L THUYT V BI TP ANKAN 1.1 c tn quc t (IUPAC) cc cht sau : a. CH3-CH(CH3)-CH2-CH3 2-metylbutan (isopentan) b. CH3-CH2-CH(CH3)-CH2-CH3 3-metylpentan c. CH3-CH(Br)-CH(C2H5)-CH3 2-brom-3-etylbutan d. CH3-CHCl-CHCl-CH(CH3)-CH2-CH3 2,3-dibrom-4-metylhexan e. CH3-CH(CH3)-CH2-CH(CH3)-CH(CH3)-CH3 2,3,5-trimetylhexan 1.2 T cc tn gi hy vit cng thc cu to ca cc cht : a. 4-etyl-2,3-imetyl hexan CH3-CH(CH3)-CH(CH3)-CH(C2H5)-CH2-CH3 d. 3,3,5-tri metyl octan CH3-CH2-C(CH3)2-CH2-CH(CH3)-CH2-CH2-CH3 b. 6-etyl -2,2-imetyl octan CH3-C(CH3)2-CH2-CH2-CH2-CH(C2H5)-CH2-CH3 e. 3-etyl-2,3-i metyl heptan CH3-CH(CH3)-C(C2H5)(CH3)-CH2-CH2-CH2-CH3 c. 1-brom-2-clo-3-metyl pentan CHBr-CHCl-CH(CH 3)-CH2-CH3 1.3 Vit cc phng trnh phn ng theo s chuyn ha sau : CH3Cl CH2Cl2 CHCl3 CCl4 a. CH3COONa CH4 C2H2 C2H6 C2H4 etanCaO ,t CH3COONa + NaOH CH4 + Na2CO3 as CH4 + Cl2 CH3Cl + HCl clometan (metyl clorua)as CH3Cl + Cl2 CH2Cl2 + HCl 0

iclo metan (mrtylen clrrua)

as CH2Cl2 + Cl2 CHCl 3 + HCl

triclometan (clorofom)

as CHCl 3 + Cl2 CCl4

+

HCl

2CH4 C2H2 +3H2Ni ,t C2H2 + 2H2 C2H6 5000 C , xt C2H6 C2H4 +H2 Ni ,t 0 C2H4 + H2 C2H6 C2H6 C2H5Cl C4H10 C4H8 n butan b. C4H10 isopropylclorua C3H6 propan n propylclorua. 500 C , xt CH3-CH2-CH2-CH3 CH3-CH3 + CH2=CH2 as CH3-CH3 + Cl2 CH3-CH2Cl +HCl t 0C , xt 2CH3-CH2Cl + 2Na CH3-CH2-CH2-CH3 + 2NaCl 5000 C , xt CH3-CH2-CH2-CH3 CH2=CH-CH2-CH3 + H2 Ni ,t 0 CH2=CH-CH2-CH3 + H2 CH3-CH2-CH2-CH3 500 C , xt CH3-CH2-CH2-CH3 CH3-CH=CH2 +CH4 Ni ,t 0 CH3-CH=CH2 + H2 CH3-CH2-CH3 CH3-CH2-CH3 + Cl2 CH3-CHCl-CH3 (isopropylclorua) + HCl CH2Cl-CH2-CH3 (propylclorua) + HCl0 0 0

15000 l ln

c. n Hecxan n butan etan etylclorua. 5000 C , xt CH3-CH2-CH2-CH2-CH2-CH3 CH3-CH2-CH2-CH3 + CH2=CH2 5000 C , xt CH3-CH2-CH2-CH3 CH2-CH3 + CH2=CH2 as CH3-CH3 + Cl2 CH3-CH2Cl +HCl 1.4. Vit phng trnh phn ng clo ha (t l 1 : 1), phn ng hiroha, phn ng nhit (cho bit sn no c u tin). a. Propan + Cl2 CH 2Cl CH 2 CH 3 + HCl as CH3-CH2-CH3 + Cl2 CH 3 CHCl CH 3 + HCl CH 2 = CH CH 3 + H 2 t 0C , xt CH3-CH2-CH3 CH 2 = CH 2 + CH 4 b. n-butan CH 2Cl CH 2 CH 2 CH 3 + HCl as CH3-CH2-CH2-CH3 + Cl2 CH 3 CHCl CH 2 CH 3 + HCl CH 2 = CH CH 3 + CH 4 t 0C , xt CH3-CH2-CH2-CH3 CH 2 = CH 2 + CH 3 CH 3 CH = CH CH CH + H 2 3 2 2

