,3 mol C 0,3 mol CO2 0,64 mol H 0,32 mol H2O => Khi lng (CO2, H2O) = 44.0,3 + 18.0,32 = 18,96 gam. p n C. Cch 3: Gi cng thc chung ca propan, propen v propin l C3Hy C 3Hy 0,1 mol 3CO2 0,3 mol 0, 64 1 0, 64 = 0, 32 mol 2

=> nC = 0,3 mol => mC = 0,3.12 = 3,6 gam. => mH(X) = 4,42 3,6 = 0,64 gam => n = nH

= 0, 64 =>

H2O

=

Vy khi lng (CO2, H2O) = 44.0,3 + 18.0,32 = 18,96 gam. V d 9: t chy hon ton 2,24 lt (ktc) hn hp kh X gm CH4, C2H4, C2H6, C3H8 v C4 H10 thu c 6,16 gam CO2 v 4,14 gam H2O. S mol C2H4 trong hn hp X l A. 0,09 Su y lun: Hn hp X gm anken C2H4 v cc ankan Vi ankan, n H 2 O > nCO 2 v n ankan = n 2O n HH2O CO 2 CO2

B. 0,01

C. 0,08

D. 0,02

t chy hn hp ankan v anken thH2O CO 2 anka n

Vi anken, n n ankan = => n

=nH 2O

n

> n

,n

=

H 2O

n

CO2

n

n

4,14 6,16 nCO2 = 18 44 = 0, 09 mol = 2, 24 22, 4 0, 09 = 0, 01 mol

anken

= n n X

ankan

p n B.

6


7 V d 10: Hn hp X gm mt ankan v mt anken. T khi ca X so vi H2 bng 11,25. t chy hon ton 4,48 lt X, thu c 6,72 lt CO2 (ktc). Cng thc ca ankan v anken ln lt l: A. CH4 v C2 H4 C. CH4 v C3 H6 Su y lun: nX = 4, 48 = 0, 2 mol 22, 4 B. C2H6 v C2 H4 D. CH4 v C4H8

M X = 11, 25.2 = 22, 5 => Ankan l CH4 (V ch c hirocacbon duy nht c M < 22,5 l CH4) 6, p dng BTKL: m H = m X m C = 22, 5.0, 72 .44 = 0, 9 gam 2 22, 4 n H2 =O

1 1 n = .0, 9 = 0, 45 mol H 2 2

=> n CH 4 = n

H2 O

n

CO 2

= 0, 45 0, 3 = 0,15 mol

=> nanken = 0,2 0,15 = 0,05 mol Gi cng thc tng qut ca anken l CnH2n (n 2) Ta c: 0,15 mol CH4 0,15mol CO2 0,15 mol CnH2n 0,15n mol CO2 => S mol CO2 = 0,15 + 0,15n = 0,3 => n = 3 Anken l C3H6. p n C. V d 11: t chy hon ton 1 lt hn hp kh gm C2H2 v hirocacbon X, sinh ra 2 lt kh CO2 v 2 lt hi H2O ( cc th tch kh v hi o cng iu kin nhit v p sut). Cng thc phn t ca X l: A. C2H6 Su y lun: C2H2 l ankin nn n < n , t chy hn hp kh cho V H2 O CO 2 C= V 2 CO Vhh = . 2 1 = 2 => X c 2 nguyn t C => X l C H2 6CO2

B. C2H4

C. CH4

D. C3H8

(Trch TTS vo cc trng i hc, Cao ng khi B, 2008)

= V => X phi l ankan.H2 O

p n A.

