Tap Trac Nghiem Ve Hidrocacbon Khong No

8/6/2019 Tap Trac Nghiem Ve Hidrocacbon Khong No

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HOI PHONG INTRODUCTION

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TP TRC NGHIM V HIROCACBON KHNG NO1.Cho 3,15 gam hn hp hai anken k tip nhau trong dy ng ng phn ng va vi 100 ml dung dch brom 0,60M.Cht kh c o iu kin tiu chun. Cng thc ca hai anken v th tch ca chng l: A. C2H4; 0,336 lt v C3H6; 1,008 lt B. C3H6; 0,336 lt v C4H8; 1,008 lt

C. C2H4; 1,008 lt v C3H6; 0,336 lt D. C4H8; 0,336 lt v C5H10; 1,008 lt

2.Chn tn ng nht trong s cc tn gi cho di y ca cht c cng thc: CH3-CH(CH3)-CH(CH3)-CH=CH-CH3

A. 4,5-imetylhex-2-en B. 45-imetylhex-2-enC. 4,5-imetylhexen-2 D.4,5-imetyl hex-2-enCh : V cch gi tn, trc y ta vn gi tn cc cht theo danh php Quc t - na Vit Nam nhng theo chng trnhci cch, tn gi cc cht theo danh php Quc t c quy nh rt cht ch nn khi vit tn cc cht cc em phi tun thiu ny.3.Axetilen c iu ch t cht no sau y? A. CH4 (1) B. (1) v (2) C. CaC2 (2) D. Al4C3

4.Axit axetic tc dng vi axetilen cho sn phm no di y? A. CHCOOCCH3 B. CH3COOCH2-CH3 C. CH3-O-CO-CH=CH2 D. CH3COOCH=CH2

Phng trnh: CH3COOH + CH CH CH3COOCH=CH2. CH3COOH cng l mt tc nhn cng dng H-A c u H-d in tch dng.5.Phn ng in hnh ca ankaien l loi phn ng no sau y?A. Phn ng th B. Phn ng hu

C. Phn ng cng v phn ng trng hp D. Phn ng oxi ho6.t chy hon ton 4 gam hi ca mt hirocacbon mch h cn va 12,8 gam oxi thy th tch CO 2sinh ra bng 3ln th tch hirocacbon. Gi s phn ng c tin hnh trong bnh kn dung tch 1 lt. Sau phn ng a bnh v 27,3oC,p sut trong bnh sau phn ng l:A. 7,392 atm B. 12,320 atm C. 7,239 atm D. 12,230 atm

Bi gii: Cng thc ca hirocacbon l C3Hy(Do th tch CO2sinh ra bng 3 ln th tch hirocacbon).C3Hy + (3 + y/4)O2 3CO2 + y/2 H2OT y tpa c h thc: 12,8(36 + y) = 4(3 + y/4).32 y = 4. Khi a bnh v 27,3oC th hi nc ngng t, sau phn ngch cn 0,3 mol CO2. Vy p sut trong bnh sau phn ng l:

.392,71

)3,27273.(273

4,22.3,0

atmV

nRTp

7.Hn hp A c th tch 896 cm3

cha mt ankan, mt anken v hiro. Cho A qua xc tc Ni nung nng phn ng xyra hon ton c hn hp B c th tch 784 cm3. Cho B qua bnh ng dung dch brom d thy dung dch brom b nhtmu mt phn v khi lng ca n tng 0,28 gam. Kh cn li c th tch 560 cm3v c t khi hi so vi hiro l 9,4.Cht kh c o iu kin tiu chun. Cng thc ca hai hirocacbon l A. C2H6 v C2H4 B. C3H8 v C3H6 C. CH4 v C2H4 D. C4H10 v C4H8

8.Dn hn hp M gm hai cht X v Y c cng thc phn t C3H6, C4H8vo dung dch brom trong dung mi CCl4thydung dch brom b nht mu v khng c kh thot ra. Ta c cc kt lun sau:

a. X v Y l 2 xicloankan vng 3 cnh.b. X v Y l mt anken v mt xicloankan vng 4 cnhc. X v Y l 2 anken ng ng ca nhau.d. X v Y l mt anken v mt xicloankan vng 3 cnhe. X v Y l mt xicloankan vng 3 cnh v mt xicloankan vng 4 cnh f. X v Y khng phi l ng ng ca nhau

Cc cu ng l A, B, C hay D?A. a, b, c, d B. a, b, d C. a, b, c, d, e D. a, c, dCh : Xicloankan vng 3 hoc 4 cnh km bn. Cc Xicloankan vng 5 cnh tr ln bn. + Xicloankan vng 3 cnh c kh nng cng hp H2v lm mt mu dung dch Br2 (cng m vng).+ Xicloankan vng 4 cnh ch c kh nng cng hp H2, c xc tc Ni, nhit .Vng 3 cnh hoc 4 cnh khng nht thit phi c 3 hoc 4 nguyn t cacbon trong phn t.

CnH2n+1

CmH2m+1

CpH2p+1

CmH2m+1

CnH2n+1

CpH2p+1

CqH2q+1 9.Hn hp A gm mt ankan v mt anken. t chyhon ton A cn 0,3675 mol oxi. Sn phm chy cho qua bnh ng

dung dch Ca(OH)2d thy sinh ra 23 gam kt ta. Bit s nguyn t cacbon trong ankan gp 2 ln s nguyn t cacbontrong anken v s mol ankan nhiu hn s mol anken. Cng thc ca hai hirocacbon lA. C3H6 v C6H14 B. C3H6 v C3H8 C. C2H4 v C3H8 D. C2H4 v C4H10


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10.Cng thc cu to ca 2,5-imetylhex-3-in l:A. CH3-CH(CH3)-C C-CH(CH3)-CH3 B. CH C-CH(CH3)-CH2-CH3

C. CH C-CH2-CH2-CH3 D. CH3-C C-CH(CH3)-CH(CH3)-CH3

11.Mt hn hp Z gm anken A v H2. T khi hi ca hn hp Z so vi hiro l 10. Dn hn hp qua bt Ni nung nngti phn ng hon ton thu c hn hp kh B c t khi so vi hiro l 15. Thnh phn % theo th tch ca A trong hnhp Z v cng thc phn t ca A l:A. 66,67% v C5H10 B. 33,33% v C5H10 C. 66,67% v C4H8 D. 33,33% v C4H8

Bi gii: Hn hp Z gm anken CnH2n x mol v y mol H2.Do thu c hn hp kh B nn trong B cn d H2.T y ta c cc h thc:

30214

20214

y

ynx

yx

ynx

30y = 20(x + y) y = 2x. Thay vo mt trong hai biu thc ta c:

.4202

2.214n

xx

xnx

Thnh phn % th tch ca A trong hn hp Z l: %.33,33100.2

%xx

xV

A

12.Khi cho axetilen hp nc c xc tc HgSO4/H2SO4 80oC thu c sn phm no sau y?