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c. CnH2n+2 as CnH2n+2 + Cl2 CnH2n+1Cl + HCl Cn H 2 n + H 2 t 0C , xt CnH2n+2 (n1+ n2 =n) Cn1 H 2 n1 + Cn 2 H 2 n 2+ 2 Cu 2. Vit cc ng phn v gi tn theo danh php quc t cc hp cht ng vi cng thc phn t sau: C5H12 * C5H12 CH3-CH2-CH2-CH2-CH3 pentan CH3-CH(CH3)-CH2-CH3 2-metylbutan (isopentan) CH3-C(CH3)2-CH3 2,2 dimetylpropan (neopentan) 1.5 Hai cht A, B cng cng thc phn t C5H12 tc dng vi Cl2 theo t l mol 1:1 th A ch to 1 dn xut duy nht cn B to 4 dn xut. Vit cng thc cu to ca A, B v cc dn xut clo ca chng. CH 2Cl CH (CH 3 ) CH 2 CH 2Cl + HCl CH Cl CH (CH ) CHCl CH + HCl 2 3 3 as CH3-CH(CH3)-CH2-CH3 + Cl2 CH 3 CCl (CH 3 ) CH 2 CH 3 + HCl CH 2Cl CH (CH 3 ) CH 2 CH 3 + HCl as CH3-C(CH3)2-CH3 + Cl2 CH2Cl-C(CH3)2-CH3 + HCl

1.6 Xc nh cng thc phn t v vit cng thc cu to ca cc hirocacbon trong mi trng hp sau : a. Ankan A c t khi hi so vi H2 bng 36. Dt CTTQ ca ankan A: Cn H 2 n + 2 (vi n>0) MA d A/ H 2 = M A = M H 2 .d A/ H 2 =2.36=72=14n+2n=2 MH2 CTPT ca A: C5H12 CTCT: CH3-CH2-CH2-CH2-CH3 pentan CH3-CH(CH3)-CH2-CH3 2-metylbutan (isopentan) CH3-C(CH3)2-CH3 2,2 dimetylpropan (neopentan)

b. Cng thc n gin nht ca B l C3H7. ankan B c CTPT c dng (C3H7)n C3n H 7 n v l ankan nn: 7n=2.3n + 2 n=2 Ta c : Cn H 2 n + 2 ( C3H7)n CTPT ca B : C6H14 c. Ankan X c %C= 80% ; Dt CTTQ ca ankan X: Cn H 2 n + 2 12n.100 = 80 n = 2 %C= 14n + 2 CTPT ca X: C2H6 CTCT ca X: CH3-CH3 d. Ankan Y c %H=25% ; Dt CTTQ ca ankan Y: Cn H 2 n + 2 (2n + 2).100 = 25 n = 1 %H= 14n + 2 CTPT ca Y: CH4 f. Ha hi 12g ankan D thy chim mt th tch bng th tch ca 5g etan o cng iu kin. Dt CTTQ ca ankan D: Cn H 2 n + 2 Vhi ca D = Vhi ca etan nD=nC2H6= MD = 6 = 0.2 mol 30

