CHUYN 3 :

HIROCACBON KHNG NO

BI TP V ANKEN Cu 1: Anken X c cng thc cu to: CH3CH2C(CH3)=CHCH3. Tn ca X l A. isohexan. B. 3-metylpent-3-en. C. 3-metylpent-2-en.Cch c tn anken SGK 11 nc 156 Tn v tr Tn nhnh tn mch chnh s v tr - en nh s th t gn ni i nht. 5 4 3 2 1 CH3 CH2 C(CH3) = CH CH3. => 3 metylpentan 2 en Ni i 2 , mch nhnh 3 ; mch chnh 5C pentan

D. 2-etylbut-2-en.

Cu 2: S ng phn ca C4H8 l A. 7. B. 4.

C. 6.

D. 5.

Ch ng phn hnh hc Xem li file vit p + cng thc + Cch xc nh p hnh hc Chuyn 1 C4H8 c k = 1 => 1 pi hoc 1 vng => ng phn dng anken hoc xicloankan Xt p anken Ch p hnh hc CH2 = CH CH2 CH3 ko c phh =>1 ; CH3 CH = CH CH3 c p hh =>2 CH2 =C(CH3) CH3 ko phh =>1 Xicloankan : Vng 3 cnh CH3 Tam gic CH3 => 1 Vng 4 cnh => 1 => Tng cng c 6 => C Mnh vit tt 3 cnh v 4 cnh hiu l tam gic v hnh vung

Cu 3: Hp cht C5H10 mch h c bao nhiu ng phn cu to ? A. 4. B. 5. C. 6.

D. 10.

C5H10 c k =1 + mch h => anken ; ng phn cu to => Khng tnh ng phn hnh hc. Xem file xc nh ng phn i thi hay b la CH2 = CH CH2 CH2 CH3 ; CH3 CH =CH CH2 CH3 ; CH2=CH CH(CH3) CH3 CH2 =C(CH3) CH2 CH3 CH3 C(CH3)=CH CH3 => Tng c 5 => B

Cu 4: Hp cht C5H10 c bao nhiu ng phn anken ? A. 4. B. 5.ng phn anken => tnh c ng phn hnh hc. Cu 3 c cht CH3 CH=CH-CH2-CH3 c p hnh hc => 6

C. 6.

D. 7.

Cu 5: Hp cht C5H10 c bao nhiu ng phn cu to ? A. 4. B. 5.

C. 6.

D. 10.

ng phn cu to => Khng tnh hnh hc C5H10 c k =1 => 1 pi Anken hoc 1 vng Xicloankan Anken => bi 3 => c 5 p cu to Xicloankan : => 10 p cu to ca C5H10 5 anken + 5 xicloankan

CH3

CH3

CH3 CH3 C2H5 => 5 p xicloankan CH3

Cu 6: Ba hirocacbon X, Y, Z l ng ng k tip, khi lng phn t ca Z bng 2 ln khi lng phn t ca X. Cc cht X, Y, Z thuc dy ng ng A. ankin. B. ankan. C. ankaien. D. anken.MZ = 2MX + X , Y , Z ng ng k tip => X , Y , Z l anken C th X l C2H4 v Z l C4H8

Cu 7: Anken X c c im: Trong phn t c 8 lin kt xch ma. CTPT ca X l A. C2H4. B. C4H8. C. C3H6. D. C5H10.Anken => c 1 lin kt pi Mo lin kt xch ma = s C + s H 1 i vi mch h - khng i vi mch vng ; Lin kt xch ma = s lin kt to gia C v H + s lin kt to gia C v C = S H + s C 1 C3H6 c s lin kt xch ma = 3 + 6 1 = 8 lin kt xch ma => C tha mn

Cu 8: Vitamin A cng thc phn t C20H30O, co cha 1 vong 6 canh va khng co cha lin kt ba. S lin kt i trong phn t vitamin A la

A. 7.

B. 6.

C. 5.

D. 4.

ADCT tnh s pi + vng = (2.20 -30 +2)/2 = 6 A cha 1 vng => s pi = 6 1 =5 pi hay 5 lin kt i V khng cha lin kt 3 => C

Cu 9: Licopen, cng thc phn t C40H56 la cht mau o trong qua ca chua, chi cha lin kt i va lin kt n trong phn t. Hiro hoa hoan toan licopen c hirocacbon C40H82. Vy licopen co A. 1 vong; 12 ni i. B. 1 vong; 5 ni i. C. 4 vong; 5 ni i. D. mach h; 13 ni i.C40H56 c tng s pi + vng = (2.40 56 + 2)/2 = 13 => Loi B v C. C40H56 ch cha lin kt i v lin kt n => loi trng hp vng ny mnh ko chc Hoc hidro ha hon ton to ra C40H82 ankan => C40H56 nu ng th c 1 vng 3 cnh cn li 12 i th mnh ngh vn ng . => D th chc chn hn , cn A c trng hp c bit 1 vng 3 cnh + 12 i th ng => D

Cu 10: Cho cc cht sau: 2-metylbut-1-en (1); 3,3-imetylbut-1-en (2); 3-metylpent-1-en (3); 3-metylpent-2-en (4); Nhng cht no l ng phn ca nhau ? A. (3) v (4). B. (1), (2) v (3). C. (1) v (2). D. (2), (3) v (4).ng phn => Cng CTPT: (1) C5H10 ; 2 ,3 ,4 u l C6H10 => 2,3,4 cng l ng phn.

Cu 11: Hp cht no sau y c ng phn hnh hc ? A. 2-metylbut-2-en. B. 2-clo-but-1-en. C. 2,3- iclobut-2-en.

D. 2,3- imetylpent-2-en.

K c p hh :R1#R2 v R3#R4 1 2 3 4 A loi v R1 v R2 u l CH3 : CH3 C(CH3) = CH CH3 1 2 3 4 B loi v - 1-en => R1 v R2 u l H : CH2 = C(Cl) CH2 CH3 1 2 3 4 C ng v tha mn k : CH3 C(Cl) =C(Cl) CH3 R1 # R2 v R3#R4 ; CH3 #Cl =>C 1 2 3 4 5 D sai v R1 ging R2 CH3: CH3 C(CH3) = C(CH3) CH2 CH3

Cu 12: Nhng hp cht no sau y c ng phn hnh hc (cis-trans) ? CH3CH=CH2 (I); CH3CH=CHCl (II); CH3CH=C(CH3)2 (III); C2H5C(CH3)=C(CH3)C2H5 (IV); C2H5C(CH3)=CClCH3 (V). A. (I), (IV), (V). B. (II), (IV), (V). C. (III), (IV). D. (II), III, (IV), (V).Thy ngay I v III u loi v R3 ging R4 => A , C , D loi => B da vo k R1#R2 v R3#R4 Dng bi ny loi p n nhanh hn l i tm ng.

Cu 13: Cho cc cht sau: CH2=CHCH2CH2CH=CH2; CH2=CHCH=CHCH2CH3; CH3C(CH3)=CHCH2; CH2=CHCH2CH=CH2; CH3CH2CH=CHCH2CH3; CH3C(CH3)=CHCH2CH3; CH3CH2C(CH3)=C(C2H5)CH(CH3)2; CH3CH=CHCH3. S cht c ng phn hnh hc l: A. 4. B. 1. C. 2. D. 3.CH2=CHCH2CH2CH=CH2 ko c v dng R CH2=CH2 CH2=CHCH=CHCH2CH3 c p hh ni i th 2. => 1 CH3C(CH3)=CHCH2 ko c v R1v R2 l CH3 CH2=CHCH2CH=CH2 ko c ging cht 1 CH3CH2CH=CHCH2CH3 c p hh => 1 CH3C(CH3)=CHCH2CH3 khng c v R1 v R2 l CH3 CH3CH2C(CH3)=C(C2H5)CH(CH3)2 c R1 # R2 hay C2H5 # CH3 ; R3#R4 hay C2H5 # C3H7 => 1 CH3CH=CHCH3 c => 1 => Tng c 4 cht.

Cu 14: p dng quy tc Maccopnhicop vo trng hp no sau y ? A. Phn ng cng ca Br2 vi anken i xng. C. Phn ng cng ca HX vo anken i xng. B. Phn ng trng hp ca anken. D. Phn ng cng ca HX vo anken bt i xng. SGK 11nc 162 => D To ra 2 sn phm ; chnh v phC sai v anken i xng nh CH2 = CH2 ch to ra 1 sn phm hoc CH3 CH=CH CH3 i xng nhau qua lin kt i

Cu 15: Khi cho but-1-en tc dng vi dung dch HBr, theo qui tc Maccopnhicop sn phm no sau y l sn phm chnh ? A. CH3-CH2-CHBr-CH2Br. C. CH3-CH2-CHBr-CH3. B. CH2Br-CH2-CH2-CH2Br . D. CH3-CH2-CH2-CH2Br.SGK 11nc 162 => sn phm chnh Halogen vo C t H cn H vo C nhiu H Hoc halogen vo C bc cao nht v H vo C cn li Bc 1 2 3 4 But 1 en : CH2 = CH CH2 CH3 + HBr => CH3 CHBr CH2 CH3 => C

Cu 16: Anken C4H8 c bao nhiu ng phn khi tc dng vi dung dch HCl ch cho mt sn phm hu c duy nht ? A. 2. B. 1. C. 3. D. 4.P to anken + HCl to ra 1 sn phm duy nht => Anken i xng

CH3 CH =CH CH3 => c p hnh hc => 2 => D

Cu 17: Cho cc cht: xiclobutan, 2-metylpropen, but-1-en, cis-but-2-en, 2-metylbut-2-en. Dy gm cc cht sau khi phn ng vi H2 (d, xc tc Ni, to), cho cng mt sn phm l: A. xiclobutan, cis-but-2-en v but-1-en. B. but-1-en, 2-metylpropen v cis-but-2-en. C. xiclobutan, 2-metylbut-2-en v but-1-en. D. 2-metylpropen, cis -but-2-en v xiclobutan. Cu 18: Cho hn hp tt c cc ng phn mch h ca C 4H8 tc dng vi H2O (H+,to) thu c ti a bao nhiu sn phm cng ? A. 2. B. 4. C. 6. D. 5P vi H2O => OH vo C bc cao v H vo C cn li C=C AD Cu 2 => CH2 = CH CH2 CH3 => sp OH-CH2 CH2 CH2 CH3 hoc CH3 CH(OH)-CH2 CH3 CH3 CH = CH CH3 => sp : CH3 CH(OH) CH2 CH3 CH2 =C(CH3) CH3 => sp : CH3 (OH)C(CH3) CH3 hoc OH CH2 CH(CH3) CH3 Gp 3 trng hp => c 4 sn phm TH1 v TH2 cng CH3 CH(OH) CH2 CH3

=> B

Cu 19: C bao nhiu anken th kh (kt) m khi cho mi anken tc dng vi dung dch HCl ch cho mt sn phm hu c duy nht ? A. 2. B. 1. C. 3. D. 4.SGK 11 nc 159 => Anken th kh t C2 ti C4 Vi C2H4 => to ra 1 cht anken i xng C3H6 => C = C C => to ra 2 sn phm anken bt i xng C4H6 => C - C = C C => mi ng phn hnh hc to ra 1 sn phm => 2 cht tng l 3 : C2H4 ; cis C4H6 ; trans C4H6 => C mnh Khng dm khng nh cis v trans Cu ny khng chc p n.

Cu 20: Hirat ha 2 anken ch to thnh 2 ancol (ru). Hai anken l A. 2-metylpropen v but-1-en (hoc buten-1). B. propen v but-2-en (hoc buten-2). C. eten v but-2-en (hoc buten-2). D. eten v but-1-en (hoc buten-1). 2anken to thnh 2 ancol => mi anken to thnh 1 ancol => anken i xng.A,D loi v cha but 1 en : C = C C C to ra 2 ancol => cht cn li = 1 => 3 ancol B loi v Propen to ra 2 ancol + but - 2 en to ra 1 ancol (i xng ) C. Eten v but 2 en u mch i xng => mi cht to ra 1 ancol duy nht => C

Cu 21: Anken thch hp iu ch ancol sau y (CH3 CH2)3C-OH l A. 3-etylpent-2-en. B. 3-etylpent-3-en. C. 3-etylpent-1-en.D. 3,3- imetylpent-1-en.(CH3 CH2)3C-OH ; CH3 CH2 (CH3CH2)C(OH) CH2 CH3 1 2 3 4 5 => anken iu ch : CH3 CH2 = (CH3CH2)C CH2 CH3 Ni i v tr 2 ; etyl v tr 3 ; mch chnh c 5 C => pent => A .3 etylpent 2 en =>A Xem li cch vit danh php anken

Cu 22: Hirat ha hn hp X gm 2 anken thu c ch thu c 2 ancol. X gm A. CH2=CH2 v CH2=CHCH3. B. CH2=CH2 v CH3CH=CHCH3. C. B hoc D. D. CH3CH=CHCH3 v CH2=CHCH2CH3.Cu ny t B hoc D => Chn B hoc D cng c m B chc chn ng ri Bi 20 C bao qut => Cu ny p n khng hp l D khng tha mn Sa p n . C . B v D B ng

Cu 23: S cp ng phn cu to anken th kh (kt) tho mn iu kin: Khi hirat ho to thnh hn hp gm ba ancol l A. 6. B. 3. C. 5. D. 4.Anken th kh => C2 ti C4 SGK 11 nc 159 Hidrat ha l p anken + H2O => Ru SGK 11 nc 161

Cu 24: S cp ng phn anken th kh (kt) tho mn iu kin: Khi hirat ho to thnh hn hp gm ba ancol l: A. 6. B. 7. C. 5. D. 8. Cu 25: Hp cht X c CTPT C3H6, X tc dng vi dung dch HBr thu c mt sn phm hu c duy nht. Vy X l: A. propen. B. propan. C. isopropen. D. xicloropan.C3H6 => k = 1 => 1pi hoc 1 vng => loi B . Khng c p n C . i vi Cht c 3C khng c iso. => A v D . M A to ra 2 sn phm => D ng Hoc thy ngay p SGK bi xicloankan

Cu 26: Hai cht X, Y c CTPT C3H6 v C4H8 v u tc dng c vi nc brom. X, Y l A. Hai anken hoc xicloankan vng 3 cnh. C. Hai anken hoc xicloankan vng 4 cnh. B. Hai anken hoc hai ankan. D. Hai anken ng ng ca nhau.Anken lun p vi dd Br2. Xicloankan c vng 3 cnh lun p vi dd Br2 => A B sai v ankan ; C sai v vng 4 cnh ko p vi dd Br2 ; D thiu trng hp Xicloankan

Cu 27: C hai ng nghim, mi ng cha 1 ml dung dch brom trong nc c mu vng nht. Thm vo ng th nht 1 ml hexan v ng th hai 1 ml hex-1-en. Lc u c hai ng nghim, sau yn hai ng nghim trong vi pht. Hin tng quan st c l:

A. C s tch lp cc cht lng c hai ng nghim. B. Mu vng nht vn khng i ng nghim th nht C. ng nghim th hai c hai lp cht lng u khng mu D. A, B, C u ng. Cu 28: Trng hp eten, sn phm thu c c cu to l: A. (-CH2=CH2-)n . B. (-CH2-CH2-)n . C. (-CH=CH-)n. D. (-CH3-CH3-)n . Eten : C2H4 => trng hp => (-CH2-CH2-)n => B P SGK 11 nc 162 Cu 29: Oxi ho etilen bng dung dch KMnO4 thu c sn phm l: A. MnO2, C2H4(OH)2, KOH. C. K2CO3, H2O, MnO2. B. C2H5OH, MnO2, KOH. D. C2H4(OH)2, K2CO3, MnO2.P SGK 11 nc 162 : 3C2H4 + 2KMnO4 + 4H2O => 3C2H4(OH)2 etylenglicol + 2MnO2 + 2KOH => A

Cu 30: X la h n h p gm 2 hirocacbon. t cha y X c nCO2 = nH2O. X co th gm A. 1xicloankan + anken.B. 1ankan + 1ankin. C. 2 anken. D. A hoc B hoc C.X c th : A ng v c 2 cht u c k = 1 ; B c th : v nu nankan = nankin C ng v k = 1 => D Nu phn vn B cha bit th ta thy A v C ng => D ng

Cu 31: iu ch etilen trong phng th nghim t C2H5OH, (H2SO4 c, 170oC) thng ln cc oxit nh SO2, CO2. Cht dng lm sch etilen l: A. dd brom d. B. dd NaOH d. C. dd Na2CO3 d.D. dd KMnO4 long d.Lm sch etilen tc l lm mt i SO2 v CO2 trong kh etilen. Xt A dd Br2 d => Etilen v SO2 u lm mt mu => khng th loi c B ng v ch c SO2 v CO2 p => cn li etilen => B SO2 + NaOH d => Na2SO3 + H2O ; CO2 + NaOH d => Na2CO3 + H2O C sai v khng cht no p D sai v Etilen v SO2 u p.