A. CH3COOH B. CH3CHO C. C2H5OH D. HCHO13.Anken Z th lng. Ho hi 1,4 gam Z trong bnh kn dung tch 0,5 lt 273oC. Sau khi ho hi ht p sut bnh oc 1,792 atm. Cng thc phn t ca anken Z lA. C3H6 B. C2H4 C. C4H8 D. C5H10

Bi gii: S mol ca anken Z l: 02,0)273273(

273

4,22

5,0.792,1

RT

PVn .

Khi lng mol (phn t khi ca anken Z) l: 1,4/0,02 = 70. Vy CTPT ca Z l C5H10.Ch : Ngoi ra ta da vo d kin Z th lng ( k thng) ta cng c th d on Z l C5H10 m khng cn tnh ton.

Nhng hirocacbon c 4 nguyn t cacbon tr xung th th kh iu kin thng, nhng HC c t 5 n khong 18nguyn t C k thng l cht lng, cn li l cht rn.14.Khi iu ch etilen trong phng th nghim t ancol etylic vi xc tc axit sunfuric c nhit trn 170 oC th kh

etilen thu c thng c ln cc oxit nh CO2 v SO2. lm sch etilen phi dng ho cht no di y? A. dung dch natri cacbonat B. dung dch bromC. dung dch NaOH D. dung dch kali pemanganat longCh : NaOH c tnh kim mnh nn hp th rt tt nhng oxit axit nh CO2 v SO2. bi ny khng phi l phn bit CO2 v SO2 nn ta khng dung Br2 v KMnO4.15.Cht no sau y lm mt mu dung dch nc brom?A. but-1-en (2) B. 2-metylpropen (3) C. C (1), (2) v (3) D. but-2-en16.Cho bng d liu sau:

I. Tn cht II. Cng thc cu to1 Hexan a CH2=CH-CH=CH22 But-2-en b CH3(CH2)4CH3

3 But-1-in c CH3-CH=CH-CH3

4 Buta-1,3-ien d CH C-CH2-CH3

5 XiclohexanKhi ghp tn cc hp cht hu c ct (I) vi cc cng thc cu to ph hp ct (II) bn hc sinh a ra cc kt qudi y. Hi kt qu no chnh xc?A. 1-b, 2-c, 3-d, 4-a B. 1-b, 2-d, 3-c, 4-a C. 1-b, 2-a, 3-d, 4-c D. 5-b, 2-c, 3-d, 4-a

Ch : Xiclohexan c cng thc dng vng no 6 cnh (CH2)6.17.iu kin anken c ng phn hnh hc l:A. mi nguyn t cacbon lin kt i lin kt vi hai nguyn t hoc nhm nguyn t khc nhau B. mi nguyn t cacbon lin kt i lin kt vi hai nguyn t hoc nhm nguyn t ging nhauC. mi nguyn t cacbon lin kt i lin kt vi hai nguyn t hoc nhm nguyn t bt kD. bn nguyn t hoc nhm nguyn t lin kt vi hai nguyn t cacbon lin kt i phi khc nhau. Ch : V ng phn hnh hc cis trans.

C = C

A

B

C

D


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iu kin cht trn c ng phn hnh hc l: A B v C D. (hay ni cch khc, mi nguyn t C lin kt i linkt vi hai nguyn t hoc nhm nguyn t khc nhau).

Nu A > B v C > D v nguyn (phn) t khi th ta c ng phn dng cis-.Nu A > B v C < D v nguyn (phn) t khi th ta c ng phn dng trans-.Cch phn bit ng phn cis-trans trn ch phhp vi ho ph thng (n s khng cn ng vi mt s cht). phn

bit ng phn cis-trans chnh xc th ta phi dng n h danh php Z-E.ng phn trans- th bn hn ng phn cis-, chnh v iu ny m nhng cht c trong thin nhin u dng trans-, inhnh l cao su thin nhin: (-CH2-C(CH3)=CH-CH2-)n.

18.Hn hp X gm hai anken k tip nhau trong dy ng ng. Trn mt th tch hn hp X vi mt lng va khoxi c mt hn hp Y ri em t chy hon ton th thu c sn phm khv hi Z. T khi ca Y so vi Z l744:713. (Cc th tch kh o cng iu kin nhit v p sut). Cng thc phn t ca 2 anken l: A. C5H10 v C6H12 B. C3H6 v C2H4 C. C4H8 v C5H10 D. C3H6 v C4H8

Bi gii: t cng thc chung cho X l CnH2n. CnH2n + 3n/2O2 nCO2 + nH2O.

Ta c h thc sau: 4,2713

744

2/31n

n

nn. Vy hai anken l C2H4 v C3H6.

19.C hn hp X gm 3 hirocacbon A, B, C. Khi t chy hon ton ln lt A, B, C th trong c 3 trng hp th tchCO2thu c u bng 2 ln th tch ca mi hirocacbon cng iu kin. A, B, C c th l: A. l ng phn ca nhau B. l ng ng ca nhauC. l ng khi ca nhau D. c cng s nguyn t cacbon.

20. Cho 2,6 gam C2H2hp th ht trong 100 ml dung dch brom 1,8M thy dung dch brom b mt mu hon ton. Cc snphm thu c sau phn ng gm:

A. CHBr=CHBr v CHBr2-CHBr2 B. CHBr=CHBrC. CHBr2-CHBr2 D. CHBr=CHBr hoc CHBr2-CHBr2Bi gii: s mol C2H2l 0,1. S mol Br2l 0,18. phn ng ht vi C2H2ta cn 0,2 mol Br2.Nh vy cc sn phm thuc gm CHBr=CHBr v CHBr2-CHBr2.21.C mt hn hp gm 11 gam ankan A v 20 gam ankin B c th tch 16,8 lt. Bit rng chng c cng s nguyn tcacbon v A c s nguyn t hiro nhiu hn. Cht kh c o iu kin tiu chun. Cng thc phn t ca A v B l: A. C2H6 v C2H2 B. C4H10 v C4H6 C. C5H12 v C5H8 D. C3H8 v C3H4

Bi gii: Cng thc ca A v B l CnH2n+2 v CnH2n-2.

T y ta c h thc:4,22

8,16

214

20

214

11

nnn = 3.

Vy cng thc ca A v B l C3H8 v C3H4.

22.Trong chui phn ng: butilen X Y Z T axetilen. Cc cht X, Y, Z, T trong chui phnng trn ln lt c tn gi:A. but-2-en, butan, propen, metan B. butan, etan, cloetan, icloetanC. butan, but-2-en, propen, metan D. butan, propan, etan, metan

Ch : T butan khng c cch no iu ch trc tip ra propan, t propan cng khng c cch no iu ch trc tip ra etan, 23.Trong cc ng phn cu to dng anken ca C4H8, cht c ng phn hnh hc l:A. but-2-en B. but-1-en v but-2-en C. 2-metylpropen D. but-1-en

24.C bn bnh ng kh: CH4, C2H2, C2H4 v CO2. Dng cc no trong cc cch sau y c th nhn ra 4 kh trn (tinhnh theo ng trnh t):A. t chy, dng nc vi trong d B. Dng nc vi trong d, dng dung dch bromC. Dng dung dch brom D. Dng qu tm m, t chy, dng nc vi trong d 25.Mt hn hp gm 3 hirocacbon c s mol nh nhau. Tng khi lng phn t ca 3 hirocacbon ny l 70. Hai trong

ba hirocacbon ca hn hp l:A. C4H4 v C2H4 B. CH4 v C2H2 C. C3H4 v CH4 D. C2H6 v C2H4

Bi gii: Do ba hirocacbon c s mol nh nhau nn phn t khi trung bnh ca hn hp l: 70/3 = 23,33. Vy phi c 1HC c phn t khi nh hn 23,33, l CH4.