8.8 = 44 =14n+2n=3 0, 2 CTPT ca D: C3H6 CTCT ca D: CH3-CH2-CH3 1.7 Xc nh cng thc phn t v vit cng thc cu to ca cc hirocacbon trong mi trng hp sau : a. t chy hon ton 1 lt ankan A sinh ra 3 lt CO2. Cc th tch o cng iu kin. Dt CTTQ ca ankan A: Cn H 2 n + 2 Cn H 2 n + 2 +

1lTa c :

(3n+ 1) /2 O2 nCO2 + (n + 1)H2O 3l

1 n = n=3 1 3 CTPT ca A: C3H6 CTCT ca A: CH3-CH2-CH3 b. t chy hon ton 1 ankan B vi lng O2 va th thy tng s mol cc cht trc phn ng bng tng s mol cc cht sau phn ng. Dt CTTQ ca ankan A: Cn H 2 n + 2

(3n+ 1) /2 O2 nCO2 + (n + 1)H2O 1 + (3n+ 1) /2 = n + n + 1 n= 1CTPT ca B: CH4 d. t chy hon ton mt hirocacbon Y thu c 17,6 g CO2 v 0,6 mol H2O. Dt CTTQ ca ankan Y: Cn H 2 n + 2 nCO 2 = 17, 6 = 0, 4 mol 44

Cn H 2 n + 2 +

Cn H 2 n + 2 + Ta c:

(3n+ 1) /2 O2 nCO2 + (n + 1)H2O0,4 0.6

n n +1 = n=2 0, 4 0.6 CTPT ca Y: C2H6 CTCT ca Y: CH3-CH3 e. t chy hon ton mt hirocacbon Z thu c CO2 v H2O theo t l Vco2 :Vhi nc = 3 : 4. V nH 2O > nCO2 hirocacbon Z l ankan Dt CTTQ ca ankan Z: Cn H 2 n + 2 Cn H 2 n + 2 + (3n+ 1) /2 O2 Ta c:

nCO2 + (n + 1)H2O3 4

n n +1 = n=3 3 4 CTPT ca Y: C3H8 CTCT ca Y: CH3-CH2- CH3 1.8 t chy hon ton 7,4 gam hn hp X gm etan v propan. Cho ton b sn phm chy vo dung dch Ca(OH)2 d sau phn ng thy khi lng bnh ng Ca(OH)2 tng ln 34,6 gam. Tnh % khi lng mi kh trong hn hp X v dX/H2 = ? C2 H 6 x mol + O2 Ca ( OH )2 du 7, 4 g hhX sp (CO2 + H 2O) bnh tng 34,6g= mCO2 + mH 2 o C3 H 8 y mol mhh = mC2 H6 + mC3 H8 = 30 x + 42 y = 7, 4 g (1)

C2H6 + 7/2O2 2 CO2 + 3H2O x 2x 3x C3H8 + 5 O2 3CO2 + 4 H2O y 3y 4ybnh tng 34,6g= mCO2 + mH 2O = nCO2 .44 + nH 2O .18 = (2x + 3y).44 + (3x + 4y).18=142x + 204y=34,6g (2) mC2 H 6 = 30 x + 42 y = 7, 4 x = ....... t (1) v (2) ta c h pt: 142 x + 204 y = 34, 6 y = ........ mC3H8 = %mC2 H 6 = MX = mC2 H 6 .100 mhh = ........100 = .......... 7, 4 d X / H2 = %mC3 H8 = mC3 H8 .100 mhh = ........100 =........... 7, 4

mhhX 7, 4 7, 4 = = = ......... nhhX x + y .... + .....

M X ........ = = M H2 2

1.9. t chy hon ton a gam hn hp gm etan v butan. Cho ton b sn phm chy ln lt qua bnh I ng dd H2SO4 v bnh II ng dd Ca(OH)2 d. Sau th nghim thy khi lng bnh I tng 7,2 gam v bnh II c 30 gam kt ta. Tnh % khi lng v % theo s mol mi kh trong hn hp ban u. H 2 SO4 dac b I t ng 7, m g= H 2O nh 2 C3 H 8 x mol + O2 a g hhX sp (CO2 + H 2O) Ca ( OH ) du 2 C4 H10 y mol bnh II c 30 g =mCaCO3 = mCO2 mCaCO3 ...... mH 2O ...... nH 2O = = = .... nCO 2 = nCaCO3 = = = 18 ....... 18 .......