Cu 32: Sn phm chnh ca s ehirat ha 2-metylbutan-2-ol l cht no ? A. 3-Metylbut-1-en. B. 2-Metylbut-1en. C. 3-Metylbut-2-en.ehidrat ha tc l p tch nhm H2O t ancol to thnh anken SGK 11 nc 227 Quy tc Zaixep OH tch cng vi H bc cao bn cnh sn phm chnh Sn phm ph ngc li cng H bc thp bn cnh I II 2 metylbutan 2 ol : CH3 (CH3)C (OH) CH2 CH3 => tch cng H bc II 1 2 3 4 => CH3 C(CH3)=CH CH3 => 2 metylbut 2 en => D

D. 2-Metylbut-2-en.

Cu 33: Khi tch nc t ru (ancol) 3-metylbutanol-1 (hay 3-metylbutan-1-ol), sn phm chnh thu c l: A. 2-metylbuten-3 (hay 2-metylbut-3-en). B. 3-metylbuten-2 (hay 3-metylbut-2-en). C. 3-metylbuten-1 (hay 3-metylbut-1-en). D. 2-metylbuten-2 (hay 2-metylbut-2-en). 1 2 3 4AD 32 : 3-metylbutan-1-ol : OH CH2 CH2 CH(CH3) CH3 1 2 3 4 => CH2=CH2 CH(CH3) CH3 => 3 metyl but 1 en => C

Cu 34: Hp cht 2-metylbut-2-en l sn phm chnh ca phn ng tch t cht no ? A. 2-brom-2-metylbutan.B. 2-metylbutan -2- ol. C. 3-metylbutan-2- ol.D. Tt c u ng.2 metylbut 2 en : CH3 - C(CH3) = CH CH3 A. 2-brom-2-metylbutan P tch HX SGK 11 nc 214 Quy tc Zai Xp ; Nguyn t halogen X u tin tch cng vi H C bc cao bn cnh Bc: I II CH3 (Br)C(CH3) CH2 CH3 => tch cng C bc II => CH3 C(CH3) = CH CH3 Tha mn P vi kim KOH c xc tc C2H5OH , nhit B. 2-metylbutan -2- ol. AD bi 32 I II CH3 (OH)C(CH3) CH2 CH3 => Tch cng C bc 2 CH3 C(CH3)=CH CH3 Tha mn A, B ng => D I III Xt C. 3 metylbutan 2 ol ; CH3 CH(OH) CH(CH3) CH3 => tch cng C bc 3 => CH3 CH=C(CH3) CH3 Tha mn Ngc li

Cu 35: Khi lng etilen thu c khi un nng 230 gam ru etylic vi H2SO4 m c, hiu sut phn ng t 40% l: A. 56 gam. B. 84 gam. C. 196 gam. D. 350 gam.Phn ng tch H2O SGK 11 nc 227 Ru etylic C2H5OH => C2H5OH => C2H4 etilen + H2O 5 mol => 5 mol

=> mC2H4 theo PT = 140g . CT tnh H% ; H% p = mPT . 100% / mTT ; H%Sp = mTT.100% / mPT mPT l m phng trnh Tnh theo PT ; mTT l m thc t thu c hoc ban u sp l sn phm ; p l phn ng C th thay khi lng bng th tch hay s mol - mPT v mTT ca cng mt cht C2H4 l sn phm => H%sp = mTT.100% / mPT 40% = mTT.100% / 140 mTT = 140.40/100 = 56g => A mC2H4 thc t thu c

Cu 36: Cho 3,36 lt hn hp etan v etilen (ktc) i chm qua qua dung dch brom d. Sau phn ng khi lng bnh brom tng thm 2,8 gam. S mol etan v etilen trong hn hp ln lt l: A. 0,05 v 0,1. B. 0,1 v 0,05. C. 0,12 v 0,03. D. 0,03 v 0,12.C lin kt pi => c phn ng cng Br2 => EtilenC2H4 p vi Br2 cn etanC2H6 khng p. SGK 11 nc 160 ; anken p cng Br2 Tng qut : X + kBr2 => XBr2k X l cht hu c mch h c k 1 k = 0 l ankan ko c p cng P cng xut pht t lin kt pi Tng qut vi k = 1 => CnH2nOz ; k =2 => CnH2n-2Oz k=1 c gc hidrocacbon ging Anken; k = 2 c gc hidrocabon ging Ankin VD: C2H4 + Br2 => C2H4Br2 V C2H4 c k = pi C3H6O2 + Br2 => C3H6O2Br2 v C3H6O2 c k = 1 C5H8 + 2Br2 => C5H8Br4 V c k =2 Cng Br2 nh cng X2 , H2 ; X l halogen m bnh tng = mAnken cho vo V anken b hp th - BT khi lng => mC2H4 = 2,8 g => nC2H4 = 0,1 mol => nC2H6Etan = nhh nC2H4 = 0,15 0,1 = 0,05 => A

Cu 37: 2,8 gam anken A lam mt mau va u dung dich cha 8 gam Br2. Hirat hoa A chi thu c mt ancol duy nht. A co tn la: A. etilen. B. but - 2-en. C. hex- 2-en. D. 2,3-dimetylbut-2-en.Hidrat ha A p anken + H2O => ancol => thu c 1 sn phm => Mch i xng Cc p n A,B,D i xng A. CH2 = CH2 ; B . CH3 CH=CH-CH3 ; C. CH3-CH=CH-CH2-CH3 Ko i xng D. CH3 C(CH3) =C(CH3)-CH3 A , B , C , D u l anken ui en => nAnken = nBr2 = 0,05 mol => M anken = 2,8 / 0,05 = 56 = 14n n = 4 CnH2n => C4H8 => B Ch c B c 4C

Cu 38: 0,05 mol hirocacbon X lam mt mau va u dung dich cha 8 gam brom cho ra san phm co ham lng brom at 69,56%. Cng thc phn t cua X la: A. C3H6. B. C4H8. C. C5H10. D. C5H8.Ta c nX = nBr2 = 1 =k => Hidrocabon c CT : CnH2n Bi 36 P : CnH2n + Br2 => CnH2nBr2 %Br = 160.100% / (14n+160) = 69,56% n = 5 Cch bm nh chuyn 1 ; ly 160.100%/69,56 - 160 sau ly kt qu chia 14 => n = 5 => C5H10 =>C

Cu 39: Dn t t 8,4 gam hn hp X gm but-1-en v but-2-en li chm qua bnh ng dung dch Br 2, khi kt thc phn ng thy c m gam brom phn ng. m c gi tr l: A. 12 gam. B. 24 gam. C. 36 gam. D. 48 gam.But 1 en ; But 2 en l ng phn ca C4H8 => n hn hp = 8,4 / 56 = 0,15 mol = nBr2 V k =1 :anken => mBr2 p = 24g => B

Cu 40: Dn 3,36 lt (ktc) hn hp X gm 2 anken l ng ng k tip vo bnh nc brom d, thy khi lng bnh tng thm 7,7 gam. Thnh phn phn % v th tch ca hai anken l: A. 25% v 75%. B. 33,33% v 66,67%. C. 40% v 60%. D. 35% v 65%.Anken p vi Br2 => m bnh tng = mAnken p = 7,7 g V anken p vi dd Br2 Gi CT ca hn hp : CnH2n => M =

mhh 7,7 = = 51,33 = 14n n = 3,67 => n=3 v n = 4 2 anken kt tip nhau nhh 0,15

Xem li cch xc nh % th tch nhanh Bi 47 chuyn 2 hoc trong file pp gii nhanh ha hu c %C4H8 = 67% Hay 66,67 mnh lm trn % C ln = s sau du , => % C3H6 = 100 % - %S ln = 33,33 % => B

Cu 41: Hn hp X gm 2 anken l ng ng lin tip c th tch 4,48 lt ( ktc). Nu cho hn hp X i qua bnh ng nc brom d, khi lng bnh tng ln 9,8 gam. % th tch ca mt trong 2 anken l: A. 50%. B. 40%. C. 70%. D. 80%. Tng t Bi 40 => n = 3,5 => %C4H8 = 50% => %C3H6 = 50% => A Cu 42: Dn 3,36 lt (ktc) hn hp X gm 2 anken l ng ng k tip vo bnh nc brom d, thy khi lng bnh tng thm 7,7 gam. CTPT ca 2 anken l:

A. C2H4 v C3H6. B. C3H6 v C4H8.

C. C4H8 v C5H10.

D. C5H10 v C6H12.

Tng t bi 40 => n = 3,67 => n = 3 (C3H6) v n = 4 (C4H8) => B v lin tip

Cu 43: Mt hn hp X c th tch 11,2 lt (ktc), X gm 2 anken ng ng k tip nhau. Khi cho X qua nc Br 2 d thy khi lng bnh Br2 tng 15,4 gam. Xc nh CTPT v s mol mi anken trong hn hp X. A. 0,2 mol C2H4 v 0,3 mol C3H6. B. 0,2 mol C3H6 v 0,2 mol C4H8. C. 0,4 mol C2H4 v 0,1 mol C3H6. D. 0,3 mol C2H4 v 0,2 mol C3H6. Tng t bi 40 => n = 2,2 => n = 2C2H4 v n =3 C3H6 2 anken lin tip Xem li bi 47 chuyn 2 . Tm t l s mol 2 cht lin tip t n

0,2nC2H4 = 0,8nC3H6 nC2H4 = 4nC3H6 => chn nC3H6 = x mol => nC2H4 = 4xmol M nC2H4 + nC3H6 = 0,5 mol => x = 0,1 => nC2H4 = 0,4 ; nC3H6 = 0,1 => C

Cu 44: Mt hn hp X gm ankan A v anken B, A c nhiu hn B mt nguyn t cacbon, A v B u th kh ( ktc). Khi cho 6,72 lt kh X (ktc) i qua nc brom d, khi lng bnh brom tng ln 2,8 gam; th tch kh cn li ch bng 2/3 th tch hn hp X ban u. CTPT ca A, B v khi lng ca hn hp X l: A. C4H10, C3H6 ; 5,8 gam. B. C3H8, C2H4 ; 5,8 gam. C. C4H10, C3H6 ; 12,8 gam. D. C3H8, C2H4 ; 11,6 gam.Ch c anken p vi Br2 => Th tch cn li = 2/3 th tch hh ban u = V ankan V ankan = 2.6,72/3 = 4,48 lt => nankan = 0,2 mol => nAnken = nhhX nAnkan = 0,3 0,2 = 0,1 mol m bnh tng = mAnken = 2,8 g => Manken = 28 =14n => n =2 => C2H4 B A l C3H8 V A c C ln hn B 1 C v A c dng CnH2n+2 M hn hp X = mC3H8 + mC2H4 = 0,2.44 + 0,1.28 = 11,6 g => D

Cu 45: Mt hn hp X gm ankan A v mt anken B c cng s nguyn t C v u th kh ktc. Cho hn hp X i qua nc Br2 d th th tch kh Y cn li bng na th tch X, cn khi lng Y bng 15/29 khi lng X. CTPT A, B v thnh phn % theo th tch ca hn hp X l A. 40% C2H6 v 60% C2H4. B. 50% C3H8v 50% C3H6 C. 50% C4H10 v 50% C4H8. D. 50% C2H6 v 50% C2H4Bi 44 => Th tch Y = V ankan = VX /2 => Vankan = V anken = VX / 2 => %theo th tch = 50% Loi A. A v B cng s C => A c CT : CnH2n+2Ankan => CnH2n l CT Banken Chn nAnkan = 1 mol => nAnken = 1mol

=>

mY mAnkan 15 14n+2 15 = = = n = 2 => C2H6 vaf C2H4 mhhX mAnkan+mAnken 29 14n+2 + 14n 29

Mo p n A v D cng cng thc ; p n B,C,D cng % V => Ly C t A hoc D ; Ly % t B,C,D => Cch ny dng cho bn khng lm c khi i thi

Cu 46 : Hn hp X gm metan v 1 olefin. Cho 10,8 lt hn hp X qua dung dch brom d thy c 1 cht kh bay ra, t chy hon ton kh ny thu c 5,544 gam CO2. Thnh phn % v th tch metan v olefin trong hn hp X l: A. 26,13% v 73,87%. B. 36,5% v 63,5%. C. 20% v 80%. D. 73,9% v 26,1%.Metan CH4 v Olefinanken CnH2n 1 cht kh l CH4 V anken b Br2 d hp th nCH4 = nCO2 to ra = 0,126 mol Bt nguyn t C VCH4 = 2,8224 => %CH4 = 2,6133 .100% / 10,8 = 26,13% => % Anken = 73,87% => A

Cu 47: Cho 8960 ml (ktc) anken X qua dung dch brom d. Sau phn ng thy khi lng bnh brom tng 22,4 gam. Bit X c ng phn hnh hc. CTCT ca X l: A. CH2=CHCH2CH3. B. CH3CH=CHCH3. C. CH3CH=CHCH2CH3. D. (CH3)2C=CH2.X c ng phn hnh hc => Loi A v D Xem li k ng phn hnh hc Tng t bi 40 => n = 4 => C4H8 => B v C c 5C

Cu 48: a. Cho hirocacbon X phn ng vi brom (trong dung dch) theo t l mol 1 : 1, thu c cht hu c Y (cha 74,08% Br v khi lng). Khi X phn ng vi HBr th thu c hai sn phm hu c khc nhau. Tn gi ca X l: A. but-1-en. B. but-2-en. C. Propilen. D. Xiclopropan.p n => u c th p cng Br2 v u c 1 pi hoc 1 vng => Cng thc : CnH2n PT : CnH2n + Br2 => CnH2nBr2 => %Br = 160.100% / (14n + 160) = 74,08 n= 4 => Loi C,D X p vi HBr thu c 2 sn phm => X l anken khng i xng => A :CH2=CH-CH2-CH3 b. Hirocacbon X cng HCl theo ti l mol 1:1 tao san phm co ham lng clo la 55,04%. X co cng thc phn t l: A. C4H8. B. C2H4. C. C5H10. D. C3H6. p n => X c CT : CnH2n

P : CnH2n + HCl => CnH2n+1Cl => %Cl = 35,5.100% / (14n + 1 + 35,5) = 55,04% n = 2 => B .C2H4

Cu 49: Hn hp X gm metan v anken, cho 5,6 lt X qua dung dch brom d thy khi lng bnh brom tng 7,28 gam v c 2,688 lt kh bay ra (ktc). CTPT ca anken l: A. C4H8. B. C5H10. C. C3H6. D. C2H4Metan CH4 v Anken CnH2n Tng t bi 44 => V kh bay ra = nAnkan = 2,688 lt => V anken = Vhh Vankan = 5,6 2,688 = 2,912 lt => nAnken = 0,13 mol + mAnken = m bnh tng = 7,28 => M anken = 7,28/0,13 =14n =>n = 4 =>A