Phn t khi trung bnh ca hai HC cn li l: .272

1670Vy mt HC trong hn hp phi l C2H2.

Hai trong ba HC ca hn hp l CH4 v C2H2. 26.Etilen d tham gia phn ng cng v l do no sau y?A. Etilen l cht c nm lin kt trong phn tB. Etilen c phn t khi bC. Phn t etilen c mt lin kt i (gm mt lin kt v mt lin kt )D. Etilen l cht kh khng bn

27.Hn hp X gm 1 ankin A v 1 anken B, trong s nguyn t hiro trong A bng s nguyn t cacbon trong B. Hnhp X lm mt mu va dung dch cha 0,08 mol brom. Mt khc, khi cho hn hp X phn ng ht vi hiro th thuc hn hp Y gm 2 ankan. t chy hon ton hn hp Y thu c 9,68 gam CO 2 v 5,04 gam H2O. Cht kh c o iu kin tiu chun. Cng thc cu to A, B v th tch ca chng l:


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A. C2H2; 0,448 lt v C3H6; 0,668 lt B. C3H4; 0,896 lt v C4H8; 0,448 lt


Bi gii: Cng thc ca A v B l CnH2n-2 v C2n-2H4n-4.t x, y ln lt l s mol A v B. Ta c h phng trnh2x + y = 0,08

06,044

68,9

18

04,5yx

Gii ra ta c x = 0,02 v y = 0,04.S mol CO2 thu c l 0,02.n + 0,04.(2n-2) = 0,22 n = 3.Vy cng thc A, B v th tch ca chng l: C. C3H4; 0,448 lt v C4H8; 0,896 lt28.C cc cu sau ni v ankin:1. Ankin l phn cn li sau khi ly i 1 nguyn t hiro t phn t ankan2. Ankin l hirocacbon mch h c cng thc phn t CnH2n-2 ( 2n )3. Ankin l hirocacbon khng no c mt lin kt ba C C4. Ankin l hirocacbon mch h c mt lin kt ba C C5. Ankin l nhng hp cht c cng thc chung l R1-C C-R2vi R1, R2l nguyn t hiro hoc gc hirocacbon no,mch h.Cc cu ng l:A. 4 v 5 B. 4 v 1 C. 4 v 2 D. 4 v 3

Ch : Cc cu sai sa li l

1. Ankin l phn cn li sau khi ly i hainguyn t H t phn t ankan.2. Ankin l hirocacbon mch h c mt lin kt ba trong phn t v c cng thc phn t l CnH2n -2 ( 2n ).3. Nh cu 2.4. v 5. ng.29. lm mt mu 200 gam dung dch brom nng 20% cn dng 10,5 gam anken X. Cng thc phn t ca X l: A. C4H8 B. C5H10 C. C2H4 D. C3H6

Bi gii: S mol anken cn dng l 200*20/(100*160) = 0,25. Phn t khi ca anken l: 10,5/0,25 = 42. Vy X l C3H6.

30.X v Y l hai hirocacbon c cng cng thc phn t l C5H8. X l mt monome dng trng hp thnh cao suisoprene; Y c mch cacbon phn nhnh v to kt ta khi cho phn ng vi dung dch AgNO 3/NH3. Cng thc cu to caX v Y ln lt l:A. CH2=C(CH3)-CH=CH2 v CH3-CH(CH3)-C CH

B. CH3-CH=CH-CH=CH2 v CH3-CH(CH3)-C CHC. CH2=C(CH3)-CH=CH2 v CH2(CH3)-CH2-C CH

D. CH3-CH=CH-CH=CH2 v CH2(CH3)-CH2-C CH

31.Cho cc d kin lin quan n mt s ankaien nh sau:1. T khi hi ca ankaien A so vi amoniac l 4.2. Trn ln mt ankaien B th kh vi etan theo t l th tch 1: 2 c hn hp kh c t khi so vi hiro bng 19.3. Trong phn t ankaien D c 6 lin kt .4. Ankaien E c tn gi: 2,3-imetylbuta-1,3-ien.A, B, D, E c cng thc phn t ln lt l:A. C5H8, C3H4, C4H6, C6H10 B. C5H8, C3H4, C6H10, C4H6

C. C5H8, C6H10, C4H6, C3H4 D. C5H8, C4H6, C3H4, C6H10

Bi gii:1. Cng thc ankaien l CnH2n-2.Nh vy 14n 2 = 4*17 = 5. Cng thc ca A l: C5H8.

2. Ta c: 2*1921

2.30M M = 54n = 4. Cng thc ca B l: C4H6.

n y ta c th kt lun c p n ng D.3. to ra mt lin kt cng ho tr cn 2 electron. n nguyn t C c 4n electron, 2n -2 nguyn t H c 2n-2 electron.Tng s electron trong phn t ankaien l 6n-2 to ra 3n-1 lin kt cng ho tr, trong c 3n-3 lin kt xch ma (trong c 2 lin kt PI).Vy 3n 3 = 6n = 3. Cng thc ca D l : C3H4.4. E c cng thc: CH2=C(CH3)-C(CH3)=CH2. Cng thc phn t ca E l C6H10. 32.Cao su buna l sn phm ca phn ng trng hp monome no di y?A. Isopren B. Vinyl clorua C. ivinyl D. EtilenCh : vinyl c cng thc l CH2=CH-, ivinyl tc l CH2=CH-CH=CH2 (tc buta-1,3-ien) dng sn xut cao su buna.Isopren c cng thc: CH2=C(CH3)-CH=CH2 l monomer ca cao su thin nhin. 33.t chy hon ton 0,014 mol hn hp 2 ankin k tip nhau trong dy ng ng. Sn phm chy cho hp th ht v odung dch cha 0,03 mol Ca(OH)2thy to ra 2 gam kt ta trng. Cng thc phn t ca 2 ankin v th tch ca chng(ktc) lA. C2H2; 0,2688 lt v C3H4; 0,0448 lt B. C2H2; 0,0448 lt v C3H4; 0,2688 lt


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34.C 0,896 lt hn hp A gm 2 hirocacbon mch h lm mt mu va 100 ml dung dch brom 0,5M. Sau phn ngthy cn 0,336 lt kh khng b hp th. T khi hi ca A so vi hiro l 19. Cht kh c o iu kin tiu chun. Cngthc phn t ca hai hirocacbon l:A. C2H2 v C4H10 B. C2H2 v C3H8 C. C3H4 v C4H10 D. C3H4 v C3H8

35.ng vi cng thc phn t C5H8c bao nhiu ng phn dng ankin? A. 2 B. 3 C. 4 D. 5

Ch : bi hi ng phn dng ankin. Ta c ba ng phn sau:C C-C-C-C, C C-C(C)2, C-C C-C-C.