C3H8 + 5O2 3 CO2 + 4H2O x 3x 4x

C4H10 + 13/2 O2 4CO2 + 5H2O y 4y 5ynH 2O = 4 x + 5 y = .......... nCO2 = 3x + 4 y = .....

ta c h pt:1.11 a) Hn hp X gm hai ankan c dX/H2 = 11,5. Xc nh hai ankan ni trn v tnh % theo th tch ca hh X. Bit hai ankan l lin tip trong dy ng ng. * t CT chung hhX l: Cn H 2 n + 2 d X / H2 = MX M X = d X / H 2 .M H 2 = 11,5.2 = 23 = 14n + 2 n = 1.5 M H2 = 2 n 0,5 1 = = n 1 0,5 1

Vy hai ankan l lin tip trong dy ng ng l CH4 v C2H6* nCH 4 nC2 H6 %VCH 4 = 50% %VC2 H6 = 50%

c) Mt hh 2 parafin k cn trong dy ng ng c t khi hi i vi khng kh bng 2,3. Xc nh ctpt ca 2 parafin ny v tnh % mi cht v th tch. * t CT chung hh 2 parafin l: Cn H 2 n + 2 M d X / kk = X M X = d X / kk .M kk = 2,3.29 = 66.7 = 14n + 2 n = 4.6 M kk

Vy hai parafin l k cn trong dy ng ng l C4H10 v C5H12 *nC4 H10 nC5 H12 = 5 n 0, 4 = %VC H = 40% 4 10 n 4 0, 6 %VC5 H12 = 60%

d) Mt hn hp 2 ankan th kh ktc c t khi i vi C2H4 bng 0,875. Xc nh cng thc phn t v % th tch hn hp. * t CT chung hn hp 2 ankan th kh l: Cn H 2 n + 2 MX d X / C2 H 4 = M X = d X / H 2 .M C2 H 4 = 0,875.28 = 24,5 = 14n + 2 n = 1.6 M C2 H 4

Vy 2 ankan th kh l CH4 v C2H4 *nCH 4 2 n 0, 4 = = %VCH = 40% 4 nC2 H6 n 1 0, 6 %VC2 H 6 = 60%

1.12. t chy hon ton 19,8 gam hn hp X gm hai ankan sau phn ng thu c 57,2 gam CO2. a. Tnh khi lng nc to thnh v s mol O2 phn ng. b. Nu 2 ankan trn l ng ng lin tip. Hy xc nh 2 ankan v tnh % theo khi lng mi ankan. mCO2 57, 2 mC = .12 = .12 = 15.6 g 44 44 mH = 19,8 15, 6 = 4, 2 g nH = 4.2 mol nH = nH 2O .2 = 4, 2 nH 2O = 2,1mol mH 2O = 2,1.18 = 37,8 mhhX + mO2 = mCO2 + mH 2O mO2 = 57, 2 + 37,8 19,8 = 75.2 g * t CT chung hn hp 2 ankan th kh l: Cn H 2 n + 2 Cn H 2 n + 2 + 3n + 1 O2 2 nCO2 + ( n + 1) H 2O

1,3Ta c:

2.1

n n +1 = n = 1.25 1,3 2.1 Vy 2 ankan ng ng lin tip l CH4 v C2H6 nCH 4 2 n 0, 75 = = %VCH = 75% %VC2 H 6 = 25% 4 nC2 H6 n 1 0, 25