Cu 50: Dn 3,36 lt (ktc) hn hp X gm 2 anken l vo bnh nc brom d, thy khi lng bnh tng thm 7,7 gam. CTPT ca 2 anken l: A. C2H4 v C4H8. B. C3H6 v C4H8. C. C4H8 v C5H10. D. A hoc B. Tng t bi 40 => n = 3,67 => A v B ng V n = 3,67 nm gia 2 v 4 A ; 3 v 4 B C sai v n = 3,67 < 4 ; D Cu 51: Cho 10 lt hn hp kh (54,6oC; 0,8064 atm) gm 2 olefin li qua bnh dung dch brom d thy khi lng bnh brom tng 16,8 gam. CTPT ca 2 anken l (Bit s C trong cc anken khng vt qu 5) A. C2H4 v C5H10.B. C3H6 v C5H10. C. C4H8 v C5H10. D. A hoc B.ADCT : nhn hp = PV/T.0,082 = (0,8064.10/((273+54,6).0,082) = 0,3 mol => M=

16,8 =14n n = 4 => A , B tha mn V n nm gia 2 v 5 A ; 3 v 5B 0,3

C sai v n = 4 khng nm gia 4 v 5 => D

Cu 52: Mt hirocacbon X cng hp vi axit HCl theo t l mol 1:1 to sn phm c thnh phn khi lng clo l 45,223%. Cng thc phn t ca X l: A. C3H6. B. C4H8. C. C2H4. D. C5H10.Tng t b bi 48 => n = 3 => C3H6 => A

Cu 53: Cho hn hp X gm etilen v H2 c t khi so vi H2 bng 4,25. Dn X qua bt niken nung nng (hiu sut phn ng 75%) thu c hn hp Y. T khi ca Y so vi H2 (cc th tch o cng iu kin) l: A. 5,23. B. 3,25. C. 5,35. D. 10,46.Dng bi anken p vi H2 cho M hn hp trc v H% => Tm M hn hp sau Hoc cho M trc ; M sau tm H% ; hoc H% v M sau tm M trc Phng php mo: Lun chn nAnken = 1 mol => nH2 t M trc Hoc Gi x, y l s mol Anken v H2 t M => t l gia x v y sau chn Tm ra s khi lng trc p . Bo ton khi lng => m trc = m sau VD: bi trn etilen C2H4 Cch 1: chn 1mol C2H4 => M trc = (mC2H4 + mH2) / (nC2H4 + nH2) = 8,5 (I) (28 + 2x) / (1+x) = 8,5 => x = 3 mol X l nH2 m trc = mC2H4 + mH2 = 28 + 3.2 = 34 g = m sau Cch 2 : (I) (28x + 2y) / (x + y) = 8,5 3x = y => chn x = 1 => y = 3 Cch ny l ng cho Mnh hay dng cch chn => bm my tnh Tip theo p vo Phng trnh: gi x , y ln lt l s mol CnH2n ; H2 PT : CnH2n + H2 => CnH2n+2 Ban u:x mol y mol P x mol => x mol => xmol Sau p y x mol x mol nSau p = nH2 d + nCnH2n+2 to thnh = y x + x = y mol Chnh bng s mol H2 - Nh M sau = m trc / nH2 Cng thc Vi H% = 100% , Anken p ht Nu c H% => nCnH2n p = x.H% PT: CnH2n + H2 => CnH2n+2 Ban u x mol ymol P x.H% => x.H% mol => x.H% mol Sau p x x.H% y x.H% x.H% n sau p = nCnH2n d + nH2 d + nCnH2n+2 to thnh = (x x.H%) + y x.H% + x.H% = x + y - x.H% M sau = m trc / (x+y x.H%) M sau= M trc . n trc / (x+y x.H%) = Mtrc . (x+y)/(x+y x.H%) Cng thc tng qut i vi H% Cng thc p dng gii nhanh vi dng ny => ch thi c. Nu khng ni n H% => coi nh 100% AD bi trn => m trc = 34 g ; x = 1mol ; y = 3mol ; H% = 75%

=> M sau = 34 / (1 + 3 1.0,75) = 10,46 => T khi vi H2 = 10,46 / 2 = 5,23 => A

Cu 54: Cho H2 v 1 olefin c th tch bng nhau qua Niken un nng ta c hn hp A. Bit t khi hi ca A i vi H2 l 23,2. Hiu sut phn ng hiro ho l 75%. Cng thc phn t olefin l A. C2H4. B. C3H6. C. C4H8. D. C5H10.Ta c H2 v olefin Anken CnH2n c th tch bng nhau => Chn nH2 = nOlefin = 1 mol ADCT bi trn M sau = m trc / (x + y x.H%) vi x = y = 1 M sau = 23,2.2 = 46,4 t l vi H2 = 23,2 m trc = mAnken + mH2 => 46,4 = (14n + 2) / (1 + 1 1.0,75) n = 4 => C

Cu 55: Hn hp kh X gm H2 v mt anken c kh nng cng HBr cho sn phm hu c duy nht. T khi ca X so vi H2 bng 9,1. un nng X c xc tc Ni, sau khi phn ng xy ra hon ton, thu c hn hp kh Y khng lm mt mu nc brom; t khi ca Y so vi H2 bng 13. Cng thc cu to ca anken l: A. CH3CH=CHCH3. B. CH2=CHCH2CH3. C. CH2=C(CH3)2. D. CH2=CH2.Anken cng HBr to ra sp duy nht => anken i xng => Loi B v C Y khng lm mt mu nc Brom => Anken ht => H% = 100% AD Cng thc bi trn => M sau = Mtruoc. (x+y) / (x + y x.1) 13.2 = 9,1.2 .(x+y) / y 3,9y = 9,1x 3y = 7x => Chn x = 3 mol => y = 7 mol (Mo chn t l i nhau 3y => chn x = 3 ; 7x => chn y = 7) M trc = (mAnken + mH2) / (nAnken + nH2) 9,1.2 = (3.14n + 7.2) / (3+7) n = 4 => A (Hoc c th ly t l 3,9y = 9,1x => chn x =3,9 => y = 9,1 => l => a v nguyn nh trn co d

Cu 56: Cho hn hp X gm anken v hiro c t khi so vi heli bng 3,33. Cho X i qua bt niken nung nng n khi phn ng xy ra hon ton, thu c hn hp Y c t khi so vi heli l 4. CTPT ca X l: A. C2H4. B. C3H6. C. C4H8. D. C5H10.T khi vi Heli He2; M =4 Tng t bi 55 => 0,67y = 3,33x 67y = 333x => chn y = 333 => x =67 M trc = (67.14n + 333.2)/(67 + 333) = 3,33.4 n ~ 5 => D :C5H10 S liu hi l

Cu 57: Hn hp kh X gm H2 v C2H4 c t khi so vi He l 3,75. Dn X qua Ni nung nng, thu c hn hp kh Y c t khi so vi He l 5. Hiu sut ca phn ng hiro ho l: A. 20%. B. 25%. C. 50%. D. 40%.Chn 1 mol C2H4 => M trc = (28 + 2y) / (1 + y) = 3,75.4 y = 1 ADCT : M sau = M trc . (x + y)/(x + y x.H%) Vi x = y = 1 mol M trc = 15 ; M sau = 20 => H% = 50% => C

Cu 58: t chy hon ton a gam hn hp eten, propen, but-2-en cn dng va b lt oxi ( ktc) thu c 2,4 mol CO 2 v 2,4 mol nc. Gi tr ca b l: A. 92,4 lt. B. 94,2 lt. C. 80,64 lt. D. 24,9 lt.BT nguyn t Oxi : 2nO2 = 2nCO2 + nH2O nO2 = 3,6 mol => V = 80,64 lt => C

Cu 59: t chy hon ton V lt (ktc) hn hp X gm CH4, C2H4 thu c 0,15 mol CO2 v 0,2 mol H2O. Gi tr ca V l: A. 2,24. B. 3,36. C. 4,48. D. 1,68.BT nguyn t C , H vi x,y ln lt l s mol CH4 , C2H4 x + 2y = nCO2 ; 4x + 4y = 2nH2O x + 2y = 0,15 ; 4x + 4y = 0,4 x = y = 0,05 => nhh = x + y = 0,1 => V = 2,24 lt => A

Cu 60: t chy hon ton 0,1 mol hm hp gm CH 4, C4H10 v C2H4 thu c 0,14 mol CO2 v 0,23mol H2O. S mol ca ankan v anken trong hn hp ln lt l: A. 0,09 v 0,01. B. 0,01 v 0,09. C. 0,08 v 0,02. D. 0,02 v 0,08. Hn hp gm ankan v anken => nhh ankan = nH2O nCO2 = 0,09 mol => nAnken = n hh nankan = 0,1 0,09 = 0,01 => A Cu 61: Mt hn hp kh gm 1 ankan v 1 anken c cng s nguyn t C trong phn t v c cng s mol. Ly m gam hn hp ny th lm mt mu va 80 gam dung dch 20% Br2 trong dung mi CCl4. t chy hon ton m gam hn hp thu c 0,6 mol CO2. Ankan v anken c cng thc phn t l: A. C2H6 v C2H4. B. C4H10 v C4H8. C. C3H8 v C3H6. D. C5H12 v C5H10.mBr2 Cht tan = mdd . C% / 100% = 16 g => nAnken = nBr2 = 0,1 mol = nankan v ankan v anken cng s mol Mt khc Ankan v anken cng C => CT : ankan : CmnH2m+2 => anken : CmH2m BTNT C : m. nCnH2n+2 + m. nCnH2n = nCO2 m.0,1 +m.0,1 = 0,6 m = 3 => C3H8 v C3H6 => C

Cu 62: t chy hon ton 10ml hirocacbon X cn va 60 ml kh oxi, sau phn ng thu c 40 ml kh cacbonic. Bit X lm mt mu dung dch brom v c mch cacbon phn nhnh. CTCT ca X A. CH2=CHCH2CH3. B. CH2=C(CH3)2. C. CH2=C(CH2)2CH3. D. (CH3)2C=CHCH3.

X c mch C phn nhnh => Loi A . ADCT phn chuyn i cng CT : CxHy => x = VCO2 / VX = 40/10 = 4 => B v B c 4C Hoc tm y : x + y/4 = VO2 / VX 4 + y/4 = 6 y = 8 ci ny t => p n => CT : CnH2n 1 lin kt pi

Cu 63: Cho 0,2 mol hn hp X gm etan, propan v propen qua dung dch brom d, thy khi lng bnh brom tng 4,2 gam. Lng kh cn li em t chy hon ton thu c 6,48 gam nc. Vy % th tch etan, propan v propen ln lt l: A. 30%, 20%, 50%. B. 20%, 50%, 30%. C. 50%, 20%, 30%. D. 20%, 30%, 50%.Etan C2H6 ; propan C3H8 propen C3H6 => m tng = mC3H6= 4,2 g V ch c C3H6 p n C2H6 + nC3H8 = nHn hp nC3H6 = 0,2 0,1 = 0,1 => %VC3H6 = nC3H6.100%/ nhh = 50% kh i l C2H6 v C3H8 => BTNT H : => 6nC2H6 + 8nC3H8 = 2nH2O = 0,72 Gii h => nC2H6 = 0,04 ; nC3H8 = 0,06 => %V = n / nhn hp => D

Cu 64: Mt hn hp X gm 2 hirocacbon A, B c cng s nguyn t cacbon. A, B ch c th l ankan hay anken. t chy 4,48 lt (kc) hn hp X thu c 26,4 gam CO2 v 12,6 gam H2O. Xc nh CTPT v s mol ca A, B trong hn hp X. A. 0,1 mol C3H8 v 0,1 mol C3H6. B. 0,2 mol C2H6 v 0,2 mol C2H4. C. 0,08 mol C3H8 v 0,12 mol C3H6. D. 0,1 mol C2H6 v 0,2 mol C2H4.Ta c A,B cng s C => CT: CmH2m+2A ; CmH2mB A,B l ankan v anken m = nCO2 / nhh = 0,6 / 0,2 = 3 => C3H8 v C3H6 => B v D sai Th p n A vo tha mn iu kin 8nC3H8 + 6nC3H6 = 2nH2O => A ng Nu A sai => C ng Hoc c th gii h BTNTC , H => 3x + 3y = nCO2 ; 8x + 6y = 2nH2O => x = y = 0,1

Cu 65: Mt hn hp X gm 1 anken A v 1 ankin B, A v B c cng s nguyn t cacbon. X c khi lng l 12,4 gam, c th tch l 6,72 lt. Cc th tch kh o ktc. CTPT v s mol A, B trong hn hp X l: A. 0,2 mol C2H4 v 0,1 mol C2H2. B. 0,1 mol C3H6 v 0,1 mol C3H4. C. 0,2 mol C3H6 v 0,1 mol C3H4. D. 0,1 mol C2H4 v 0,2 mol C2H2.Anken v Ankin c cng C => CT A:CnH2n ; B: CnH2n 2 A l anken ; B l ankin Gi x , y l s mol ca A,B => m hn hp = 14n.x + (14n 2).y = 12,4 14n(x+y) 2y = 12,4 n hn hp = x + y = 0,3 mol => Th vo trn => 4,2n 2y = 12,4 4,2n = 12,4 + 2y n > 12,4/4,2 =2,95 hay n > 2,95 => p n => n = 3 => C3H6 v C3H4 => Loi A v D B sai v nhn hp = 0,2 # 0,3 => C

Cu 66: Mt hn hp A gm 2 hirocacbon X, Y lin tip nhau trong cng dy ng ng. t chy 11,2 lt hn hp X thu c 57,2 gam CO2 v 23,4 gam H2O. CTPT X, Y v khi lng ca X, Y l: A. 12,6 gam C3H6 v 11,2 gam C4H8. B. 8,6 gam C3H6v 11,2 gam C4H8. C. 5,6 gam C2H4 v 12,6 gam C3H6. D. 2,8 gam C2H4 v 16,8 gam C3H6. p n => CT dng : CnH2n Hoc da vo nCO2 = nH2O = 1,3 molCch 1.m tng p n => Ph hp n hn hp = 0,5 mol ; a . nCaH2a + b . nCbH2b = nCO2 BT nguyn t C => C ng Cch 2. Gii nhanh : ta c n = nCO2 / nhh = 1,3 / 0,5 = 2,6 => C2H4 v C3H6 => Loi A , B V 0,6nC2H4 = 0,4nC3H6 = 3nC2H4 = 2nC3H6 Xem phn chuyn 1 v cch xc nh t l da vo n => Chn nC2H4 = 2x => nC3H6 = 3x => n hn hp = 2x + 3x = 0,5 => x = 0,1 => nC2H4 = 0,2 ; nC3H6 = 0,3 => m => C ng

Cu 67: t chy hon ton 0,05 mol mt anken A thu c 4,48 lt CO2 (ktc). Cho A tc dng vi dung dch HBr ch cho mt sn phm duy nht. CTCT ca A l: A. CH2=CH2. B. (CH3)2C=C(CH3)2. C. CH2=C(CH3)2. D. CH3CH=CHCH3.Anken : CnH2n => n = nCO2 / nAnken = 4 => C4H8 ; A p vi HBr => to ra 1 sn phm => A l anken i xng => D tha mn

Cu 68: Hn hp X gm propen v B l ng ng theo t l th tch 1:1. t 1 th tch hn hp X cn 3,75 th tch oxi (cng k). Vy B l: A. eten. B. propan. C. buten. D. penten.T l th tch = t l s mol => nPropen = nB p n => A,C,D u l anken ui en => Xt Trng hp B l anken nu khng ng => B ng v A,C,D sai B c CT : CnH2n ; T l th V = t l s mol => Chn nX = 1 mol => nO2 = 3,75 mol ; nC3H6 = nB = 0,5 mol T l 1 : 1 Ta lun c nO2 = (x + y/4).nCxHy => 3,75 = (3+6/4).nC3H6 + (n+2n/4).nCnH2n 3,75 = 2,25 + (3n/2) . 0,5 n = 2 => C2H4 => eten => A

Cu 69: em t chy hon ton 0,1 mol hn hp X gm 2 anken l ng ng k tip nhau thu c CO2 v nc c khi lng hn km nhau 6,76 gam. CTPT ca 2 anken l: A. C2H4 v C3H6. B. C3H6 v C4H8. C. C4H8 v C5H10. D. C5H10 v C6H12. Ta lun c nCO2 = n .nAnken ; nH2O = n . nAnken C n H2 n + O2 => n CO2 + n H2O Hoc bo ton nguyn t => mCO2 mH2O = 44. n .0,1 18. n .0,1 = 6,76 => n = 2,6 => n = 2 v n =3 Lin tip =>A

Cu 70: X, Y, Z la 3 hirocacbon k tip trong day ng ng, trong o MZ = 2MX. t chay hoan toan 0,1 mol Y ri hp thu toan b san phm chay vao 2 lit dung dich Ba(OH)2 0,1M c mt lng kt tua la: A. 19,7 gam. B. 39,4 gam. C. 59,1 gam. D. 9,85 gam.Ta thy X,Y,Z => C2H4 ; C3H6 v C4H8 xem li cc bi trn 0,1 mol YC3H6 => 0,3 mol CO2 BTNT C ADCT : nCO32- BaCO3= nOH- Bazo - nCO2 = 2nBa(OH)2 nCO2 = 2.0,2 0,3 = 0,1 Xem phn chuyn 1 C CT cu 65 m kt ta = 19,7 g =>A

Cu 71: Chia hn hp gm C3H6, C2H4, C2H2 thnh hai phn u nhau. Phn 1: t chy hon ton thu c 2,24 lt CO2 (ktc). Phn 2: Hiro ho ri t chy ht th th tch CO2 thu c (ktc) l bao nhiu ? A. 1,12 lt. B. 2,24 lt. C. 4,48 lt. D. 3,36 lt.Phn 1 => nC(trong hn hp) = nCO2 thu c Phn 2=> nC (trong hn hp) = nCO2 thu c => nCO2 phn 1 = nCO2 phn 2 => V1 = V2 = 2,24 lt =>B Khi hidro ha ch lm thay i H P cng H2 => khng lam thay i C => BT nguyn t C khng i.