36.Hn hp A (gm 2 anken k tip nhau trong dy ng ng v mt ankan) c t khi hi so vi hiro l 14,25. Cho1,792 lt hn hp A qua dung dch brom d thy c 0,448 lt kh khng b brom hp th. Sau phn ng khi lng bnhng dung dch brom tng 1,96 gam. Cht kh c o iu kin tiu chun. Cng thc phn t ca hirocacbon v thnh

phn % theo th tch ca ankan trong hn hp A lA. C4H8, C3H6 v CH4; 25% B. C4H8, C3H6 v CH4; 75%

C. C2H4, C3H6 v CH4; 25% D. C2H4, C3H6 v CH4; 75%

37.Cho cc cht sau: metan, etilen, but-2-in v axetilen. Khi ni v kh nng phn ng ca cc cht ny th nhn nh nosau y lng?A. khng c cht no lm nht mu dung dch KMnO4B. c ba cht c kh nng lm mt mu dung dch brom.C. c hai cht to kt ta vi dung dch AgNO3 trong NH3.D. c bn cht u c kh nng lm mt mu dung dch brom.Ch : Metan l ankan nn khng c kh nng phn ng vi dung dch Br2 v KMnO4.Cc ankin c ni ba u mch cho phn ng th vi AgNO3 trong dung dch NH3 to kt ta vng. 38.Cho s phn ng sau:

CaC2 A

B

E

D

polietilen

cao su buna

Cc cht A, B, D, E ln lt c cng thc cu to l: A. CH CH, CH2=CH-CH=CH2, CH C-CH=CH2, CH2=CH2

B. CH2=CH2, CH C-CH=CH2, CH2=CH-CH=CH2, CH CH

C. CH2=CH2, CH2=CH-CH=CH2, CH C-CH=CH2, CH CH

D. CH CH, CH C-CH=CH2, CH2=CH-CH=CH2, CH2=CH2

Ch : t axetilen ta tin hnh nh hp (ime) to thnh vinylaxetilen trong iu kin xc tc CuCl, NH4Cl v nhit . 2CH CH CH C-CH=CH2.

CaC2 phn ng vi H2O (hoc axit) to thnh C2H2. Al4C3 phn ng vi H2O (hoc axit) to thnh CH4.Mg2C3 phn ng vi H2O (hoc axit) to thnh C3H4. 39.Hn hp A gm mt ankan v mt ankin. t chy hon ton hn hp ny c 12,6 gam H2O, khi lng oxi cn

dng cho phn ng ny l 36,8 gam v th tch CO2sinh ra bng3

8th tch hn hp A. Ly 5,5 gam A cho qua dung dch

AgNO3 trong NH3(dng d) th khi lng kt ta thu c nh hn 15 gam. Cng thc phn t ng ca 2 hirocacbonl:

A. C4H10 v C3H6 B. C2H6 v C3H4 C. C2H6 v C2H2 D. C4H10 v C2H2

Bi gii: p dng nh lut bo ton mol nguyn t oxi, ta c th tnh c s mol CO2.

8,02

18/6,1216/8,36S mol A l 0,3.

t cng thc ca hai cht l CnH2n+2 (x mol) v CmH2m-2 (y mol). Ta c:nx + my = 0,8

(n+1)x + (m-1)y = 0,7

x + y = 0,3.

T y ta c th tnh c x = 0,1 v y = 0,2. Nh vy n + 2m = 8. Ta c cc trng hp sau: (n phi chn).TH1: n = 2 m = 3

TH2: n = 4 m = 2

Da vo d kin cui l khi lng kt ta ta s loi c TH2. CH C-CH3 + AgNO3 + NH3 AgC C-CH3 + NH4NO3

Khi lng kt ta thu c l: 0,1.147 = 14,7 < 15 (Trong 5,5 gam A c 0,1 mol ankin v 0,05 mol ankan).HC CH + 2AgNO3 + 2NH3 AgC CAg + 2NH4NO3

Khi lng kt ta l 0,1.240 = 24 > 15 (loi).

Vy cng thc phn t ng ca hai HC l C2H6 v C3H4.

40. Cho cc cu sau:


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1. Ankaien l nhng hirocacbon khng no, mch h c hai lin kt i trong phn t.2. Nhng hirocacbon khng no c hai lin kt i trong phn t l ankaien.3. Nhng hirocacbon c kh nng cng hp vi hai phn t hiro thuc loi ankaien.4. Ankaien l nhng hirocacbon c cng thc chung l CnH2n-2 ( 3n ).S cu ng l:A. 3 B. 2 C. 4 D. 1Ch : Ch c 1 ng, cc khc sai.2. Cha y , c th c mch vng.3. Ankin cng c kh nng cng hp vi hai nguyn t H. 4. Cng thc CnH2n-2 (n>=3) c th l ankin41.Mt bnh kn dung tch 2 lt 27,3oC cha 0,03 mol C2H2; 0,015 mol C2H4 v 0,04 mol H2. Trong bnh trn c snmt t bt Ni (th tch khng ng k), nung bnh nhit cao phn ng xy ra hon ton, sau a bnh v nhit

ban u c hn hp kh A gm 3 hirocacbon c p sut p2. p2nhn kt quA. 0,6 atm B. 1,6 atm C. 1,2 atm D. 1,0 atm

42.Cho butaien tc dng vi hiro c kim loi Ni lm xc tc c th thu c: A. isobutilen B. butilen C. butan D. isobutan

43.Chn nh ngha ng v anken.A. Anken l nhng hirocacbon ng vi cng thc CnH2n ( 2n ).B. Anken l nhng hirocacbon khng no l phn t cha mt lin kt i C=C.C. Anken l nhng hirocacbon m phn tcha mt lin kt i C=C.

D. Anken l nhng hirocacbon khng no c cng thc CnH2n ( 2n ).Ch :A. ng vi cng thc phn t CnH2n cn c xicloankan.B. C th c mch vng khng no, ch cha 1 ni i trong phn t.C. Nh B.D. ng vi cng thc phn t CnH2n c th l anken (khng no) hoc xicloankan (no).