1.13. t chy V(lt) hn hp hai ankan k tip trong dy ng ng. Dn sn phm ln lt qua bnh 1 ng CaCl2 khan ri bnh 2 ng dung dch KOH. Sau th nghim khi lng bnh 1 tng 6,43gam v bnh 2 tng 9,82 gam. a. Lp cng thc hai ankan. b. Tnh % theo s mol cc ankan trong hn hp, tnh V (kc). CaCl2 khan b 1 t ng 6, 43gam= H 2O nh m Cn H 2 n + 2 + O2 V (l ) hhX sp (CO2 + H 2O) dung dich KOH nh g= Cn +1 H 2( n +1) + 2 b 2 t ng 9,82 m = CO2 t CT chung hn hp 2 ankan th kh l: Cn H 2 n + 2 6, 43 = 0.357 mol 18 3n + 1 Cn H 2 n + 2 + O2 2 nH O =2

nCO2 = nCO2

9,82 = 0, 223 mol 44 + ( n + 1) H 2O

0,223Ta c:

0,357

n n +1 = n = 1, 67 0.223 0,357 Vy 2 ankan ng ng lin tip l CH4 v C2H6 nCH 4 2 n 0, 23 = = % nCH = 0.23 mol % nC2 H 6 = 0, 67 mol 4 nC2 H6 n 1 0, 67 1.14. t chy 20,4 gam mt hn hp 2 hirocacbon no mch h cn dng 51,52 lt oxi (ktc). a. Tnh th tch kh CO2 (ktc) v khi lng nc to thnh. b. Xc nh ctpt v tnh % theo th tch mi hirocacbon trong hh. Bit 2 hirocacbon u l cht kh iu kin thng. 1.15. Khi brom ha 22 gam propan ngi ta thu c 33,948 gam isopropyl bromua v 2,952 gam n-propyl bromua. Tnh hiu sut tng sn phm v hiu sut chung ca phn ng. 1.16. t chy hon ton 29,2 gam hn hp 2 ankan k tip nhau trong dy ng ng. Hp th ton b sn phm vo bnh Ba(OH)2 thy khi lng bnh tng 134,8 gam. Tnh khi lng CO2 v H2O to thnh v tm ctpt ca 2 ankan. 1.17. Hn hp X gm ankan A v B c khi lng phn t hn km nhau 28 (.v.C). t chy hon ton m(g) hh X cho hn hp sn phm kh v hi sau phn ng i qua bnh 1 ng dung dch H 2SO4 v bnh 2 ng dung dch KOH th khi lng bnh 1 tng m1(g) v bnh 2 tng m2(g). a. Nu m1 = 25,2 v m2 = 44. Xc nh cng thc phn t v % theo s mol ca A, B trong hh X, tnh m?

b. Nu m1 = 32,4 v m2 = 61,6. Xc nh cng thc phn t ca A, B v tnh m = ? Bit A, B u l chtkh kt. 1.18. t chy mt hn hp gm 2 hirocacbon ng ng k tip A, B thu c CO 2 v H2O theo t l s mol ln lt l 11 : 14. Tm cng thc phn t v % theo th tch ca hai hirocacbon ny. 1.19*. t chy hon ton a gam hn hp hai ng ng ca cc hirocacbon no, mch h c thnh phn hn km nhau k nguyn t cacbon th thu c b gam CO2.

a. Tm khong xc nh ca s nguyn t C trong hirocacbon theo a, b, k. b. Cho a = 2,72 (g) ; b = 8,36 (g) v k = 2. Tm cng thc ca cc hirocacbon v tnh % theo khi lng ca chng trong hn hp. 1.20** Mt hn hp X gm hirocacbon (A) v O2 d em t chy hon ton thu sn phm lm lnh th th tch gim 50 %. Nu cho kh cn li qua KOH d th tch gim i 83,3 % s cn li. a. Xc nh cng thc phn t v vit cng thc cu to cc ng phn ca A. b. Tnh thnh phn % v th tch ca A v oxi trong hn hp X. c. ng phn no ca A khi phn ng th vi Cl2 cho mt sn phm duy nht.