Cu 72: t chy hon ton 20,0 ml hn hp X gm C3H6, CH4, CO (th tch CO gp hai ln th tch CH4), thu c 24,0 ml CO2 (cc th tch kh o cng iu kin nhit v p sut). T khi ca X so vi kh H2 l: A. 12,9. B. 25,8. C. 22,2. D. 11,1T l th tch = t l s mol => 20 ml hn hp X => 24 ml CO2 20 mol X => 24 mol CO2 nCO = 2nCH4 Gi x,y l mol C3H6 v CH4 => nCO = 2y => n hn hp = nC3H6 + nCH4 + nCO = x + y + 2y = x + 3y = 20 BTNT C => 3nC3H6 + nCH4 + nCO = nCO2 3x + y + 2y = 24 Gii h => x = 2 ;y = 6 nC3H6 = 2 ; nCH4 = 6 ; nCO = 12 => m hn hp = mC3H6 + mCH4 + mCO = 2.42 + 6.16 + 12.26 = 516 M hn hp = m hn hp / n hn hp = 516 / 20 = 25,8 => T khi vi H2 = 25,8/2 = 12,9 =>A

Cu 73: t chy hon ton 0,1 mol anken X thu c CO 2 v hi nc. Hp th hon ton sn phm bng 100 gam dung dch NaOH 21,62% thu c dung dch mi trong nng ca NaOH ch cn 5%. Cng thc phn t ng ca X l: A. C2H4. B. C3H6. C. C4H8. D. C5H10.Anken => CT : CnH2n => nCO2 = 0,1.n mol CO2 p vi NaOH sau p thy d NaOH => CO2 ht ; NaOH d ( bi) P : CO2 + 2NaOH => Na2CO3 + H2O => nNaOH p = 2nCO2 = 0,2n => mNaOH p = 8n (g) m NaOH d = mNaOH ban u - mNaOH p = 100.21,62/100% - 8n = 21,62 8n Ta c m dd sau p => mH2O + mCO2 + mddNaOH = 1,8 n + 4,4n + 100 (V cho ton b sn phm vo) C% NaOH sau p = mNaOH d / mdd sau p 5% = (21,62-8n).100%/(4,4n + 1,8n +100) n = 2 => C2H4 (Chc chn n = 2 v 21,62 8n >0 => n < 2,7 n = 2 (Hoc c th th p n => A Dng ny c CT : C% = (mNaOH ban u 8 n ) / (6,2 n + mddNaOH)

Cu 74: X la hn hp gm hirocacbon A va O2 (ti l mol tng ng 1:10). t chay hoan toan X c hn hp Y. Dn Y qua binh H2SO4 c d c hn Z co ti khi so vi hiro la 19. A co cng thc phn t la: A. C2H6. B. C4H8. C C4H6. D. C3H6.T l mol 1:10 => Chn nA = 1 mol => nO2 = 10 mol t X => Y => cho Y qua H2SO4 c => H2O b H2SO4 hp th => hn hp Z l CO2 v O2 d V Nu ch c CO2 => M = 44 m M = 38 Gi a , b l mol CO2 v O2 d => M = m hn hp / n hn hp = (44a + 32b) / (a+b) = 38 6a = 6b a = b ; Ta lun c x . nCxHy = nCO2 BT nguyn tt C x = nCO2 => x = a = b Ta lun c nO2 p = (x + y/4) .nCxHy nO2 ban u nO2 d = (x + y/4).nCxHy 10 x = (x +y/4) 2x + y/4 = 10 => Th p n => B tha mn x = 4 v y = 8

Cu 75: m gam hn hp gm C3H6, C2H4 v C2H2 chy hon ton thu c 4,48 lt kh CO2 (ktc). Nu hiro ho hon ton m gam hn hp trn ri t chy ht hn hp thu c V lt CO2 (ktc). Gi tr ca V l: A. 3,36. B. 2,24. C. 4,48. D. 1,12.Xem bi 71 => C

Cu 76: Dn 1,68 lt hn hp kh X gm hai hirocacbon vo bnh ng dung dch brom (d). Sau khi phn ng xy ra hon ton, c 4 gam brom phn ng v cn li 1,12 lt kh. Nu t chy hon ton 1,68 lt X th sinh ra 2,8 lt kh CO 2. Cng thc phn t ca hai hirocacbon l (bit cc th tch kh u o ktc) A. CH4 v C2H4. B. CH4 v C3H4. C. CH4 v C3H6. D. C2H6 v C3H6.p n => Hn hp X cha ankan => cn li 1,12 lt = Vankan V ankan ko p => nAnkan = 0,05 mol Vcht cn li X = V hn hp Vankan = 1,68 1,12 = 0,56 lt => n X = 0,025 mol Ta c nX = nBr2 = 0,025 mol => X c k = 1 hay c CT : CnH2n ; Y l ankan : CmH2m+2 Da vo cn li 1,68 lt => 2,8 lt => m.nAnkan + n.nanken = 0,125 m.0,05 + n.0,025 = 0,125 2m + n = 5 => m = 1 v n = 3 hoc m = 2 ; n =1 Loi n = 1 v khng c CnH2n no c n =1 ; n 2 => m = 1 ; n = 3 => CH4 v C3H6 => C

Cu 77: Hn hp X gm C3H8 va C3H6 co ti khi so vi hiro la 21,8. t chay ht 5,6 lit X (ktc) thi thu c bao nhiu gam CO2 va bao nhiu gam H2O ? A. 33 gam va 17,1 gam. B. 22 gam va 9,9 gam. C. 13,2 gam va 7,2 gam.D. 33 gam va 21,6 gam.Gi x , y l mol C3H8 v C3H6 => nhn hp = nC3H8 + nC3H6 = x + y = 5,6/22,4 = 0,25 mol m hn hp = mC3H8 + mC3H6 = 44x + 42y = M hn hp . nhn hp = 21,8.2.0,25 GIi h => x = 0,2 ; y = 0,05 BTNT C => 3nC3H8 + 3nC3H6 = nCO2 = 3.0,2 + 3.0,05 = 0,75 mol => mCO2 = 33 g BTNT H => 8nC3H8 + 6nC3H6 = 2nH2O => . => mH2O = 17,1 g => A

Cu 78: Hin nay PVC c iu ch theo s sau: C2H4 CH2ClCH2Cl C2H3Cl PVC. Nu hiu sut ton b qu trnh t 80% th lng C2H4 cn dng sn xut 5000 kg PVC l: A. 280 kg. B. 1792 kg. C. 2800 kg. D. 179,2 kg.PVC : C2H3Cl SGK 11 nc 163 BT NT C => 2nC2H4 = 2nC2H3Cl nC2H4 = nC2H3Cl = 80mol mC2H4 theo PT = 80.28 = 2240 mol H% p = mPT . 100% / mTT 80% = 2240.100% / m TT => mTT = 2800 m Thc tt => C Xem li H% bi 35

Cu 79: Thi 0,25 mol kh etilen qua 125 ml dung dch KMnO4 1M trong mi trng trung tnh (hiu sut 100%) khi lng etylen glicol thu c bng A. 11,625 gam. B. 23,25 gam. C. 15,5 gam. D. 31 gam.PT: SGK11 nc 162 3C2H4 + 2KMnO4 + 4H2O => 3C2H4(OH)2 + 2MnO2 + 2KOH 0,25 0,125 => nC2H4 d tnh theo nKMnO4 v 0,25.2 > 0,125.3 => nC2H4(OH)2 = 3nKMnO4 /2 = 0,1875 mol => m = 0,1875.62 = 11,625 g => A

Cu 80: kh hon ton 200 ml dung dch KMnO4 0,2M to thnh cht rn mu nu en cn V lt kh C2H4 ( ktc). Gi tr ti thiu ca V l: A. 2,240. B. 2,688. C. 4,480. D. 1,344.Thm cht rn mu nu en MnO2 PT bi 79 => nC2H4 = 3nKMnO4 /2 = 0,4.3/2 = 0,6 mol => V = 1,344 lt => D

Cu 81: 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 Z, 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.X,Y , Z k tip nhau + MZ = 2MX => X,Y,Z c cng thc : CnH2n v n = 2;3;4 Duy nht => Z l C4H8 ; its 0,1 mol C4H8 => 0,4 mol CO2 BTNT C => nCaCO3 kt ta = nCO2 = 0,4 mol => m = 40 g => B

Cu 82: 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. 18,60 gam. B. 18,96 gam. C. 20,40 gam. D. 16,80 gam.Propan C3H8 propen C3H6 ; Propin C3H4 => Nhn thy cng s C => CT : C3Hy Ta c MY = 21,2.2 = 12.3 + y y = 6,4 PT : C3H6,4 + O2 => 3CO2 + 3,2H2O => nCO2 = 0,3 mol ; nH2O = 0,32 mol => Tng khi lng = mCO2 + mH2O = 0,3.44 + 0,32.18 = 18,96g

Cu 83: X la hn hp C4H8 va O2 (ti l mol tng ng 1:10). t chay hoan toan X c hn hp Y. Dn Y qua binh H2SO4 c d c hn Z. Ti khi ca Z so vi hiro la A.18. B. 19. C. 20. D. 21.Ngc li bi 74 : T l 1 : 10 => chn nC4H8 = 1 mol => nO2 = 10 mol nO2 p = (4 + 8/4) . nC4H8 = 6 mol => nO2 d = 4 mol nCO2 to thnh = 4nC4H8 = 4mol => M sau khi hp th = (mCO2 + mO2 d) / (nCO2 + nO2 d) = 38 T khi vi H2 = 19 => B

Cu 84: Hn hp X gm 2 anken kh phn ng va vi dung dch cha 48 gam brom. Mt khc t chy hon ton hn hp X dng ht 24,64 lt O2 (ktc). Cng thc phn t ca 2 anken l: A. C2H4 v C3H6. B. C2H4 v C4H8. C. C3H6 v C4H8. D. A v B u ng.Anken p vi Br2 => nhn hp Anken = nBr2 = 0,3 mol k=1 Xem li CT: nBr2 = k.nX vi k = s pi Gi cng thc trung ca hn hp 2 anken l C n H2 n

nO2 = ( n + 2n /4) . nC n H2 n nO2 = (x +y).nCxHy 1,1 = 3 n .0,3/ 2 n = 2,44 => A v B ng n nm gia 2 s C ca 2 cht => DCu 85: t chy mt s mol nh nhau ca 3 hirocacbon K, L, M ta thu c lng CO 2 nh nhau v t l s mol nc v CO2 i vi s mol ca K, L, M tng ng l 0,5 ; 1 ; 1,5. CTPT ca K, L, M (vit theo th t tng ng) l: A. C2H4, C2H6, C3H4. B. C3H8, C3H4, C2H4. C. C3H4, C3H6, C3H8. D. C2H2, C2H4, C2H6.

1C 11C 21A 31B 41A 51D 61C 71B 81B

2C 12B 22C 32D 42B 52A 62B 72A 82B

3B 13A 23C 33C 43C 53A 63D 73A 83B

4C 14D 24B 34D 44D 54C 64A 74B 84D

5D 15C 25D 35A 45D 55A 65C 75C 85D

6D 16A 26A 36A 46A 56D 66C 76C

7C 17A 27D 37B 47B 57C 67D 77A

8C 18B 28B 38C 48AB 58C 68A 78C

9D 19C 29A 39B 49A 59A 69A 79A

10D 20C 30D 40B 50D 60A 70A 80D

BI TP V ANKAIEN -TECPEN - ANKIN Cu 1: S ng phn thuc loi ankaien ng vi cng thc phn t C5H8 l A. 4. B. 5. C. 6. D. 7.

Ankandien => Ch ng phn hnh hc; C5H8 c k = 2 Ankandien => c 2 lin kt i hay 2 pi Ankandien lin hp v khng lin hp SGK 11 nc 166 ng phn: CH2 = C = CH CH2 CH3 ; => ko c p hnh hc => 1 CH2 = CH CH = CH CH3 ; => c p hnh hc ni i th 2=> 2 CH2 = CH CH2 CH =CH2 ; => ko c p hh => 1 CH3 CH=C=CH CH3 ; => ko c p hh => 1 CH2 = C =C(CH3)-CH3 => ko c p hh => 1 CH2=C(CH3)-CH=CH2 => ko c p hh => 1 => Tng c 7 => D

Cu 2: C5H8 c bao nhiu ng phn ankaien lin hp ? A. 2. B. 3.Lin hp => 2 ni i gn nhau . Cu 1 => CH2 = C = CH CH2 CH3 ; => ko c p hnh hc => 1 CH3 CH=C=CH CH3 ; => ko c p hh => 1 CH2 = C =C(CH3)-CH3 => ko c p hh => 1 => 3 p => B

C. 4.

D. 5.

Cu 3: Trong cc hirocacbon sau: propen, but-1-en, but-2-en, penta-1,4- ien, penta-1,3- ien hirocacbon cho c hin tng ng phn cis - trans ? A. propen, but-1-en. B. penta-1,4-dien, but-1-en. C. propen, but-2-en. D. but-2-en, penta-1,3- ien.But 1 en CH2=CH-CH2-CH3 ko c p hnh hc => Loi A v B Propen khng c ng phn hnh hc : CH2 = CH CH3 ko c => Loi C => D But 2n : CH3 CH=CH CH3 ; Penta 1,3 ien : CH2 = CH CH = CH CH3 p hnh hc lin kt pi th 2=> D

Cu 4: Cng thc phn t ca buta-1,3-ien (ivinyl) v isopren (2-metylbuta-1,3-ien) ln lt l A. C4H6 v C5H10.B. C4H4 v C5H8. C. C4H6 v C5H8. D. C4H8 v C5H10.Thy c 2 cht u c ui ien => k = 2 2 lin kt pi => CnH2n 2 => C tha mn Buta 1,3 ien : CH2 = CH CH = CH2 ; 2 metylbuta 1,3 ien => C4H6 CH2 = C(CH3) CH = CH2 => C5H8