44.Trong s cc ng ng ca etilen th cht no c thnh phn % nguyn t cacbon l 85,71%? A. C2H4 B. C3H6 C. Tt c cc anken D. C6H12Ch : Tt c cc anken u c cng thc chung (CH2)n nn thnh phn % v khi lng cc nguyn t l ging nhau. y mt s bn th trng hp A thy ng chn lun p n A, tt nhin l vn ng nhng p n ng nht l p nC. Khi chn p n cc em th trng hp th cng nn xem cc trng hp cn li trnh nhm ln. 45.Hn hp X gm mt ankan v mt anken. Cho X tc dng vi 3,136 lt hiro ti phn ng hon ton thu c hn hp

Y gm 2 kh trong c hiro d v mt hirocacbon. t chy hon ton hn hp Y ri dn hn hp kh v hi sinh ravo bnh ng dung dch Ca(OH)2d thy khi lng bnh tng 13,52 gam ng thi c 16 gam kt ta c to thnh.Cht kh c o iu kin tiu chun. Cng thc ca hai hirocacbon l: A. C3H8 v C3H6 B. C5H12 v C5H10 C. C2H6 v C2H4 D. C4H10 v C4H8

Bi gii: S mol hiro l 0,14. Cng thc ca ankan v anken l CnH2n+2 v CnH2n ( n >=2) Hirocacbon trong Y l CnH2n+2 x mol v y mol H2. Ta c:Khi t chy hn hp Y ta thu c: nx mol CO2 v (nx + x + y) mol H2O. T y ta c:nx = 16/100 = 0,16

Khi lng bnh tng l tng khi lng ca CO2 v H2O.44nx + 18(nx + x + y) = 13,52

x + y = 0,2

S mol ankan ban u l: 0,2 0,14 = 0,06.0,6 < x < 0,2. Vy 67,206,0

16,0

2,0

16,08,0 n

Vy n = 2. Cng thc hai hirocacbon l C2H6 v C2H4.

46.C 3 hirocacbon A, B, D c cng s nguyn t cacbon, trong t l mol nguyn t hiro v cacbon ln lt l1:1,2:1, 3:1. A, B ln lt c cng thc phn t:A. C4H4, C4H8 B. C3H4, C3H6 C. C2H2, C2H4 D. C6H6, C6H12

47.C hn hp X gm 3 hirocacbon A, B, C. Khi t chy ln lt A, B, C th trong c ba trng hp th tch CO 2 thuc u bng hai ln th tch ca mi hirocacbon cng iu kin. Trong hn hp X, nu t chy hon ton A v C th

s mol CO2 v H2O sinh ra bngnhau, cn nu t chy hon ton A v B th t l s mol H2O v CO2thu c l14

17. A,

B, C v thnh phn % mi cht trong hn hp X l:A. C2H6, 30% - C2H4, 30% - C2H2, 40% B. C3H8, 30% - C3H6, 30% - C3H4 - 40%

C. C2H6, 30% - C2H4, 40% - C2H2, 30% D. C2H6, 40% - C2H4, 30% - C2H2, 20%

Bi gii: Khi t chy ln lt A, B, C th trong c ba trng hp, th tch CO2 thu c u bng hai ln th tch ca miHC cng iu kinA, B, C u c 2 nguyn t C trong phn t. Cng thc phn t A, B, C c th l C2H2, C2H4,C2H6.Khi t chy A v C th s mol CO2 v H2O sinh ra bng nhau, vy A, C c th l C2H6, C2H2 v c s mol bng nhau.


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Convert by TVDT

Khi t chy A v B th thu c CO2 v H2O vi H2O c s mol ln hn s mol CO2, nh vy trong A, B phi c 1 chtl ankan, Vy A l C2H6, B l C2H4 v C l C2H2. T chn lng cht, s mol ankan A l 17 14 = 3 = s mol ankin C. S mol anken l 14/2 3 = 4.Vy p n C ng.48.t chy hon ton 1 hirocacbon A thy s mol CO2sinh ra bng 2 ln s mol H2O. Bit A l mt ankin, cng thcca A l:A. C3H4 B. C2H2 C. C5H8 D. C4H6

49.Khi cho isopren tc dng vi HCl (t l mol 1:1) to ra sn phm chnh c cng thc cu to l: A. CH2Cl-CH(CH3)-CH=CH2 B. CH2=C(CH3)-CH2-CH2Cl

C. CH3-CH(CH3)-CCl=CH2 D. CH3-CCl(CH3)-CH=CH2

Ch : Phn ng cng vo ni i hoc ni ba vi tc nhn dng H A tun theo quy tc cng Maccopnhicop, viankaien i xng ta c cc sn phm cng -1,2 v -1,4; sn phm chnh trong trng hp ny ph thuc vo nhit .Vi cc ankaien bt i xng nh Isopren ta c cc sn phm cng -1,2; -1,4 v -3,4 v tt yu snphm chnh trongtrng hp ny cng ph thuc vo nhit . Nhng vi cc p n trn th sn phm chnh y l sn phm p n D.

50.t chy hon ton 1 hirocacbon thu c CO2 v H2O, trong th tch CO2bng 2 ln th tch hirocacbon cngiu kin. Trong phn t ca hirocacbon ny nht thit phi c

a. 2 nguyn t C b. 2 nguyn t hiroc. 4 nguyn t hiro d. 6 nguyn t hiro

Nhn nh ng l:A. a, c v d B. a, b, c v d C. a v b D. a

Ch : cu ny, bi ch ni n th tch CO2 bng 2 ln th tch HC cng iu kin nn ta ch suy ra c HC trnc 2 nguyn t cacbon trong phn t p n D ng.51. C 2,24 lt hn hp A gm hai anken k tip nhau trong dy ng ng v hiro. t chy ht A cn 6,944 lt oxi. Sn

phm chy cho qua bnh (1) ng P 2O5thy khi lng bnh (1) tng 3,96 gam. Cht kh c o iu kin tiu chun.Cng thc cu to hai anken v% th tch ca hiro trong hn hp A lA. C3H6, C4H8 v 80% B. C2H4, C3H6 v 80%

C. C2H4, C3H6 v 20% D. C3H6, C4H8 v 20%

52.Vinylaxetilen c to ra t hp cht v trong iu kin no sau y? A. Trng hp axetilen 100oC c xc tc CuCl, NH4ClB. Trng hp axetilen 600oC c bt thanC. Trng hp butaien vi xc tc Na kim loiD. Trng hp isopreneCh : Vinylaxetilen c cng thc l CH C-CH=CH2 c sinh ra khi tin hnh nh hp axetilen iu kin p n A. 53. Polietilen v polietilen-propilen c to ra t phn ng no di y?A. Phn ng tch nc ca ancol.B. Phn ng cng vi hiro.C. Phn ng trng hp etilen v phn ng ng trng hp etilen -propilen.D. Phn ng cng vi HCl.54.Mt bnh kn dung tch 2 lt cha 0,03 mol C2H2; 0,015 mol C2H4; 0,04 mol H2v mt t bt Pd (c th tch khng ngk). Nung nng bnh n phn ng hon ton ri a v nhit 27,3oC th c p sut p1. p1 nhn gi tr A. 0,70 B. 1,20 C. 0,68 D. 1,00

Bi gii: Khi thchin phn ng cng H2 vi xc tc Pd, ta ch thu c sn phm c cha cc ni i trong phn t Ch c C2H2 phn ng, sau phn ng c 0,45 mol C2H4 v 0,1 mol H2.