7


2

2

2

2

2

8 V d 12: t chy hon ton hn hp M gm mt ankan X v mt ankin Y thu c s mol CO2 bng s mol H2O. Thnh phn % v s mol ca X v Y trong hn hp M ln lt l A. 75% v 25% Su y lun: n ankan = nH =O

B. 20% v 80%

C. 35% v 65%

D. 50% v 50%

(Trch TTS vo cc trng Cao ng, 2008) n CO ; n ankin n CO n H

O

Khi t chy hon ton hn hp M cho n H p n D. Bi ton 3: Ta c s sau: Hirocacbon cha no H2 Ni, 0 t

O

= n CO => nankan = nankin

(X)

(Y)

Cc hirocacbon H2 (c th d)

O2, t

0

CO2 H2 O

Theo BTNT, (C, H) trong (X) chuyn thnh (C, H) trong (Y) chuyn thnh (C, H) trong (CO2, H2O) => t chy hon ton hn hp (Y) ging nh t chy hon ton (X) ban u => nn tnh ton theo (X) s n gin hn. Phn ng hiro ha khng lm thay i mch C, nn t chy hn hp (Y) hay (X) cho cng s mol CO2. V d 13: Chia hn hp gm C3H6, C2H4 thnh 2 phn u nhau: - t chy phn 1 thu c 2,24 lt CO2 (ktc) - Hiro ha phn 2 ri t chy ht sn phm th th tch CO2 (ktc) thu c l A. 2,24 lt p n A. D qu! Bi ton 4: Sau khi hiro ha hon ton hirocacbon khng no ri t chy th thu c s mol H2O nhiu hn so vi khi t lc cha hiro ha. S mol nc tri hn chnh l s mol H2 tham gia phn ng hiro ha CnH2n+2-2k + O2 + kH2 + O2 CnH2n+2 + O2 = (n + 1) H2O B. 1,12 lt C. 3,36 lt D. 4,48 lt

(n + 1 k) H2O + kH2O

Ta thy, hiu [(n + 1) (n + 1 k)] = k (chnh l s mol nc tri hn ca ankan do H2 to ra)

8


9 V d 14: t chy hon ton 0,1 mol ankin c 0,2 mol H2O. Nu hiro ha hon ton 0,1 mol ankin ny ri t chy th s mol H2O thu c l A. 0,3 Su y lun: Ankin cng hp vi H2 theo t l mol 1 : 2. Khi cng hp c 0,2 mol H2 phn ng nn s mol H2O thu c cng thm l 0,2 mol, do s mol H2O thu c l 0,2 + 0,2 = 0,4 mol p n B. Bi ton 5: Sau khi Crackinh ankan ri em t chy hn hp hirocacbon thu c: Ankan: CnH2n+2 (n 3)Crackinh Hoc tch H2

B. 0,4

C. 0,5

D. 0,6

(X)

(Y)

Anke n Anken H2

O2, t

0

CO2 H2 O

Theo BTNT, t chy hn hp (Y) ging nh t chy hon ton (X) ban u. V d 15: Tin hnh Crackinh nhit cao 5,8 gam butan. Sau mt thi gian thu c hn hp kh X gm CH4, C2H6, C2H4 v C4 H10. t chy hon ton X trong kh O2 d, ri dn ton b sn phm sinh ra qua bnh ng H2 SO4 c. tng khi lng ca bnh ng H2SO4 c l: A. 9,0 gam Su y lun: S phn ng: C4H10 C4H10 0,1 mol p n A.Crackin h

B. 4,5 gam (X)

C. 18,0 gamO2, 0 t

D. 13,5 gam

H2 O

t chy hon ton X ging nh t chy hon ton 5,8 gam butan ban u 4CO2 + 5H2O 0,5 mol

tng khi lng bnh ng H2SO4 c chnh l khi lng H2O = 0,5.18 = 9 gam.