Cu 5: Hp cht no trong s cc cht sau c 9 lin kt xch ma v 2 lin kt ? A. Buta-1,3-ien. B. Penta-1,3- ien. C. Stiren. D. Vinyl axetilen.Xem li cu 7 phn anken => Cng thc tnh lin kt xch ma ; 2 lin kt pi => CT : CnH2n-2 Xem li phn tm CT 2 cch chuyn 1 Lin kt xch ma = s C + s H 1 = 9 s C + s H = 8 = n + 2n 2 = 10 n = 4 => C4H6 => A D c 3 lin kt pi :CH2=CH-C=*CH Ch =* l ni 3 Vinyl : CH2=CH

Cu 6: Hp cht no trong s cc cht sau c 7 lin kt xch ma v 3 lin kt ? A. Buta-1,3-ien. B. Tuloen. C. Stiren. D. Vinyl axetilen.Tng t bi 5: 3 lin kt pi => CT : CnH2n 4 ; => S lin kt xch ma = n + 2n 4 - 1 = 7 n = 4 => D:C4H4

Cu 7: Cho phn ng gia buta-1,3-ien v HBr -80oC (t l mol 1:1), sn phm chnh ca phn ng l A. CH3CHBrCH=CH2. B. CH3CH=CHCH2Br. C. CH2BrCH2CH=CH2.D. CH3CH=CBrCH3. nhit -80 oC => Br C bc cao ; v sn phm cng v tr 1,2 SGK 11nc 167 1(I) 2(III) CH2=CH CH=CH2 + HBr => CH3-CHBr CH=CH2 => A 1 2 3 4 Ch cch cng : R C = CR CR = C R => cng vo v tr 1,2 hoc 1 ,4 v cht phi c dng nh zy R c th l H hoc hidrocabon ; halogen VD: CH3 C=C(CH3) C(C2H5)=C-C3H7 ; CH2=CH CH =CH2 ; . Mnh hiu l cng vo v tr 1,2 ni i 1 hoc 1,4 ni i 1 v ni i 4 a lin kt i vo trong 1,2,3,4 l v tr C cha lin kt i Ch khng phi nh s th t C

Cu 8: Cho phn ng gia buta-1,3-ien v HBr 40oC (t l mol 1:1), sn phm chnh ca phn ng l A. CH3CHBrCH=CH2. B. CH3CH=CHCH2Br. C. CH2BrCH2CH=CH2.D. CH3CH=CBrCH3. nhit 40 oC => Br C bc cao v sn phm ng v tr 1,4 v chuyn ni i vo trongSGK 11nc 167 CH2=CH CH=CH2 + HBr => CH3-CH=CH-CH2Br => B

1 2 3 4 Ch c p cng 1,4 => Cht c dng R C =C C =C R 2 lin kt i cch nhau 1 v tr Cu 9: 1 mol buta-1,3-ien c th phn ng ti a vi bao nhiu mol brom ? A. 1 mol. B. 1,5 mol. C. 2 mol. D. 0,5 mol.

Buta 1,3 ien => 2 lin kt pi => ADCT : nBr2 = k.nX => nBr2 = 2nX = 2mol k l tng s pi =>C

Cu 10: Isopren tham gia phn ng vi dung dch Br2 theo t l mol 1:1 to ra ti a bao nhiu sn phm ? A. 4. B. 1. C. 3. D. 2.Isopren : CH2 =C(CH3)-CH=CH2 SGK 11nc 168 V cho l ti a => Br2 cng vo ni i 1 => 1 Br2 cng vo ni i 2 => 1 Br2 cng vo c 2 ni i => 1 Br2 cng vo v tr 1,4 =>1 CH2Br C(CH3)=CH-CH2Br => 4 sn phm Bi ny mnh ang thc mc => c trng hp phn ng th vi gc CH3 khng nh - khng bo l phn ng cng ch c t l 1 : 1

Cu 11: Isopren tham gia phn ng vi dung dch HBr theo t l mol 1:1 to ra ti a bao nhiu sn phm cng ? A. 8. B. 5. C. 7. D. 6.Phn ng vi dung dch HBr theo t l 1 :1 => sn phm cng => Ch cng 1 HBr CH2 =C(CH3)-CH=CH2 => 1 ni i => 2 sn phm chnh v ph => 2 ni i c 4 sn phm Cng theo quy tc macopnhicop SGK 11nc 161 Cng vo v tr 1,2 ging 2 trng hp trn=> khng Cng vo v tr 1,4 c 2 sn phm ; BrCH2-C(CH3)=CH-CH3 ng phn hnh hc CH3 C(CH3)=CH-CH2Br => 3 => Tng = 4 + 3 = 7 =>C

Cu 12: Cht no sau y khng phi l sn phm cng gia dung dch brom v isopren (theo t l mol 1:1) ? A. CH2BrC(CH3)BrCH=CH2. B. CH2BrC(CH3)=CHCH2Br. C. CH2BrCH=CHCH2CH2Br. D. CH2=C(CH3)CHBrCH2Br. Cu 13: Ankaien A + brom (dd) CH3C(CH3)BrCH=CHCH2Br. Vy A l A. 2-metylpenta-1,3-ien. B. 2-metylpenta-2,4-ien. C. 4-metylpenta-1,3-ien. D. 2-metylbuta-1,3-ien.Sn phm CH3 CBr(CH3)CH = CH CH2Br sn phm cng v tr 1,4V ni i gia 5 4 3 2 1 cht A ; CH3 C(CH3)=CH CH=CH2 => 4 metylpenta 1,3 ien cch gi tn s ch v tr mch nhnh mch chnh s ch v tr ien Vi cch nh s C gn lin kt i nht ien ch c 2 lin kt i tr ln Iso pren c nhnh => C khng tha mn

Cu 14: Ankaien B + Cl2 CH2ClC(CH3)=CH-CH2Cl-CH3. Vy A l A. 2-metylpenta-1,3-ien. B. 4-metylpenta-2,4-ien. C. 2-metylpenta-1,4-ien. D. 4-metylpenta-2,3-ien.1 2 3 4 5 Ging 13 cng vo 1,4 => CH3 =C(CH3)-CH=CH-CH3 => 2 metyl penta 1,3 ien => A

Cu 15: Cho 1 Ankaien A + brom(dd) 1,4-ibrom-2-metylbut-2-en. Vy A l A. 2-metylbuta-1,3-ien. C. 3-metylbuta-1,3-ien. B. 2-metylpenta-1,3-ien. D. 3-metylpenta-1,3-ien.1 2 3 4 1,4 ibrom 2 metylbut 2 en => CH2Br C(CH3)=CH CH2Br cng vo v tr 1,4 1 2 3 4 => A : CH2 = C(CH3)-C=CH2 => 2 metylbuta 1,3 ien => A

Cu 16: Trng hp ivinyl to ra cao su Buna c cu to l ? A. (-C2H-CH-CH-CH2-)n. B. (-CH2-CH=CH-CH2-)n. C. (-CH2-CH-CH=CH2-)n. D. (-CH2-CH2-CH2-CH2-)n.ivinyl hay 2vinyl Vinyl : CH2 =CH- => ivinyl : CH2=CH-CH=CH2 trng hp => (-CH2-CH=CH-CH2-)n => B Trng hp l tch ht ni i thnh ni n ri ni vo nhau VD: CH2 = CH2 => tch ni i : -CH2 - CH2 => -CH2 CH2CH2=CH-CH = CH2 ; tch ; CH2=CH => -CH2-CH - .Tch CH = CH2 => - CH CH2 - => ni vi nhau => -CH2 CH = CH CH2 -

Cu 17: ng trng hp ivinyl v stiren thu c cao su buna-S c cng thc cu to l ( P SGK 11 nc 195 ) A. (-CH2-CH=CH-CH2-CH(C6H5)-CH2-)n. B. (-C2H-CH-CH-CH2-CH(C6H5)-CH2-)n. C. (-CH2-CH-CH=CH2- CH(C6H5)-CH2-)n. D. (-CH2-CH2-CH2-CH2- CH(C6H5)-CH2-)n .Stiren SGK 11 nc 194 C6H5-CH=CH2 ; ivinyl : CH2=CH-CH=CH2 P SGK 11 nc 195

P ng trng hp Tch ht cc lin kt i ban u ri ni vi nhau => A

Cu 18: ng trng hp ivinyl v acrylonitrin (vinyl xianua) thu c cao su buna-N c cng thc cu to l A. (-C2H-CH-CH-CH2-CH(CN)-CH2-)n. B. (-CH2-CH2-CH2-CH2- CH(CN)-CH2-)n. C. (-CH2-CH-CH=CH2- CH(CN)-CH2-)n. D. (-CH2-CH=CH-CH2-CH(CN)-CH2-)n .ivinyl : CH2=CH2-CH=CH2 ; Vinyl xiannua : CN - CH=CH2 => ng trng hp => (-CH2-CH2=CH-CH2-CH(CN)-CH2-)n => D

Cu 19: Trng hp isopren to ra cao su isopren c cu to l A. (-C2H-C(CH3)-CH-CH2-)n . B. (-CH2-C(CH3)=CH-CH2-)n.

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

SGK 11nc 198 =>B Isopren : CH2=CH(CH)3-CH=CH2 => (-CH2-CH(CH3)=CH-CH2-)n => B

Cu 20: Tn gi ca nhm hirocacbon khng no c cng thc chung l (C5H8)n (n 2) l A. ankaien. B. cao su. C. anlen. D. tecpen.SGK 11 nc 171.

Cu 21: Caroten (licopen) l sc t mu ca c rt v c chua chn, cng thc phn t ca caroten l A. C15H25. B. C40H56. C. C10H16. D. C30H50.SGK 11 Nng cao-171

Cu 22: Oximen c trong tinh du l hng qu, limonen c trong tinh du chanh. Chng c cng cng thc phn t l A. C15H25. B. C40H56. C. C10H16. D. C30H50.SGK Ha hc 11 Nng cao-171

Cu 23: C4H6 c bao nhiu ng phn mch h ? A. 5. B. 2.

C. 3.

D. 4.

(CH2=C=CH-CH3; CH2= CH-CH=CH2; CHC-CH2-CH3 ; CH3-CC-CH3. CT CxHyOzNtCluNav khng no=(2x-y+t-u-v+2)/2. khng no ca C4H6 l 2.==> TH1:0 vng,2 lk i; TH2:0 vng,1 lk ba;TH3:1 vng,1 lk i;v l mch h nn ch xy ra TH1 v TH2,sau dch chuyn v tr ca cc ni i, ba to ra ng phn.)

Cu 24: C bao nhiu ankin ng vi cng thc phn t C5H8 ? A. 1. B. 2. C. 3. D. 4 (CHC-CH2-CH2-CH3; CH3-CC-CH2-CH3; CH3-CH2-CC-CH3) Cu 25: Ankin C4H6 c bao nhiu ng phn cho phn ng th kim loi (phn ng vi dung dch cha AgNO3/NH3) A. 4. B. 2. C. 1. D. 3.(CH3-CH2-CCH .Nguyn t H nh vo C mang lk ba linh ng hn rt nhiu so vi H nh vi C mang lk i, n,nn c th b thay th bng nguyn t KL.Nhng ch xy ra cc ankin co lk ba u mch R-CH)

Cu 26: C bao nhiu ng phn ankin C5H8 tc dng c vi dung dch AgNO3/NH3 to kt ta A. 3. B. 2. C. 4. D. 1.Nh cu 25 => cht c dng R CH (CH3-CH2-CH2-CCH , CH3-CH(CH3)-CCH )

Cu 27: Ankin C6H10 c bao nhiu ng phn phn ng vi dung dch AgNO3/NH3 ? A. 3. B. 4. C. 5.Cu 25: (CH3-CH2-CH2-CH2-CCH ; CH3 CH(CH3) CH2 CCH ; CH3 CH2 CH(CH3) CCH ; CH3 (CH3)C(CH3) CCH)

D. 6.

Cu 28: Trong phn t ankin X, hiro chim 11,111% khi lng. C bao nhiu ankin ph hp A. 1. B. 2. C. 3. D. 4(CT ankin CxHy.% C = 100-11,111=88,889%.x:y=88,889/12 : 11,111/1=2:3==> (C2H3)n => n =2 hay C4H6 ;

Cu 29: Cho ankin X c cng thc cu to sau : Tn ca X l A. 4-metylpent-2-in. B. 2-metylpent-3-in.

CH3C C CH CH3 CH3C. 4-metylpent-3-in. D. 2-metylpent-4-in.

S ch v tr Tn nhnh / tn mch chnh/ - s ch v tr in . Mch chnh l mch c lk i, di nht v c nhiu nhnh nht.nh s C mch chnh bt u t pha gn lk i hn.S ch v tr lk i ghi ngay trc ui in)

Cu 30: Cho phn ng : C2H2 + H2O A l cht no di y A. CH2=CHOH. B. CH3CHO. P SGK 11 nc - 177

A C. CH3COOH. D. C2H5OH.

(p cng H2O ca ankin:H2O cng vo lk ba to ra hp cht trung gian khng bn v chuyn thnh andehit hoc xeton) Quy tc h bin ca ru c OH gn vi C lin kt i c dng R CH=CH-OH , R- C(OH)=CH2 s b chuyn thnh andehit hoc xeton . R-CH=CH-OH => R CH2 CHO ; R- C(OH)=CH2 => R C(O) CH3

Cu 31: Cho s phn ng sau: X c cng thc cu to l?

CH3-CCH + AgNO3/ NH3 X + NH4NO3

A. CH3-CAgCAg.

B. CH3-CCAg.

C. AgCH2-CCAg.

D. A, B, C u c th ng.

(p th bng ion KL ca ankin:nguyn t H nh vo C mang lk ba b thay th bng nguyn t KL Ag) Ag ch th vo H lin kt vi C ni 3 u mch

Cu 32: Trong s cc hirocacbon mch h sau: C4H10, C4H6, C4H8, C3H4, nhng hirocacbon no c th to kt ta vi dung dch AgNO3/NH3 ? A. C4H10 ,C4H8. B. C4H6, C3H4. C. Ch c C4H6. D. Ch c C3H4.(RH c th to kt ta vi dd AgNO3/NH3 l ankin loi C4H10 (ankan) v C4H8 (anken hoc xicloankan) >>> B

Cu 33: Hn hp A gm hiro v cc hirocacbon no, cha no. Cho A vo bnh c niken xc tc, un nng bnh mt thi gian ta thu c hn hp B. Pht biu no sau y sai ? A. t chy hon ton hn hp A cho s mol CO2 v s mol nc lun bng s mol CO2 v s mol nc khi t chy hon ton hn hp B. B. S mol oxi tiu tn t hon ton hn hp A lun bng s mol oxi tiu tn khi t hon ton hn hp B. C. S mol A - S mol B = S mol H2 tham gia phn ng. D. Khi lng phn t trung bnh ca hn hp A bng khi lng phn t trung bnh ca hn hp B.A ng v bo ton nguyn t C v H trc v sau p . Ta c hn hp A p to thnh hn hp B => Tng s H , C trong hn hp A = Tng s H,C trong hn hp B V hn hp A ch c H v C , nh H2 l H , hidrocacbon no , ko n cng cha H v C B ng . nu ta gi CT tng qut ca hn hp A l CxHy V thnh phn ch cha C , H => hn hp B cng l CxHy Bo ton nguyn t trc v sau p => u t chy cng mt lng O2. >>>>>>>>>>.C ng.