p sut trong bnh sau phn ng l: 776,62

)3,27273.(273/4,22.55,01

V

nRTp

55. Anken C5H10 c bao nhiu ng phn cu to?A. 3 ng phn B. 4 ng phn C. 7 ng phn D. 5 ng phn

Ch : Anken C5H10 ch c 5 ng phn cu to (khng tnh ng phn hnh hc). 56.Tnh cht quan trng nht ca cao su l tnh cht no? A. Tan trong dung mi hu c B. Khng dn in, khng dn nhitC. Khng tan trong nc D. C tnh n hiCh : Cao su chc chn phi c tnh n hi (t hi phc hnh dng ban u khi khng chu tc dng ca ngoi lc) Cht do l cht khi thi chu tc dng ca ngoi lc th n vngi nguyn hnh dng khi b bin dng.Cao su c tt c cc tnh cht A, B, C, D. 57.t chy hon ton 1 mol hirocacbon A thy s mol CO2sinh ra bng 2 ln s mol H2O. Cng thc n gin nht caA l:

A. C3H4 B. C2H2 C. CH2 D. CH

Ch : C2H2 khng phi l cng thc n gin nht. Cc cht c cng thc n gin nht l CH gm: C2H2, C4H4,

C6H6, C8H8, 58.Mt hn hp X gm ankan A v anken B u th kh.- t chy hon ton 2,24 lt hn hp X gm a mol A v b mol B th khi lng CO2 sinh ra nhiu hn khi lng H2O l7,6 gam.


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Convert by TVDT

Bi gii: Kh b brom hp th l etilen vi s mol l 8/160 = 0,05.Th tch kh b brom hp th l 0,05.22,4 = 1,12 lt.71. Cho cc ankin: pent-2-in; 3-metylpent-1-in; 2,5-imetylhex-3-in v pent-1-in. Trong cc ankin ny, s cht c kh nngtc dng c vi dung dch AgNO3 trong NH3 l:A. 2 B. 1 C. 3 D. 4

72.Trong phn ng t chy mt hirocacbon A, a l s mol CO2, b l s mol H2O.

a.Nu A l ankan th t l

a

bm tho mn l:

A. 2m1 B. 1m2

1C. 2m1 D. 1m

Bi gii: Vi ankan: CnH2n+2 + (3n+1)/2 O2 nCO2 + (n+1) H2O. ( n > =1)

21

11

1nn

n

a

bm

b.Nu A l ankin th t la

bm tho mn l:

A. 2m1 B. 1m2

1C. 2m1 D. 1m0

c.Nu A l aren, ta c th kt lun nh th no v t la

bm

A. 2m1 B. 1m2

1 C. 2m1 D.2

1m0

Bi gii: Vi A l Aren CnH2n-6 (n >=6).

13

13

2

1

nn

n

a

bm

d.Nu A l ankaien, ta c th kt lun nh th no v t la

bm

A. 2m1 B. 1m3

2C. 2m1 D.

3

1m0

73.Trong phn ng t chyhn hp A gm hai hirocacbon, a l s mol CO2, b l s mol H2O.

Nu hn hp A gm 1 ankan v 1 anken th t la

bm tho mn l:

A. 2m1 B. 1m2

1C. 2m1 D. 1m

Bi gii: Hn hp gm x mol ankan CnH2n+2 v y mol CmH2m khi t chy ta c:

21)1(

1mynx

x

mynx

mynx

a

bm (n>=1; m>=2, x, y >0).

74.t chy hn hp X gm 2 hirocacbon (ch thuc nhm ankan hoc anken hoc ankin hoc aren) ta thu c hn hpsn phm kh v hi l CO2 v H2O vi s mol bng nhau. Kt lun no sau y l sai: A. Hn hp X c th gm 2 anken B. Hn hp X c th gm 1 ankan v 1 ankenC. Hn hp X c th gm 1 ankan v 1 aren D. Hn hp X c th gm 1 anken v 1 aren75. Hnhp kh X gm H2 v mt anken c khnngcng HBr cho snphmhucduy nht. Tkhica X so vi H2

bng 9,1. un nng X c xc tc Ni, sau khi phnngxy ra hon ton, thu c hn hp kh Y khng lm mt munc brom; tkhica Y so vi H2 bng 13. Cng thccu to ca anken l

A. CH2=C(CH3)2. B. CH2=CH2. C. CH2=CH-CH2-CH3. D. CH3-CH=CH-CH3.

Bi gii: Hn hp Y khng lm mt mu dung dich Br nn Y cha ankan + H2 d.; x l s mol anken, y l s mol H2 banu. Ta c:

26214

2,18214

y

ynx

yx

ynx

26y = 18,2(x + y) 7,8y = 18,2xy = 7x/3. Thay vo mt trong hai biu thc, ta c

n = 4. Cng thc phn t ca anken l C4H8. Do Anken phn ng cng vi HBr cho sn phm hu c duy nht nn ankenphi c cu to i xngCng thc cu to ca anken l: CH3-CH=CH-CH3.

76. Cho hn hp X gm CH4, C2H4 v C2H2. Ly 8,6 gam X tc dng ht vi dung dch brom (d) th khi lng

bromphnng l 48 gam. Mt khc, nu cho 13,44 lt (ktc)hnhp kh X tc dng vi lng d dung dch AgNO3trong NH3, thu c 36 gam kt ta. Phn trm th tch ca CH4 c trong X lA. 20%. B. 50%. C. 25%. D. 40%.


10/13


Convert by TVDT

Bi gii: x, y, z ln lt l s mol ca 3 hirocacbon trong 8,6 gam hn hp.Ta c: 16x + 28y + 26z = 8,6 v y + 2z = 48/160 = 0,3

a, b, c ln lt l s mol ca 3 hirocacbon trong 13,44 lt hn hp X. Ta c:a + b + c = 13,44/22,4 = 0,6

c = s mol AgC=-CAg = 36/240 = 0,15.Nh vy: zzyx

c

cba

z

zyx44

15,0

6,0

Gii h 3 n ta thu c x = 0,2; y = z = 0,1.Phn trm th tch ca CH4 l 0,2/0,4 = 0,5 (50%).

77. Hn hp X gm axit Y n chc v axit Z hai chc (Y, Z c cng s nguyn t cacbon). Chia X thnh hai phn

bng nhau. Cho phn mt tc dng ht vi Na, sinh ra 4,48 lt kh H2 (ktc). t chy hon ton phn hai, sinh ra 26,4gam CO2. Cng thccu to thu gn vphn trmvkhi lng ca Z trong hn hp X ln lt lA. HOOC-CH2-COOH v 70,87%. B. HOOC-CH2-COOH v 54,88%.

C. HOOC-COOH v 60,00%. D. HOOC-COOH v 42,86%.

Bi gii: t x, y ln lt l s mol Y, Z trong hn hp, n l s nguyn t C trong Y hoc Z.