9


10 III. Bi tp t luyn Cu 1: t chy hon ton 0,01 mol hn hp hai ankan thu c 0,72 gam nc. Cho sn phm t chy i qua bnh ng dung dch nc vi trong d th khi lng kt ta thu c l A. 0,3 gam B. 3,0 gam C. 0,6 gam D. 6,0 gam Cu 2: t chy hon ton mt lng hirocacbon X cn ti thiu 7,68 gam O2. Ton b sn phm chy c dn qua bnh (1) ng H2 SO4 c, d, sau qua bnh (2) ng dung dch Ca(OH)2 d. Kt thc th nghim thy bnh (1) tng 4,32 gam, bnh (2) thu c m gam kt ta. Cng thc phn t ca X v gi tr m ln lt l A. C2H6 v 10 B. C2H4 v 11 C. C3H8 v 9 D. CH4 v 12 Cu 3: Mt hn hp X gm hai hirocacbon k tip nhau trong dy ng ng. t chy hon ton hn hp, sau dn sn phm chy qua bnh (1) ng H2SO4 c, sau qua bnh (2) ng Ba(OH)2 d thy khi lng cc bnh tng ln lt l: 16,2 gam v 30,8 gam. Cng thc phn t ca hai hirocacbon v % v th tch l A. C3H8: 50% v C4H10 : 50% C. C2H6: 50% v C3H8 : 50% B. CH4: 50% v C2H6: 50% D. C3H8: 40% v C4H10: 60%

Cu 4: t chy hon ton hn hp hai hirocacbon c khi lng phn t hn km nhau 28 vc thu c 4,48 lt CO2 (ktc) v 5,4 gam H2O. Cng thc phn t ca hai hirocacbon ln lt l A. C2H4 v C4H8 C. C2H6 v C4H10 B. CH4 v C3H8 D. C2H2 v C4H6

Cu 5: t chy hon ton 7,0 mg hp cht X thu c 11,2 ml kh CO2 (ktc) v 9,0 mg H2O. T khi hi ca X so vi nit bng 2,5. Khi clo ha X vi t l s mol 1 : 1 ch thu c mt dn xut monoclo duy nht. X c tn gi l A. metylxiclobutan C. 1,2-imetylxiclopropan B. xiclopentan D. xiclohexan

Cu 6: t chy hon ton 0,1 mol hn hp X hm propan v xiclopropan th thu c 0,35 mol H2O. Thnh phn % theo th tch propan trong hn hp X l A. 50% B. 40% C. 30% D. 25% Cu 7: t chy hon ton 6,72 lt (ktc) hn hp gm hia hirocacbon X v Y (MX > MY), th thu c 11,2 lt CO2 (ktc) v 10,8 gam H2O. Cng thc ca X l A. C2H6 B. C2H4 C. CH4 D. C2H2 (Trch TTS vo cc trng Cao ng, 2010)

10


11 Cu 8: Hn hp X gm hai olefin. t chy 7 th tch X cn 31 th tch O2 (ktc). Bit rng olefin cha nhiu cacbon hn chim 40 50 th tch ca X. Cng thc phn t hai olefin l A. C2H4 v C4H8 B. C2H4 v C3H6 C. C3H6 v C4H8 D. C2H4 v C5H10 Cu 9: t chy hon ton 0,15 mol hirocacbon X thu c 16,8 lt kh CO2 (ktc) v 13,5 gam H2O. Tng s ng phn cu to ca X l A. 9 B. 11 C. 10 D. 5 Cu 10: t chy hon ton 6,72 lt hn hp kh X (ktc) gm hai olefin k tip nhau trong dy ng ng, sau dn sn phm chy ln lt qua bnh (1) ng CaCl2 khan, d, bnh (2) ng dung dch KOH c, d. Kt thc th nghim thy khi lng bnh (2) tng nhiu hn khi lng bnh (1) l 29,25 gam. Cng thc phn t ca hai olefin v % khi lng ca cc cht trong hn hp ban u l A. C2H4: 25% v C3H6 : 75% C. C4H8: 67% v C5H10 : 33% B. C3H6: 20% v C4H8 : 80% D. C5H10: 35% v C6H12: 65%