Cu 34: Cht no trong 4 cht di y c th tham gia c 4 phn ng: Phn ng chy trong oxi, phn ng cng brom, phn ng cng hiro (xc tc Ni, to), phn ng th vi dd AgNO3 /NH3 A. etan. B. etilen. C. axetilen. D. xiclopropan.(mi cht hu c u tham gia p chy trong O2.ankan khng tham gia p cng Br loi A.etilen v xiclopropan khng tham gia p th vi AgNO3//NH3 loi B,D.ch c axetilen c th tham gia c 4 p :C)

Cu 35: Cu no sau y sai ? A. Ankin c s ng phn t hn anken tng ng. B. Ankin tng t anken u c ng phn hnh hc. C. Hai ankin u dy khng c ng phn. D. Butin c 2 ng phn v tr nhm chc. (ankin khng c ng phn hnh hc v k c lk i khng phi lk ba) Cu 36: Cho cc phn ng sau:(1) C H4 + C2 l as kt 1:1

(2) C2H4 + H2

(3) 2 CHCH

+ H2O (4) 3 CHCH (5) C2H2 + Ag2O (6) Propin S phn ng l phn ng oxi ho kh l: A. 2. B. 3. C. 4. D. 5.(1) (2) (3) (4) (5) (6)CH4 + Cl2 CH3Cl + HCl C2H4 + H2 C2H6 2 CHCH CH2=CH-CCH 3 CHCH C6H6 C2H2 + Ag2O AgCCAg+H2O Propin + H2O C2H5CHO Xem p no c s thay i s OXH l p OXH-K => 1 , 2 ,3 , 6 => C : 4 Cch xc nh s oxi ha C trong cht hu c. (Cc s oxi ha ca cc cht O , H , N , halogen th vn vy) + Trong hp cht hu c th tch ring tng nhm Cn ra tnh VD : CH3 CH2 CH(CH3) CH3 => CH3 | CH2 | CH | CH3 | CH3 => -3|-2|-1|-3|-3 + Nu nhm chc khng cha C (halogen , -OH , -O-,NH2) thnh tnh s Oxihoa C gn c nhm chc. VD: CH3 CH(Br)-CH3 => CH3 | CHBr | CH3 => -3 | 0 | -3 CH3 CH2 CH2OH => CH3 | CH2 | CH2OH => -3 | -2 | -1 CH3 O CH2 CH3 => CH3 O| O CH2 | CH3 => -2 | -1 | -3 + Nu nhm chc c C th tnh ring. VD : CH3 CHO => CH3 | CHO => -3 | +1 CH3 COOH => CH3 | COOH => -3 | +3 P 1 thy Cl2 => HCL Cl0 + e => Cl-1 ; C-4 -2e => C-2 CH3CL P 2 thy H2 => C2H6 => H0 - e => H+1 ; C-2 + e=> C-3

P 3 thy 2CHCH => CH2=CH-CCH : C-1 + e => C-2 ; C-1 - e => C0 P 6 thy propin CH3-CCH => -3 | 0 | -1 => C2H5CHO => CH3 CH2 CHO => -3 | -2 | +1 ; C0 + 2e => C-2 ; C-1 - 2e => C+1 => 4 p. => C

Cu 37: Cho dy chuyn ho sau: CH4 A B C Cao su buna. Cng thc phn t ca B l A. C4H6. B. C2H5OH. C. C4H4. D. C4H10.Caosu buna => (-CH=CH CH=CH-)n => C l C4H6 => Loi A Thm mt st pAl2 O3 2C2 H5 OH CH 2 =CH-CH=CH 2+2H 2O+H 2 450o CPd CH 2 =CH-C=*CH+H 2 CH 2 =CH-CH=CH 2

Na, C4H10 CH =CH-CH=CH +2H2 t 2 2

0

Ch ny phi linh hot cht khng b la nu cho c 3 p n. Da vo A => B ; to thnh C2H5OH => A l C2H4 hoc C2H5X X l halogen => to thnh C2H4 t CH4 khng c p no to thnh C4H4 vinylaxetilen => A l C2H2 hoc C4H8 M t 2CH4 => C2H2 + 3H2 P 1500 0C lm lnh nhanh => C4H4 ng => C C4H10 => A l C4H6 , C4H8 nhng CH4 khng th iu ch c.

Cu 38: C chui phn ng sau: KH O B HCl N + H2 D E (spc) D Xc nh N, B, D, E bit rng D l mt hidrocacbon mch h, D ch c 1 ng phn. A. N : C2H2 ; B : Pd ; D : C2H4 ; E : CH3CH2Cl. B. N : C4H6 ; B : Pd ; D : C4H8 ; E : CH2ClCH2CH2CH3. C. N : C3H4 ; B : Pd ; D : C3H6 ; E : CH3CHClCH3. D. N : C3H4 ; B : Pd ; D : C3H6 ; E : CHCH2CH2Cl. bi => D ch c 1 ng phn => Loi B v D: C4H8 to ra do p E p => CH2=CH-CH2 CH2 ; CH3 CH = CH CH3 ng phn hnh hc) E l sn phm chnh .=> Loi D v p CH2 = CH CH3 (C3H6) + HCL => spc CH3 CHCL CH3 , sn phm ph l CH2CL CH2 CH3. Loi A v E l sn phm chnh nu D l C2H4 s to ra mt sn phm. P C2H4 + HCL => C2H5CL C ng

Cu 39: Cht no sau y khng iu ch trc tip c axetilen ? A. Ag2C2. B. CH4. C. Al4C3.Ag2C2 + 2HCl C2H2 + 2AgCl 2CH4 C2H2 + 3 H2 k 1500 oC, lm lnh nhanh CaC2 + 2H2O Ca(OH)2 + C2H2 Al4C3 +12H2O 4Al(OH)3 + 3CH4 2CH4 C2H2 + 3 H2 => C

D. CaC2.

Cu 40: lm sch etilen c ln axetilen ta cho hn hp i qua dd no sau y ? A. dd brom d. B. dd KMnO4 d. C. dd AgNO3 /NH3 d. D. cc cch trn u ng.(anken,ankin tham gia p cng halogen(Br), p OXH (KMnO4):lm mt mu thuc tm loi A,B,D.Ankin c th t/d vi dd AgNO3/NH3 d cn an ken th khng :C)

Cu 41: nhn bit cc bnh ring bit ng cc kh khng mu sau y: SO2, C2H2, NH3 ta c th dng ho cht no sau y A. Dung dch AgNO3/NH3.B. Dung dch Ca(OH)2 C. Qu tm m. D. Dung dch NaOH Cu 42: X l mt hirocacbon kh ( ktc), mch h. Hiro ho hon ton X thu c hirocacbon no Y c khi lng phn t gp 1,074 ln khi lng phn t X. Cng thc phn t X l A. C2H2. B. C3H4. C. C4H6. D. C3H6.p n => A,B,C u l c dng CnH2n-2 Xt A,B,C nu sai th => D ng PT : CnH2n-2 + 2nH2 => CnH2n+2 Hidro ha l p cng H2 vo lin kt pi + Thu c hidrocabon no Chn 1 mol CnH2n-2 => to thnh 1 mol CnH2n+2 => 14n + 2 = 1,074(14n-2) n = 4 => C4H6 => C

Cu 43: Cht hu c X c cng thc phn t C6H6 mch thng. Bit 1 mol X tc dng vi AgNO3 d trong NH3 to ra 292 gam kt ta. CTCT ca X c th l A. CH CCCCH2CH3. C. CHCCH2CH=C=CH2. B. CHCCH2CCCH3. D. CHCCH2CH2CCH.C6H6 c k = (2.6 -6 +2)/2 = 4 pi => cc p n u tha mn iu kin to ra kt ta => X c dng RCH + [Ag(NH3)2]OH => RCAg + 2H2O + 2NH3 P SGK 11nc 177

Thc t l th Ag vo H Ta lun c nRCH = nRCAg Mo gii nhanh V M kt ta = MX + 108 1 = MX + 107 TH1 th 1H M kt ta = MX + 2.108 2 = MX + 214 TH2 Th 2H Ta c M C6H6 = 78 V M kt ta = 292 => Tha mn TH2 => Th 2H => X c dng HC C RCH Tng qut l c 2 ni 3 C u v cui => D tha mn :CHCCH2CH2CCH

Cu 44: Mt hirocacbon A mch thng c CTPT C 6H6. Khi cho A tc dng vi dung dch AgNO 3/NH3 d thu c hp cht hu c B c MB - MA=214 vC. Xc nh CTCT ca A ? A. CHCCH2CH2CCH. B. CH3C CCH2CCH. C. CHCCH(CH3)CCH. D. CH3CH2CCCCH.A l mch thng => Loi C Ta c MB MA = 214 =>TH2 => C 2 ni 3 C u v cui => A

Cu 45: A la hirocacbon mach h, th khi (kt), bit A 1 mol A tac dung c ti a 2 mol Br2 trong dung dich to ra hp cht B (trong B brom chim 80 % v khi lng. Vy A co cng thc phn t la A. C5H8. B. C2H2. C. C4H6. D. C3H4.1 mol p ti a 2 molBr2 => k =2 => CT A: CnH2n-2 Hoc t p n PT : CnH2n-2 +2Br2 => CnH2n-2Br2 => %Br = 160.100% / (14n 2 + 160) = 80% n = 3 => C3H4 =>D

Cu 46: 4 gam mt ankin X c th lm mt mu ti a 100 ml dung dch Br2 2M. CTPT X l A. C5H8 . B. C2H2. C. C3H4. D. C4H6.Ankin :CnH2n-2 => k= 2=> 2nX = nBr2 nX = 0,1 mol => MX = 40 = 14n 2 n = 3 => C3H4

Cu 47: X l mt hirocacbon khng no mch h, 1 mol X c th lm mt mu ti a 2 mol brom trong nc. X c % khi lng H trong phn t l 10%. CTPT X l A. C2H2. B. C3H4. C. C2H4. D. C4H6.1 mol p 2mol Br2 =>CT X : CnH2n-2 => %H = (2n-2).100%/(14n-2) = 10% n = 3 =>C3H4

Cu 48: X l hn hp gm 2 hirocacbon mch h (thuc dy ng ng ankin, anken, ankan). Cho 0,3 mol X lm mt mu va 0,5 mol brom. Pht biu no di y ng A. X c th gm 2 ankan. B. X c th gm2 anken. C. X c th gm1 ankan v 1 anken. D. X c th gm1 anken v mt ankin.Gi k1,k2 ln lt l s pi ca cht A v B v x , y ln lt l s mol ca A ,B x + y = 0,3 ; k1.x + k2.y = 0,5 Xt A. A,B u l ankan => k1,k2 = 0 => Sai v k1.x + k2.y = 0,5 Xt B. Gm 2 anken => k1 = k2 = 1 => Gii h v nghim => loi Xt C. A l ankan , B l anken => k1 = 0 ; k2 = 1 ; gii h v nghim => loi Xt D. Anken v ankin => k1 = 1 ; k2 = 2 ; gii h => x , y => tha mn =>D Ngoi ra 2 ankin cng loi

Cu 49: Hn hp X gm 1 ankin th kh v hiro c t khi hi so vi CH 4 l 0,425. Nung nng hn hp X vi xc tc Ni phn ng hon ton thu c hn hp kh Y c t khi hi so vi CH4 l 0,8. Cho Y i qua bnh ng dung dch brom d, khi lng bnh tng ln bao nhiu gam ? A. 8. B. 16. C. 0. D. Khng tnh c. Cu 50: Hn hp A gm C2H2 v H2 c dA/H2 = 5,8. Dn A (ktc) qua bt Ni nung nng cho n khi cc phn ng xy ra hon ton ta c hn hp B. Phn trm th tch mi kh trong hn hp A v dB/H2 l A. 40% H2; 60% C2H2; 29. B. 40% H2; 60% C2H2 ; 14,5. C. 60% H2; 40% C2H2 ; 29. D. 60% H2; 40% C2H2 ; 14,5.Xem bi 53 phn anken => Chn 1 mol C2H2 = x => M hn hp A = mA/n hn hp A 5,8.2 = (26 + 2y) / (1+y) y = 1,5 mol ;H% = 100 P hon ton => %C2H2 = x / (x+y) = 1 / (1 + 1,5) = 40% => %H2= 60% PT : C2H2 + 2H2 => C2H6 Ban u 1 mol 1,5mol P 0,75mol 0,75 mol Sau p 0,25mol 0,75 mol n sau p = nC2H2 d + nC2H6 to thnh = 0,25 + 0,75 = 1 mol m trc = msau = mC2H2 + mH2 = 26 + 1,5.2 = 29 g M sau = 29 => T khi vi H2 = 29/2 = 14,5 =>D

Cu 51: Mt hn hp gm etilen v axetilen c th tch 6,72 lt (ktc). Cho hn hp qua dung dch brom d phn ng xy ra hon ton, lng brom phn ng l 64 gam. Phn % v th tch etilen v axetilen ln lt l A. 66% v 34%. B. 65,66% v 34,34%.

C. 66,67% v 33,33%.

D. Kt qu khc.

Etilen C2H4 ; k = 1 Axetilen C2H2 ; k=2 Gi x , y ln lt l s mol C2H4 ; C2H2 n hn hp = x + y = 0,3 mol ; nBr2 = nC2H4 + 2nC2H2 x + 2y = 0,4 Gii h =>x =0,2 ; y = 0,1 => %VC2H4 = x / (x+y) = 0,2 / 0,3 = 66,67% => %C2H2 = 33,33% =>C

Cu 52: Cho 10 lt hn hp kh CH4 v C2H2 tc dng vi 10 lt H2 (Ni, to). Sau khi phn ng xy ra hon ton thu c 16 lt hn hp kh (cc kh u o cng iu kin nhit p sut). Th tch ca CH4 v C2H2 trc phn ng l A. 2 lt v 8 lt. B. 3 lt v 7 lt. C. 8 lt v 2 lt. D. 2,5 lt v 7,5 lt.Gi x , y ln lt l V CH4 v C2H2 => x + y = 10 ; Ch c C2H2 mi p vi H2 C2H2 + 2H2 => C2H6 Ban u y lt 10 lt P y => 2y => y Sau p 10 2y y lt V sau khi p => x + 10 2y + y = 16 x y = 6 Gii h => x = 8 v y = 2 => C

Cu 53: Cho 28,2 gam hn hp X gm 3 ankin ng ng k tip qua mt lng d H2 (to, Ni) phn ng xy ra hon ton. Sau phn ng th tch th tch kh H2 gim 26,88 lt (ktc). CTPT ca 3 ankin l A. C2H2, C3H4, C4H6. B. C3H4, C4H6, C5H8. C. C4H6, C5H8, C6H10. D. C A, B u ng.Th tch H2 gim = th tch H2 p = 26,88 lt => nH2 p = 1,2 mol => nhn hp = nH2 / 2 = 0,6 mol M hn hp = mhn hp / n hn hp 28,2/ 0,6 = 47 = 14 n - 2 C n H2 n -2 ankin n = 3,5 => Loi C v c 3 cht u c s C > 3,5 ; A, B tha mn nm gia => D

Cu 54: Hn hp X gm propin v mt ankin A c t l mol 1:1. Ly 0,3 mol X tc dng vi dung dch AgNO3/NH3 d thu c 46,2 gam kt ta. A l A. But-1-in. B. But-2-in. C. Axetilen. D. Pent-1-in.n Propin = nA = 0,15 mol T l 1 :1 + nX = 0,3 mol Xem li bi 43. C3H4 ; CH3-CCH => CH3-CCAg => m kt ta to thnh do p A = 46,2 mCH3-CCAg = 46,2 0,15.147 = 24,15 g => M kt ta = 24,15/0,15 = 161 = MX + 107 TH1 => MX = 54 = 14n 2 Ankin :CnH2n-2 n = 4 => C4H8 => A V tha mn iu kin to kt ta R CH CHC-CH2-CH3 : but 1 in

Cu 55: Trong bnh kn cha hirocacbon X v hiro. Nung nng bnh n khi phn ng hon ton thu c kh Y duy nht. cng nhit , p sut trong bnh trc khi nung nng gp 3 ln p sut trong bnh sau khi nung. t chy mt lng Y thu c 8,8 gam CO2 v 5,4 gam nc. Cng thc phn t ca X l A. C2H2. B. C2H4. C. C4H6. D. C3H4.nCO2 = 0,2 mol ; nH2O = 0,3 mol => Y cha ankan nH2O > nCO2 n = nCO2 / (nH2O nCO2) = 0,2 / (0,3 0,2) = 2 => C2H6 Kh Y kh duy nht Ta c p sut trc = 3 p sut sau + cng nhit + bnh kn Th tch khng i n hn hp trc = 3 n hn hp sau v n = P.V/T.0,082 M p xy ra hon ton thu c 1 sn phm => cc cht tham gia p ht Ch c Ankin : CnH2n- 2 + 2H2 => CnH2n+2 mi tha mn iu kin v xmol => 2x mol => x mol n trc = nCnH2n-2 + nH2 = 3x ; n sau = x => n trc = 3n sau ,m n = 2 => C2H2 => A