Vy ta c: x + 2y = 0,4 v nx + ny = 26,4/44 = 0,6. n(0,42y) + ny = 0,6 n = 0,6/(0,4y).Li c 0


11/13


Convert by TVDT

A. Cng thc cu to. B. S nguyn t hiro. C. S nguyn t C. D. Cng thc phn t.84. Cc cht khc nhau c cng cng thc phn t c gi lA. cc cht ng phn ca nhau. B. cc cht ng ng ca nhau.C. cc dng th hnh ca nhau. D. cc cht thuc cng dy ng ng.85.Nhng cht c cng thc phn t ging nhau, nhng khc nhau v cu to, do dn n c tnh cht khc nhau, cgi l:A. ng ng B. ng phn C. ng v D. th hnh86.Nhng hp cht ging nhau v thnh phn v cu to ha hc, nhng phn t khc nhau mt hay nhiu nhm CH2cgi lA. th hnh. B. ng v. C. ng ng. D. ng phn.87. c im chung ca cacbocation v cacbanion l A. c th d dng tch c ra khi hn hp phn ng.B. km bn v c kh nng phn ng cao.C. chng u rt bn vng v c kh nng phn ng cao.D. km bn v c kh nng phn ng rt km.Ch : Cacbocation v cacboanion l cc gc hirocacbon, chng km bn v c kh nng phn ng rt cao, cc gc nyc hnh thnh trong nhng qu trnh trung gian ca cc phn ng ca cc hirocacbon nn chng km bn v c khnng phn ng cao (cc em c th xem thm trong phn c ch phn ng trong sch gio khoa nng cao). 88.Cng thc phn t ca hp cht B ng vi cc s liu thc nghim sau: C: 39,81%, H: 6,68%, 1,36d

2B/COl:

A. C2H4O2 B. C2H4O C. C3H6O2 D. C2H5O2

89. ng vi n = 1 th cng thc nguyn no sau y s l cng thc phn t? A. (C2H6O)n B. (CnH2n+1)n C. (C3H6Cl)n D. (C3H8N)nCh : Vi n = 1 th cht A l C2H6O l mt cng thc phn t ng.Cht B l CH3, khng phi l cng thc phn t.Cht C l C3H6Cl khng phi l cng thc phn t (6 + 1 = 7 l).Cht D l C3H8N khng phi l cng thc phn t.Vi nhng hp cht c cha C, H hoc C, H, O th s nguyn t H trong phn t lun chn. Vi nhng hp cht c cha C, H, N hoc C, H, O, N th s nguyn t H trong phn t lun l. 90. Hn hp no sau y khng lm mt mu dung dch Br2?A. C2H4, SO2, CO2 B. CH4, SO2, H2S C. H2, C2H6, CO2 D. CO2, C2H2, H2

Ch :Nhng hp cht HC khng no lm mt mu dung dch Br2, SO2, H2S lm mt mu dung dch Br2. 91. Cng thc phn t no ph hp vi penten ?A. C5H8 B. C3H6 C. C5H10 D. C5H12

92. Mt bnh kn dung tch 8,40 lt c cha 4,96 gam O 2 v 1,30 gam hn hp kh A gm 2 hirocacbon. Nhit trongbnh t1 = 0oC v p sut trong bnh p1= 0,50 atm. Bt tia la in trong bnh kn th hn hp A chy hon ton. Sau phn

ng, nhit trong bnh l t2 = 136,5oC v p sut l p2atm. Dn cc cht trong bnh sau phn ng i qua bnh th nht

ng H2SO4c, sau qua bnh 2 ng dung dch NaOH (c d) th khi lng bnh th hai tng 4,18 gam. Bit rng thtch bnh khng i, gi tr ca p 2 l:A. 0,87 atm B. 0,78 atm C. 0,75 atm D. 0,90 atm

Bi gii: Tng s mol kh l 0,1875S mol O2 l: 0,155.S mol hn hp A l 0,0325.S mol CO2 l: 4,18/44 = 0,095 (CO2 b NaOH gi li).S mol H2O l: (1,3 0,095*12)/2 = 0,08 (p dng bo ton khi lng: mHC = mC + mH).S mol O2 d l: 0,155 0,0950,08/2 = 0,02.Tng s mol kh v hi sau phn ng l: 0,02 + 0,095 + 0,08 = 0,195.

T y tnh c p sut bnh sau phn ng l: 78,04,8

)5,136273.(273/4,22.195,0

2 V

nRT

p

93. Cht no khng tc dng vi dung dch AgNO3 trong amoniac ?A. But-1-in B. Etin C. Propin D. But-2-in

Ch : Nhng Hirocacbon no c ni ba u mch u cho phn ng vi AgNO3 to kt ta mu vng (H ni ba umch linh ng c kh nng phn ng cao, d b th bi Ag).

94. Dn 6,72 lt axetilen (ktc) qua ng cha than hot tnh 600oC thu c 6,24 gam benzen. Hiu sut ca phn ng l:A. 90% B. 80% C. 75% D. 85%

95. Mt bnh kn dung tch 8,40 lt c cha 4,96 gam O 2v 1,30 gam hn hp kh A gm 2 hirocacbon. Nhit trongbnh t1 = 0

0C v p sut trong bnh p1= 0,50 atm. Bt tia la in trong bnh kn th hn hp A chy hon ton. Sau phnng, nhit trong bnh l t2 = 136,5

0C v p sut l p2atm. Dn cc cht trong bnh sau phn ng i qua bnh th nhtng H

2SO

4c, sau qua bnh 2 ng dung dch NaOH (c d) th khi lng bnh th hai tng 4,18 gam. Bit rng

trong hn hp A c mt cht l anken v mt cht l ankin. Cng thc phn t ca cc cht trong hn hp A l: A. C2H4 v C4H10 B. C2H4 v C4H6 C. C2H6 v C4H6 D. C4H8 v C2H2

96. Cht no c nhit si sao nht ?


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Convert by TVDT

A. Eten B. Propen C. Pent-1-en D. But-1-en

97. t chy hon ton 1 th tch hirocacbon X cn 5,5 th tch O2v thu c 4 th tch CO2(cc th tch o cng iukin). Bit X c phn ng vi dung dch AgNO3 trong NH3. Cng thc cu to ca X lA. CH C-CH3 B. CH CH C. CH C-CH2-CH3 D. CH3-C C-CH398. Gc no l ankyl ?A. -C2H3 B. -C6H5 C. -C3H5 D. -C2H5

99. Cho cc ankin sau: pent-2-in; 3-metylpent-1-in; propin; 2, 5-imetylhex-3-in. S ankin tc dng c vi dung dchAgNO3 trong NH3 l:

A. 1 B. 3 C. 4 D. 2

Ch : Ch c propin v 3-metylpent-1-in l ankin c ni ba u mch nn mi cho phnng vi AgNO3 trong NH3. 100. Dn hn hp kh A gm propan v xiclopropan i vo dung dch brom s quan st c hin tng no:A. Mu ca dung dch khng i.B. Mu ca dung dch b nht dn, khng c kh thot ra.C. Mu dung dch mt hn v khng cn kh thot ra.D. Mu dung dch nht dn v c kh thot ra.Ch : Ch c xiclopropan cho phn ng cng m vng lm mt mu dung dch Brom, propan l ankan nn khng phnng vi Brom. Nh vy khi cho hn hp qua dung dch Brom th dung dch nht mu dn v c kh thot ra. 101. iu ch 16,8 lt kh CH4(ktc) th th tch kh C3H8(ktc) bng phn ng tch cn dng l bao nhiu? Bit hiusut phn ng t 68%A. 24,7 lt B. 28,224 lt C. 16,8 lt D. 11,424 lt