Cu 11: t 8,96 lt hn hp X gm hai anken ng ng k tip ri dn ton b sn phm chy ln lt qua bnh (1) ng P2O5 d, bnh (2) ng dung dch KOH d. Kt thc th nghim thy khi lng bnh (1) tng m gam, bnh (2) tng (m + 39) gam. Thnh phn % th tch anken c s nguyn t cacbon ln hn trong hn hp X l A. 25% B. 40% C. 60% D. 75% Cu 12: Mt hn hp X gm hai hirocacbon k tip nhau trong cng dy ng ng. t chy 0,3 mol hn hp X v cho tt c sn phm chy hp th vo dung dch Ba(OH)2 d th khi lng bnh tng thm 46,5 gam v c 147,75 gam kt ta. Cng thc phn t ca hai hirocacbon l A. C2H2; C3H4 B. C3H6; C4H8 C. C2H4; C3H6 D. C2H6; C3H8 Cu 13: Ba hirocacbon X, Y, Z k tip nhau trong dy ng ng, trong khi lng phn t Z gp i khi lng phn t X. t chy 0,1 mol cht Y, sn phm kh hp th hon ton vo dung dch Ca(OH)2 d, thu c s gam kt ta l A. 20 B. 40 C. 30 D. 10 (Trch TTS vo cc trng i hc, Cao ng khi A, 2007) Cu 14: t chy hon ton m gam hn hp gm mt ankan v mt anken. Cho sn phm chy ln lt i qua bnh (1) ng P2O5 d v bnh (2) ng KOH rng, d, sau th nghim thy khi lng bnh (1) tng 4,14 gam bnh (2) tng 6,16 gam. S mol ankan c trong hn hp l A. 0,06 mol B. 0,09 mol C. 0,03 mol D. 0,045 mol

11


12 Cu 15: t chy hon ton 0,11 mol hn hp X gm CH4, C3H8 v C2 H4 thu c 0,17 mol CO2 v 0,25 mol H2O. S mol ca hn hp anken c trong X l A. 0,02 B. 0,09 C. 0,03 D. 0,08 Cu 16: Trn x mol hn hp X (gm C2H6, C3H8) v y mol hn hp Y (gm C3H6 v C4H8) thu c 0,35 mol hn hp X ri em t chy thu c hiu s mol H2O v CO2 l 0,2 mol. Gi tr ca x v y ln lt l A. 0,1 v 0,25 B. 0,15 v 0,2 C. 0,2 v 0,15 D. 0,25 v 0,1 Cu 17: t chy hon ton x gam hn hp X gm propan, but-2-en, axetilen thu c 47,96 gam CO2 v 21,42 gam H2O. Gi tr X l A. 15,46 B. 12,46 C. 11,52 D. 20,15 Cu 18: Dn hn hp X gm 0,05 mol C2H2; 0,1 mol C3H4 v 0,1 mol H2 qua ng cha Ni nung nng mt thi gian, thu c hn hp Y gm 7 cht. t chy hon ton Y ri cho sn phm chy hp th ht vo 700 ml dung dch NaOH 1M, thu c dung dch Z. Tng khi lng cht tan trong Z l A. 35,8 B. 45,6 C. 40,2 D. 38,2

Cu 19: t chy hon ton 2 lt hn hp gm axetilen v mt hirocacbon X, thu c 4 lt CO2 v 4 lt hi nc (cc th tch o cng nhit , p sut). Cng thc phn t v thnh phn % th tch ca X c trong hn hp l A. C2H6: 50% B. C4H8: 67% C. CH4: 50% D. C4H10: 25%

Cu 20: t chy hon ton 0,14 mol hn hp gm C3H6, C2H2, C3 H4 th thu c 8,288 lt kh CO2 (ktc) v 0,26 mol H2O. S mol anken c trong hn hp l A. 0,11 B. 0,12 C. 0,04 D. 0,04

Cu 21: t chy hon ton hn hp M gm mt ankan c nhnh X v mt ankin Y thu c kh CO2 v hi nc vi s mol bng nhau. T khi ca hn hp M so vi hiro l 21. Cng thc ca X v Y ln lt l A. C4H10, C2H2 B. C3H8, C3H4 C. C5H10. C2H2 D. C5H10. C3H4