Cu 56: t chy hon ton mt ankin X th kh thu c H2O v CO2 c tng khi lng l 23 gam. Nu cho sn phm chy i qua dung dich Ca(OH)2 d, c 40 gam kt ta. Cng thc phn t ca X l A. C3H4. B. C2H2. C. C4H6. D. C5H8.nCO2 = nCaCO3 kt ta = 0,4 mol => mCO2 = 0,4.44= 17,6 g => nCO2 = 0,4 mol mH2O = 23 17,6 = 5,4 g => nH2O = 0,3 mol => n = nCO2 / (nCO2 nH2O) CnH2n-2Oz = 0,4 / (0,4 0,3) = 4 => C4H6

Cu 57: t chy hon ton 5,4 gam mt hirocacbon A ri cho sn phm chy i qua bnh 1 ng dd H 2SO4 c, d; bnh 2 ng dung dch Ba(OH)2 d thy khi lng bnh 1 tng 5,4 gam; bnh 2 tng 17,6 gam. A l cht no trong nhng cht sau ? (A khng tc dng vi dd AgNO3/NH3) A. But-1-in. B. But-2-in. C. Buta-1,3-ien. D. B hoc C.Xem li phn chuyn 1 => cho vo H2SO4 => m tng = mH2O = 5,4 g => nH2O = 0,3 mol Cho vo Ba(OH)2 => m tng = mCO2 = 17,6 g => nCO2 = 0,4 mol n = nCO2 / (nCO2 nH2O) = 4 => C4H6

Da vo A khng tc dng vi dd AgNO3/NH3 => Loi A v A c dng CHC-CH2-CH3 c lin kt 3 u mch => c p ; B , Ckhng p B c khng u mch . C c 2 lin kt i =>D

Cu 58: Hn hp X gm C2H2 v H2 ly cng s mol. Ly mt lng hn hp X cho i qua cht xc tc thch hp, un nng c hn hp Y gm 4 cht. Dn Y qua bnh ng nc brom thy khi lung bnh tng 10,8 gam v thot ra 4,48 lt kh Z (ktc) c t khi so vi H2 l 8. Th tch O2 (ktc) cn t chy hon ton hn hp Y l A. 33,6 lt. B. 22,4 lt. C. 16,8 lt. D. 44,8 lt.Y gm 4 cht => C2H2(d) ; C2H4 ; C2H6 ; H2 m bnh tng = mC2H2 + mC2H4 = 10,8g v C2H2 v C2H4 b Br2 hp th => KH thot ra l AnkanC2H6 + H2 c m = M . n hn hp Z = 8.2.0,2 = 3,2 g BT khi lng => mC2H2 + mH2 = m hn hp Y = 10,8 + 3,2 = 14 g Ta c nC2H2 = nH2 => 26x + 2x = 14 x = 0,5 = nC2H2 = nH2 VO2 cn t chy hn hp Y = VO2 cn t chy hn hp X Quy i hn hp v CxHy v thnh phn hn hp ch c C , H nO2 = (2 + 2/4).nC2H2 + nH2 /2 = 3nH2 = 1,5 mol => V = 33,6 lt P : C2H2 + 3/2O2 => 2CO2 + H2O ; 2H2 + O2 => 2H2O

Cu 59: Cho 17,92 lt hn hp X gm 3 hirocacbon kh l ankan, anken v ankin ly theo t l mol 1:1:2 li qua bnh ng dd AgNO3/NH3 ly d thu c 96 gam kt ta v hn hp kh Y cn li. t chy hon ton hn hp Y thu c 13,44 lt CO2. Bit th tch o ktc. Khi lng ca X l A. 19,2 gam. B. 1,92 gam. C. 3,84 gam. D. 38,4 gam.T l 1 : 1 : 2 => chn x l mol Ankan => x l mol anken ; 2x l mol ankin x + x + 2x = n hn hp X = 0,8 mol x = 0,2 mol => nAkan = nAnken = 0,2 ; n Ankin = 0,4 mol CH c Ankin mi p vi AgNO3/NH3 M kt ta = 96/nankin = 240 = MX + 214 TH2 Xem li bi 43 MX = 26 = 14n 2 n = 2 : C2H2 t Y thu c 13,44 lt CO2 => BT nguyn t C CnH2n+2 ankan ; CmH2m anken n . nAnkan + m. nanken = nCO2 0,2n + 0,2m = 0,6 n + m = 3 Ta lun c m 2 Anken=> n = 1 v m = 2 Duy nht => CH4 v C2H4 hn hp X c 0,2 mol CH4 ; 0,2 mol C2H4 v 0,4 mol C2H2 m hn hp = 19,2 g =>A

Cu 60: Mt hn hp gm 2 ankin khi t chy cho ra 13,2 gam CO2 v 3,6 gam H2O. Tnh khi lng brom c th cng vo hn hp trn A. 16 gam. B. 24 gam. C. 32 gam. D. 4 gam.nAnkin = nCO2 nH2O = 0,3 0,2 = 0,1 mol => nBr2 = 2nAnkin = 0,2 mol => m = 32 g => C

Cu 61: Cho canxi cacbua k thut (ch cha 80% CaC2 nguyn cht) vo nc d, th thu c 3,36 lt kh (ktc). Khi lng canxi cacbua k thut dng l A. 9,6 gam. B. 4,8 gam C. 4,6 gam. D. 12 gamP : SGK 11 nc 178 : CaC2 + 2H2O => C2H2 + Ca(OH)2 nC2H2 = 3,36 / 22,4 = 0,15 mol Kh = nCaC2 => mPT CaC2 = 0,15.64 = 9,6 g => m thc t = mPT .100% / 80% = 9,6.100% / 80 = 12 g =>D

Cu 62: C 20 gam mt mu CaC2 (c ln tp cht tr) tc dng vi nc thu c 7,4 lt kh axetilen (20 oC, 740mmHg). Cho rng phn ng xy ra hon ton. tinh khit ca mu CaC2 l A. 64%. B. 96%. C. 84%. D. 48%.Ta c 1atm = 760 mmHg => 740 mm Hg =0,9736 atm nC2H2 = P.V/T.0,082 = 0,9736.7,4 / ((20 + 273).0,082) n = 0,3 mol = nCaC2 PT bi 61 mPT CaC2 = 0,3.64 = 19,2 g => tinh khit = mPT / mBan u = 19,2 .100%/ 20 = 96%

Cu 63: 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 brom phn ng l 48 gam. Mt khc, nu cho 13,44 lt ( ktc) hn hp kh X tc dng vi lng d dung dch AgNO 3 trong NH3, thu c 36 gam kt ta. Phn trm th tch ca CH4 c trong X l A. 40%. B. 20%. C. 25%. D. 50%.Gi x , y , z ln lt l s mol CH4 , C2H4 , C2H2 => 16x + 28y + 26z = 8,6 (I) P vi Br2 d => nC2H4 + 2nC2H2 = nBr2 y + 2z = 0,3mol (II) Vi 13,44 lt (ktc) => n hn hp = nCH4 + C2H4 + nC2H2 = 0,6 n C2H2 = nKet tua AgCCAg = 0,15 mol => nC2H2 / n hn hp = z / (x + y + z) = 1 /4 x + y 3z = 0 (III) Phi chia tm ra t l v 13,44 lt khng phi ca 8,6 g Gii h I , II , III => x = 0,2 ; y = z = 0,1 => %VCH4 = nCH4 / n hn hp = x .100%/ (x + y + z) = 0,2 / 0,4 = 50%

Cu 64: Hn hp kh X gm anken M v ankin N c cng s nguyn t cacbon trong phn t. Hn hp X c khi lng 12,4 gam v th tch 6,72 lt ( ktc). S mol, cng thc phn t ca M v N ln lt l A. 0,1 mol C2H4 v 0,2 mol C2H2. B. 0,1 mol C3H6 v 0,2 mol C3H4. C. 0,2 mol C2H4 v 0,1 mol C2H2. D. 0,2 mol C3H6 v 0,1 mol C3H4.Cch 1 : th p n => D tha mn Cch 2 : Ta c M = m / n = 12,4 / 0,3 = 41,33 = 12x + y => x = 3 => y = 5,33 => loi A v B v c 3 C => Th 1 trong 2 p n B v D . Xt B sai => D ng

Cu 65: Trong mt binh kin cha hirocacbon A th khi (kt) va O2 (d). Bt tia la in t chay ht A a hn hp v iu kin ban u trong o % th tich cua CO2 va hi nc ln lt la 30% va 20%. Cng thc phn t cua A va % th tich ca hirocacbon A trong hn hp la A. C3H4 va 10%. B. C3H4 va 90%. C. C3H8 va 20%. D. C4H6 va 30%.Tao c %CO2 = 30% ; %H2O = 20% => Cn li 50% ca O2 v O2 d T l % theo th tch = t l s mol => chn nCO2 = 3 mol => nH2O = 2 mol => nO2 d = 5 mol T l nCO2 / nH2O = 3 /2 => chn nCO2 = 3 ; nH2O = 2 => n = nCO2 / (nCO2 nH2O) = 3 => C3H4 V nCO2 > nH2O => CnH2n-2 nC3H4 = nCO2 / 2 = 1 mol ; BTNT oxi : 2nO2 = 2nCO2 + nH2O = 2.3 + 2 nO2 p = 4 mol mol hn hp ban u = nC3H4 + nO2 p + nO2 d = 1 + 4 + 5 = 10 mol => %VC3H4 A hn hp = nC3H4 / n hn hp = 1 / 10 = 10% => A

Cu 66: 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 trong cng iu kin nhit v p sut). Cng thc phn t ca X l A. C2H4. B. CH4. C. C2H6. D. C3H8. Cch 1 th p n T l th tch = t l s mol 1 mol hn hp C2H2 v X => 2 mol CO2 + 2molH2O p n => B,C,D u c dng CnH2n+2 => Xt X l CnH2n+2 Gi x , y ln lt l nC2H2 v CnH2n+2 x+y=1; 2x + ny = nCO2 = 2 BTNT C 2x + y(2n+2) = 2nH2O = 4 BTNT H 2(x+y) + 2ny = 4 Th 1 vo 3 ta c ny = 1 => th vo 2 => x = 0,5 th vo 1 => y = 0,5

(1) (2) (3) >>>>>

x = y = 0,5 th vo 2 => n = 2 => C2H6 => C

Cu 67: Hn hp X c t khi so vi H2 l 21 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. 18,60 gam. B. 18,96 gam. C. 20,40 gam. D. 16,80 gam.Tng t 1 bi lm ri => Propan C3H8 Propen C3H6 Propin C3H4 => u c 3C => Gi CT chung : C3Hy Ta c M = 21.2 = 42 = 12.3 + y => y = 6 => CT : C3H6 +O2 => 3CO2 + 3H2O 0,1mol 0,3mol 0,3mol => m = 0,3.44 + 0,3.18 = 18,6 g => A

Cu 68: Cho s chuyn ha: CH4 C2H2 C2H3Cl PVC. tng hp 250 kg PVC theo s trn th cn V m3 kh thin nhin ( ktc). Gi tr ca V l (bit CH4 chim 80% th tch kh thin nhin v hiu sut ca c qu trnh l 50%) A. 224,0. B. 448,0. C. 286,7. D. 358,4.PVC : C2H3Cl ; BT NT C => nCH4 = 2nC2H3Cl = 2. 4 = 8 mol => n kh thin nhin = nCH4 . 100% / 80% = 10 mol => V = 224 lt => B

Cu 69: un nng hn hp kh gm 0,06 mol C 2H2 v 0,04 mol H2 vi xc tc Ni, sau mt thi gian thu c hn hp kh Y. Dn ton b hn hp Y li t t qua bnh ng dung dch brom (d) th cn li 0,448 lt hn hp kh Z ( ktc) c t khi so vi O2 l 0,5. Khi lng bnh dung dch brom tng l A. 1,20 gam. B. 1,04 gam. C. 1,64 gam. D. 1,32 gam.Bo ton khi lng => m trc = m sau mC2H2 + mH2 = m bnh tng + mZ V m hn hp ban u = m hn hp Y v m hn hp Y = m anken, ankin nu c cho vo dd Br2 + m hn hp Z bay ra 0,06.26 + 2. 0,04 = m bnh tng + 32.0,5.0,02 m bnh tng = 1,32 g => D

Cu 70: t chy hon ton m gam hirocacbon th kh, mch h, nng hn khng kh thu c 7,04 gam CO 2. Sc m gam hirocacbon ny vo nc brom d n khi phn ng hon ton, thy c 25,6 gam brom phn ng. Gi tr ca m l A. 2 gam. B. 4 gam. C. 10 gam D. 2,08 gamnCO2 = 0,16 = x .nCxHy BT NT C Gi k l s lin kt pi trong hidrocacbon => k .nCxHy = nBr2 nCxHy = 0,16 / k 0,16 = x. 0,16 / k x = k CT tng qut : CnH2n + 2 2k Ta c n = k => CnH2 Ta c hidrocacbon nng hn khng kh => 12n + 2 > 29 => n > 2,25 Ta c m = (12n+2).0,16/n th vo trn ta c

V th kh => C2 C4 => s C = 2 4 Ta c n > 2,25 => n = 3 v n = 4 Xt n = 3 => C3H2 Khng th vit c => Loi hoc vi k = 3 => n 4 => n = 4 => m = (12.4+2).0,16 / 4 = 2 g => A

Cu 71: t chy hon ton m gam hirocacbon th kh, mch h thu c 7,04 gam CO2. Sc m gam hirocacbon ny vo nc brom d n khi phn ng hon ton, thy c 25,6 gam brom phn ng. Gi tr ca m l A. 2 gam. B. 4 gam. C. 2,08 gam. D. A hoc C.Ging bi 71 ch khng c nng hn khng kh Ta vn c m = (12n+2).0,16/n Xt n = 2 => m = 2,08 ; xt n = 4 => m = 2 g => D V C bin thin t 2 => 4 v n = 3 khng c => ch c n =2 v n = 4

Cu 72: Dn V lt ( ktc) hh X gm axetilen v hiro i qua ng s ng bt niken nung nng, thu c kh Y. Dn Y vo lng d AgNO3 trong dd NH3 thu c 12 gam kt ta. Kh i ra khi dung dch phn ng va vi 16 gam brom v cn li kh Z. t chy hon ton kh Z c 2,24 lt kh CO2 (ktc) v 4,5 gam H2O. Gi tr ca V bng A. 11,2. B. 13,44. C. 5,60. D. 8,96.Y vn cn C2H2 d v Y p vi AgNO3 / NH3 => AgCCAg => nC2H2 d = nAgCCAg = 0,05 mol ; Kh i ra khi bnh p vi Br2 => C2H4 To thnh => nC2H4 = nBr2 = 0,1 mol t kh Z => t C2H6Ankan + H2 dNu c => nC2H6 = nCO2 / 2 = 0,05 molBTNT C nH2O to thnh t C2H6 => 6nC2H6 = 2nH2O BTNT H => nH2O = 0,15 mol nH2 d = nH2O to thnh nH2O to thnh t C2H6 = 0,25 0,15 = 0,1 nH2 = nH2O BTNT H2 Ta c cht ban u : C2H2 ; H2 => Cht sau p hn hp Y gm 0,05 mol C2H2 d ; 0,1 mol C2H4 ; 0,05 mol C2H6 v 0,1 mol H2 d BTNT C => 2nC2H2 = 2nC2H2 d + 2nC2H4 + 2nC2H6 nC2H2 = nC2H2 d + nC2H4 + nC2H6 = 0,05 + 0,1 + 0,05 = 0,2 mol BTNT H => 2nC2H2 + 2nH2 = 2nC2H2 d + 4nC2H4 + 6nC2H6 + 2nH2 d 2. 0,2 + 2nH2 = 2.0,05 + 4.0,1 + 6.0,05 + 2.0,1 nH2 = 0,3 mol => n hn hp X = nC2H2 + nH2 = 0,2 + 0,3 = 0,5 mol => V = 11,2