C3H8 CH4 + C2H4

Bi gii: Th tch C2H8 cn dng cho phn ng tch (cracking) l: 16,8*100/68 = 24,7 lt. 102. t chy hon ton a mol mt ankan Y. Dn ht sn phm ln lt qua bnh (I) cha P 2O5v bnh (II) cha KOH cth khi lng bnh (I) tng 10,8 gam v bnh (II) tng 22 gam. Hi a c gi tr bao nhiu? A. a = 0,05 mol B. a = 0,5 mol C. a = 0,15 mol D. a = 0,1 mol

Bi gii: Bnh (I) tng l khi lng ca H2O, s mol l 0,6.Bnh (II) tng l khi lng ca CO2, s mol l 0,5.S mol ankan l 0,6 0,5 = 0,1.103. Khi t chy hirocacbon no X thu c kh cacbonic v hi nc c t l th tch bng 1 : 2. Cng thc cu to caX l:

A. CH4 B. C3H8 C. CH3CH3 D. C2H6

Bi gii: CxHy + (x + y/4)O2 xCO2 + y/2H2O.Nh vy x/(y/2) = x/y = . Vy HC X l CH4.104.Np 10,15 gam mt ankan X vo bnh cha kh clo (va ), a ra nh sng khuch tn phn ng xy ra honton. Dn sn phm qua dung dch AgNO3d thu c 50,225 gam kt ta trng. Cng thc phn t ca ankan l cht nosau y?A. C2H 6 B. C3H8 C. CH4 D. C4H10

Bi gii: CnH2n+2 + Cl2 CnH2n+2-xClx + xHCl.S mol ankan l 50,225/(143,5*x) = 0,35/xKhi lng mol ankan l: 14n + 2 = 10,15x/0,35 = 29x. C x = 2 v n = 4 tho mn. Cng thc phn t ca ankan l C4H10.105. Kt lun no sau y l khng ng?A. Hu ht cc ankan c kh nng tham gia phn ng th, phn ng tch.B. Ankan v xicloankan l ng phn ca nhau.C. Hu ht cc ankan c ng phn mch cacbon.D. Trong phn t ankan v xicloankan ch c cc lin kt n.106.

Khi t chy ht 1 mol ankan A thu c khng qu 5 mol CO2. Mt khc khi 1 mol A phn ng th vi 1 mol Cl2chto ra mt sn phm th duy nht. Vy A c th l:A. (1), (2), (3) u ng. B. 2,2 - imetyl propan (2)C. metan (1) D. etan (3)

Ch : C ba cht (1), (2), (3) u c cu to i xng nn ch to ra 1 sn phm th duy nht. 107. Theo chiu tng s nguyn t cacbon, phn trm khi lng ca cacbon:A. trong phn t ankan v xicloankan u bin i khng theo quy lut.B. trong phn t ankan v xicloankan u tng dn.C. trong phn t ankan v xicloankan u gim dn.D. trong phn t ankan tng dn, trong phn t xicloankan khng i.

Ch : Phn trm khi lng C trong phn t ankan l: %C =

n

n

n

214

12

214

12. Khi n tng, 2/n gim, 14 + 2/n gim, %C

tng.Xicloankan c cng thc chung (CH2)n nn %C v khi lng khng i.108. Chn p n sai:


13/13


Convert by TVDT

A. Xiclopropan l hirocabon khng no v n c phn ng cng.B. Khi un nng mnh, propan c th b tch H2chuyn thnh xiclopropan.C. Propan khng lm mt mu dung dch KMnO4.D. Xiclopropan lm mt mu dung dch KMnO4.109. Hy chn cu ng trong cc cu sau:A. Hirocacbon khng no l hirocacbon c phn ng cng vi hiro.B. Hirocacbon no l hirocacbon c cng thc phn t l CnH2n+2.C. Hirocacbon no l hirocacbon m trong phn t ch c lin kt n.D. Hirocacbon no l hirocacbon khng c phn ng cng thm hiro.110. Hn hp X gm ba hirocacbon (trong phn t c s nguyn t cacbon ln lt l 4, 5, 6 trong hirocacbon c snguyn t cacbon l 4 v 6 c s mol bng nhau) c t khi khi hi so vi hiro l 35. t chy hon ton 0,1 mol X, tngkhi lng CO2 v H2O thu c lA. 31 gam B. 25,5 gam C. 52,5 gam D. 55,2 gam

Bi gii: Do HC c s nguyn t cacbon trong phn t l 4 v 6 c s mol bng nhau nn ta c th t cng thc chung caX l C5HyS mol CO2 sinh ra l 0,5. Khi lngh hirocacbon l 0,1.70 = 7 gam.S mol H trong phn t C5Hy l: 7 0,5*12 = 1S mol H2O l 0,5 mol.Tng khi lng CO2 v H2O l: 0,5* (44 + 18) = 31 gam.111. X mch h c cng thc C3Hy. Mt bnh c dung tch khng i cha hn hp kh X v O2d 150

oC v c p sut 2atm. Bt tia la in t chy X sau a bnh v 150oC, p sut trong bnh vn l 2 atm. Ngi ta trn 9,6 gam X vi0,6 gam hidro ri cho qua binh ng Ni nung nng (H= 100%) th thu c hn hp Y. Khi lng mol trung bnh ca Yl :

A. 52,5 B. 46,5 C. 48,5 D. 42,5

Bi gii: C3Hy + (3 + y/4) O2 3CO2 + y/2H2O.Do p sut khng i nn s mol kh trc v sau phn ng cng khng i, ta c:1 +3 + y/4 = 3 + y/2 y = 4. Cng thc hirocacbon X l C3H4.

Khi lng mol trung bnh ca hn hp l: 5,42

40

6,9

6,06,9(sau phn ng ch cn hirocacbon, H2 phn ng ht).

112.Cho 4,48 lt hn hp A gm 2 hircacbon mch h li t t qua bnh cha 1,4 lt dung dch Br20,5M. Sau phn nghon ton, s mol Br gim i 1 na cn khi lng bnh tng 6,7 gam. Tm cng thc ca 2 hidrocacbon?A. C4H8 v C2H2 B. C2H4 v C4H6 C. C2H6 v C4H6 D. C4H10 v C2H2.

Bi gii: S mol Brom phn ng l: 1,4*0,5/2 = 0,35 mol. Khi lng bnh tng 6,7 gam l khi lng ca hirocacbon .

S mol hn hp hirocacbon l 0,2.1< nBr2/nhn hp = 0,35/0,2 = 1,75 < 2 Hn hp gm 0,05 mol anken v 0,15 mol ankin.

Nh vy 0,05*14n + 0,15*(14m-2) = 6,7 n + 3m = 10

Vi n = 4 v m = 2 l ph hp. Vy hai hirocacbon l C4H8 v C2H2.

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