Cu 22: t chy hon ton 11,2 lt gm C3H6 v C2H6 thu c s mol CO2 nhiu hn s mol nc l 0,4 mol. Phn trm (%) th tch ca mi kh trong hn hp u l

12


13 A. 50% v 50% B. 30% v 70% C. 70% v 30% D. 20% v 80%

Cu 23: t chy hon ton 2,24 lt hn hp X(ktc) gm C3H8, C3H6, C3 H4 (X c t khi so vi H2 bng 21), ri dn ton b sn phm chy vo bnh ng dung dch nc vi trong th tng khi lng ca bnh l A. 4,2 gam B. 5,4 gam C. 13,2 gam D. 18,6 gam

Cu 24: t chy hon ton hn hp gm mt anken v mt ankin ri cho sn phm chy qua bnh (1) ng H2SO4 c v bnh (2) ng NaOH rn, d, sau th nghim thy khi lng bnh (1) tng thm 3,6 gam v bnh (2) tng 15,84 gam. S mol ankin c trong hn hp l A. 0,15 B. 0,16 C. 0,17 D. 0,18

Cu 25: t chy hon ton hn hp X gm hai hirocacbon mch h thu c 16,8 lt kh CO2 (ktc) v 8,1 gam H2O. Hai hirocacbon trong hn hp X thuc dy ng ng no di y? A. Ankaien B. Ankin C. aren D. ankan

Cu 26: t chy hon ton V lt (ktc) mt ankin th kh thu c CO2 v H2O c tng khi lng l 25,2 gam. Nu cho sn phm chy i qua dung dch Ca(OH)2 d th thu c 45 gam kt ta. Gi tr ca V l A. 6,72 B. 2,24 C. 4,48 D. 3,36

Cu 27: Chia hn hp hai ankin thnh hai phn bng nhau - Phn 1: em t chy hon ton thu c 1,76 gam CO2 v 0,54 gam H2O. - Phn 2: dn qua dung dch Br2 d. Khi lng Br2 phn ng l A. 2,8 gam B. 3,2 gam C. 6,4 gam D. 1,4 gam

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14 Cu 28: Crackinh 11,6 gam C4 H10 thu c hn hp kh X gm 7 cht: C4 H8, C3 H6, C2H6, C2H4, CH4, H2, C4 H10 d. t chy hon ton X cn ti thiu bao nhiu th tch khng kh ktc? (Bit oxi chim 20% th tch khng kh) A. 34,944 lt B. 145,60 lt C. 29,12 lt D. 174,72 lt

Cu 29: ung nng 11,6 gam butan mt thi gian, thu c hn hp H2, CH4, C2H6, C4H8, C3H6, C2H4, C4H10. Gi s ch c cc phn ng C4H10 C4H10 C4H10 H2 + C4H8 CH4 + C3H6 C 2H6 + C 2H4 (1) (2) (3)

t chy hon ton X ri cho sn phm chy hp th ht vo dung dch Ca(OH)2 d, thy khi lng bnh tng thm m gam. Gi tr ca m l A. 35.2 B. 53,2 C. 37,4 D. 60,2

Cu 30: t chy hon ton 0,1 mol hn hp X gm CH4, C2 H4 v C2 H6, sn phm thu c dn qua bnh (1) ng dung dch H2SO4 c v bnh (2) ng dung dch Ca(OH)2 d. Sau th nghim, bnh (2) thu c 15 gam kt ta v khi lng bnh (2) tng nhiu hn khi lng bnh (1) 2,25 gam. Thnh phn % v th tch CH4, C2 H4 v C2 H6 trong hn hp X tng ng l A. 50%, 30%, 20% C. 50%, 25%, 25% B. 30%, 40%, 30% D. 50%, 15%, 35% p n bi tp t luyn 1B 11D 21A 2D 12C 22D 3A 13C 23D 4B 14B 24B 5B 15C 25B 6A 16C 26D 7C 17A 27B 8A 18C 28B 9C 19A 29B 10B 20D 30C

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pp giẢi toÁn ĐỐt chÁy h.c

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