Cu 73: Cho 4,48 lt hn hp X (ktc) gm 2 hirocacbon mch h li t t qua bnh cha 1,4 lt dung dch Br 2 0,5M. Sau khi p hon ton, s mol Br2 gim i mt na v m bnh tng thm 6,7 gam. CTPT ca 2 hirocacbon l A. C3H4 v C4H8. B. C2H2 v C3H8. C. C2H2 v C4H8. D. C2H2 v C4H6.T p n => hidrocabon => 1 ankin v 1 anken Mol Br2 gim 1 na => P 1 na ; m bnh tng = m hn hp V c 2 cht u p Gi x , y l s mol Ankin v anken => n hn hp = x + y = 0,2 mol nBr2 p = 2x + y = 0,35 1 na x = 0,15 ; y = 0,05 => Th p n : MA . 0,15 + MB .0,05 = 6,7 => C tha mn c MA = MC2H2 = 26 ; MB = MC4H8 = 56 Gi CT ankin : CnH2n-2 ; CT anken : CmH2m m hn hp = (14n 2) . 0,15 + 14m.0,05 = 6,7 3n + m = 10 => n < 10/3 => n = 2 v n = 3 Xt n = 2 => m = 4 => C tha mn Xt n = 3 => m = 1 Khng c CH2 => Loi > C

Cu 74: Dn 1,68 lt hn hp kh X gm hai hirocacbon vo bnh ng dd brom (d). Sau khi phn ng hon ton, c 4 gam brom phn ng v cn li 1,12 lt kh. Nu t chy hon ton 1,68 lt X th sinh ra 2,8 lt kh CO2 (ktc). CTPT ca hai hirocacbon l A. CH4 v C2H4. B. CH4 v C3H4. C. CH4 v C3H6. D. C2H6 v C3H6.p n => 1 ankan Hoc da vo 1,12 lt kh khng p => V ankan = 1,12 => nCnH2n+2 = 0,05 mol V cht cn li = 1,68 1,12 = 0,56 mol => n cht = 0,025 = nBr2 4 g Br2 p => Cht l anken :CnH2n nAnkan = 0,05 ; nAnken = 0,025 ; Gi CT an kan : CnH2n+2 ; anken : CmH2m t to ra CO2 => BTNT C => n.nCnH2n+2 + m.nCmH2m = nCO2 n.0,05 + m.0,025 = 0,125 2n + m = 5 n < 5/2 => n = 1 v n = 2 Xt n = 1 => m = 3 =>C tha mn Xt n = 2 => m = 1 Loi v khng c cht CH2 => Loi => C

Cu 75: t chy hon ton hn hp M gm mt ankan X v mt ankin Y, thu c s mol CO 2 bng s mol H2O. Thnh phn % v s mol ca X v Y trong hn hp M ln lt l A. 35% v 65%. B. 75% v 25%. C. 20% v 80%. D. 50% v 50%.Gi X : CnH2n+2 ; Y : CmH2n-2 ; gi x , y l s mol X v Y BTNT C => x.n + y.m = nCO2 BTNT H => (2n+2).x + (2n-2)y = 2nH2O (n+1)x + (n-1)y = nH2O Ta c nCO2 = nH2O x.n + y.m = (n + 1)x + (n-1)y x y = 0 x= y => % mi cht = 50%

Cu 76: t chy hon ton 20,0 ml hn hp X gm C3H6, CH4, CO (th tch CO gp hai ln th tch CH4), thu c 24,0 ml CO2 (cc kh o cng iu kin to, p). T khi ca X so vi kh hiro l

A. 25,8.

B. 12,9.

C. 22,2.

D. 11,1.

T l th tch = t l s mol => 20mol hn hp X => 24 mol CO2 Gi x , y ln lt l s mol C3H6 , CH4 => nCO = 2y VCO gp 2 ln VCH4 n hn hp = nC3H6 + nCH4 + nCO = x + y + 2y = x + 3y = 20 BTNT C => 3nC3H6 + nCH4 + nCO = nCO2 3x + y + 2y = 24 3x+ 3y = 24 Gii h => x = 2 ; y = 6 => m hn hp = 2.42 + 6.16 + 12.28 = 516 g => M hh X = 516/20 = 25,8 T khi vi H2 = 12,9 Nu khng lm c c th m thy A v B gp i nhau ; C v D gp i nhau B v D kh nng ng cao hn hay cho la khng tnh t khi Tng kh nng ng.

Cu 77: Mt hn hp X gm 1 ankan A v 1 ankin B c cng s nguyn t cacbon. Trn X vi H2 c hn hp Y. Khi cho Y qua Pt nung nng th thu c kh Z c t khi i vi CO2 bng 1 (phn ng cng H2 hon ton). Bit rng VX = 6,72 lt v VH = 4,48 lt. CTPT v s mol A, B trong hn hp X l (Cc th tch kh o kc) A. 0,1 mol C2H6 v 0,2 mol C2H2. B. 0,1 mol C3H8 v 0,2 mol C3H4. C. 0,2 mol C2H6 v 0,1 mol C2H2. D. 0,2 mol C3H8 v 0,1 mol C3H4.2

Phn ng cng H2 hon ton ; MZ = 44 = 14n + 2 V to ra ankan : CnH2n+2 v t khi vi CO2 => n = 3 A , B l C3H8 v C3H4 mt khc ta c nH2 = 0,2 mol => nC3H4ankin = 0,1 mol k.X = nH2 vi k l s pi => nC3H8 = 0,2 mol => D

Cu 78: Mt hn hp X gm C2H2, C3H6, CH4. t chy hon ton 11t hn hp X thu c 12,6 gam H2O. Nu cho 11,2 lt hn hp X (ktc) qua dung dch brom d thy c 100 gam brom phn ng. Thnh phn % th tch ca X ln lt l A. 50%; 25% ; 25%. B. 25% ; 25; 50%. C.16% ; 32; 52%. D. 33,33%; 33,33; 33,33%.Gi x , y , z ln lt l s mol C2H2 ; C3H6;CH4 Sa 1 lt thnh 0,4 mol n hn hp = x + y + z = 0,4 mol BTNT H => 2nC2H2 + 6nC3H6 + 4nCH4 = 2nH2O 2x + 6y + 4z = 1,4 mol Xt mt khc ta c n hn hp = 0,5 mol V nBr2 p = 2nC2H2 + nC3H6 nBr2 / n hn hp = (2nC2H2 + nC3H6) / (nC2H2 + nC3H6 + nCH4) Chia tim ra t l gia x, y, z : khng c p dng vo h trn v s mol khc nhau 5/4 = (2x + y) / (x + y + z) 3x -y 5z = 0 Gii h 1 , 2 , 3 => x = 0,2 ; y = z = 0,1 mol => %V = n / n hn hp => A

Cu 79: Dn 4,032 lt (ktc) hn hp kh A gm C2H2, C2H4, CH4 ln lt qua bnh 1 cha dung dc AgNO3 trong NH3 ri qua bnh 2 cha dung dch Br2 d trong CCl4. bnh 1 c 7,2 gam kt ta. Khi lng bnh 2 tng thm 1,68 gam. Th tch (ktc) hn hp A ln lt l: A. 0,672 lt; 1,344 lt; 2,016 lt. B. 0,672 lt; 0,672 lt; 2,688 lt. C. 2,016; 0,896 lt; 1,12 lt. D. 1,344 lt; 2,016 lt; 0,672 lt.nC2H2 = nAgC=*CAgXem mt s bi tp trn = 0,03 mol => V C2H2 = 0,672 lt mC2H4 = m bnh 2 v C2H4 b hp th bi Br2 = 1,68 g => nC2H4 = 0,06 mol => VC2H4 = 1,344 lt => V CH4 = V hn hp VC2H2 VC2H4 = 2,016 lt => A

Cu 80: X, Y, Z l 3 hirocacbon th kh trong iu kin thng, khi phn hu mi cht X, Y, Z u to ra C v H2, th tch H2 lun gp 3 ln th tch hirocacbon b phn hu v X, Y, Z khng phi l dng phn. CTPT ca 3 cht l A. C2H6 ,C3H6 C4H6. B. C2H2 ,C3H4 C4H6. C. CH4 ,C2H4 C3H4. D. CH4 ,C2H6 C3H8.th tch H2 lun gp 3 ln th tch hirocacbon b phn hu t lun Th tch ca mi hidrocacbon phn hy => u to ra 3 th tch H2 BTNT H => y.VCxHy = 2VH2 y = 6 => ch c A tha mn 3 cht u c 6 H

Cu 81: Hn hp X gm 0,1 mol C2H2; 0,15 mol C2H4 ; 0,2 mol C2H6 v 0,3 mol H2. un nng X vi bt Ni xc tc 1 thi gian c hn hp Y. t chy hon ton hn hp Y c s gam CO2 v H2O ln lt l A. 39,6 v 23,4. B. 3,96 v 3,35. C. 39,6 v 46,8. D. 39,6 v 11,6.BTNT C ; BTNT H => A C mt bi v n ri nCO2 = 2nC2H2 + 2nC2H4 + 2nC2H6 ; nH2 = 2nC2H2 + 4nC2H4 + 6nC2H6 + 2nH2ban u

Cu 82: Hn hp ban u gm 1 ankin, 1 anken, 1 ankan v H2 vi p sut 4 atm. un nng bnh vi Ni xc tc thc hin phn ng cng sau a bnh v nhit ban u c hn hp Y, p sut hn hp Y l 3 atm. T khi hn hp X v Y so vi H2 ln lt l 24 v x. Gi tr ca x l A. 18. B. 34. C. 24. D. 32.

CT tnh s mol ; V trong 1 bnh => th tch khng i , Nhit khng thay i cho a v nhit ban uP1.V 1 nA T1.0,082 P1 = = P2.V P2 2 nB T2.0,082

=>

P1.V1 nA T1.0,082 P1 = = P2.V2 P2 nB T2.0,082

mA P1 MB P1 MA = = mB P2 MA P2 MB

V BTKL => mA = mB

Thay s MA = 24 ; p1 = 4 ; p2 = 3 => MB = 32 =>D

Cu 83: Hn hp A gm H2, C3H8, C3H4. Cho t t 12 lt A qua bt Ni xc tc. Sau phn ng c 6 lt kh duy nht (cc kh o cng iu kin). T khi hi ca A so vi H2 l A. 11. B. 22. C. 26. D. 13.T l th tch = t l s mol => Chn 12 mol A => To ra 6 mol kh duy nht C3H8 Gi x , y , z ln lt l s mol H2 , C3H8 , C3H4 x + y + z = 12 mol PT p : C3H4 + 2H2 => C3H8 z mol x mol => zmol => x = 2z v phn ng va do to ra 1 sn phm duy nht l C3H8 Ngoi ra nC3H8 = y + z = 6 v C3H8 ban u khng p Gii h => x = 6 ; y = z = 3 C th dng my tnh hoc th t x = 2z vo cc PT cn li ri gii h 2 n => m hn hp = 6.2 + 3.44 + 3.40 = 264 => M hn hp = m / n = 264/12 = 22 => T khi vi H2 = 11 =>A

Cu 84: un nng hn hp X gm 0,1 mol C3H4 ; 0,2 mol C2H4 ; 0,35 mol H2 vi bt Ni xc tc c hn hp Y. Dn ton b Y qua bnh ng dung dch KMnO4 d, thy thot ra 6,72 l hn hp kh Z (ktc) c t khi so vi H2 l 12. Bnh ng dung dch KMnO4 tng s gam l: A. 17,2. B. 9,6. C. 7,2. D. 3,1.BT khi lng => m hn hp X = m hn hp Y = m bnh tng + m hn hp kh thot ra m hn hp X = m bnh tng + m hn hp kh thot ra 0,1.40 + 0,2.28 + 0,35.2 = m bnh tng + 0,3.12.2 m bnh tng = 3,1 g =>D Bi ny c ni n mt s bi trn

Cu 85: t chy m gam hn hp C2H6, C3H4, C3H8, C4H10 c 35,2 gam CO2 v 21,6 gam H2O. Gi tr ca m l A. 14,4. B. 10,8. C. 12. D. 56,8.AD CT : m hn hp hidrocacbon = 12.nCO2 + 2.nH2O = 12g Xem Chuyn pp gii bi tp ha hu c V nC = nCO2 ; nH = 2nH2O

Cu 86: t chy 1 hirocacbon A c 22,4 lt kh CO2 (ktc) v 27 gam H2O. Th tch O2 (ktc) (l) tham gia phn ng l: A. 24,8. B. 45,3. C. 39,2. D. 51,2.BTNT Oxi => 2nO2 = 2nCO2 + nH2O => nO2 => V = 39,2 lt =>C

Cu 87: Mt hn hp X gm 1 ankin v H2 c V = 8,96 lt (kc) v mX = 4,6 gam. Cho hn hp X i qua Ni nung nng, phn ng hon ton cho ra hn hp kh Y, c t khi A. 0,16 mol; 3,6 gam; C2H2. C. 0,2 mol; 4 gam; C2H2.Pt: CnH2n-2 +2H2 =>CnH2n+2 Ban u:x mol ymol P x mol 2x mol xmol Sau p y 2x x mol => n sau p = nH2 d + nCnH2n+2 y 2x +x = y x n trc p = x + y ; m trc = m sau BT khi lng (x+y).M trc = (y x).M sau p dng bi ny=> (x+y) = (y x).2 V M sau = 2MY do dY/X = 2 y = 3x Mt khc x + y = 0,4 mol => x = 0,1 v y = 0,3 Gii h => nH2 = y = 0,3 mol m ankin + mH2 = 4,6 g 0,1 . (14n 2) + 0,3.2 = 4,6 n = 3 => C3H4 vi n = 0,1 mol=> m =4 g >B

d

Y

X

= 2. S mol H2 phn ng; khi lng; CTPT ca ankin l B. 0,3 mol; 4 gam; C3H4. D. 0,3 mol; 2 gam; C3H4.

Xem cch chng minh phn anken dng bi anken + H2 bi 53 => Phn ny tng t

Cu 88: t chy mt hirocacbon M thu c s mol nc bng s mol CO2 v s mol CO2 nh hn hoc bng 5 ln s mol M. Xc nh CTPT v CTCT ca M bit rng M cho kt ta vi dung dch AgNO3/NH3. A. C4H6 v CH3CH2C CH. B. C4H6 v CH2=C=CHCH3. C. C3H4 v CH3C CH. D. C4H6 v CH3C CCH3.nH2O : nCO2 = 3 / 4 => Chn nH2O = 3 mol => nCO2 = 4 mol => = nCO2 / (nCO2 nH2O) = 4 CnH2n-2 v nCO2 > nH2O PP gii ha hu c => C4H6 ; Mt khc M to kt ta vi AgNO3/NH3 => A v phi c lin kt 3 u mch

n

BI TP V ANKAIEN -TECPEN - ANKIN 1D 11C 21B 31B 41C 51C 61D 71D 81A 2B 12C 22C 32B 42C 52C 62B 72A 82D 3D 13C 23D 33D 43D 53D 63D 73C 83A 4C 14A 24C 34C 44A 54A 64D 74C 84D 5A 15A 25C 35B 45D 55A 65B 75D 85C 6D 16B 26B 36C 46C 56C 66C 76B 86C 7A 17A 27B 37C 47B 57D 67A 77D 87B 8B 18D 28B 38C 48D 58A 68B 78A 88A 9C 19B 29A 39C 49C 59A 69D 79A 10A 20D 30B 40C 50D 60C 70A 80A