PHƯƠNG PHÁP GIẢI BÀI TẬP HỢP CHẤT HỮU CƠ CÓ NHÓM CHỨC - PHẠM ĐỨC BÌNH

8/10/2019 PHNG PHP GII BI TP HP CHT HU C C NHM CHC - PHM C BNH

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W.DAYKEMQUYNHON.UCOZ.COM

g gp PDF bi GV. Nguyn Thanh T


2/233

PHM C BNH

Phng phpgii bi tp

M P CHT M l Cdc NHM cildc

Nh xut bn Gio dc



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3/233

Chu trch nhim xut bn:

Ch tch HQT kim Tng Gim c NG TRN i

Ph TngGim c kim Tng bin tp NGUYN QUTHAO

T chc bn tho v chu trch nhim ni dung:

Ph Tng Gim c kim Gim c NXBGD ti TP. H Ch Minh

V B HO

Bin tp ni dung:

NG CNG HIP

Trnh by ba:

H TU HNG

Bin tp k - m thut:PHM VIT DN

Sa bn in:

NHNG THNG

Ch bn:

Phng SCT - NXBGD ti TP. HCM

02-2006/CXB/57-1 844/GD M s: 81120m6-CPH



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4/233

^ I N I U

gp phn nng cao cht lng hc tp b mn Ho hc, gip ccem hc sinh lp 1 0, 1 1, c bit l cc em hc sinh lp 12 n tp ch thng kin thc c bn, chuyn su, luyn thi tt nghip phthng, thi tuyn vo i hc v thi hc sinh gii Ho 10, 11, 12 tkt qu cao, chng ti bin son b sch Phng php gii bi tp ho hc 10, 11, 12 gm 5 cun. y l b sch c vit theo tngchuyn . Trong mi cun chng ti u tuyn chn, phn loi, giithiu y :

- Phng php gii v cc ch quan trng kh gii bi tp.

- Bi tp cng c, nng cao, luyn thi i hc.

B sch nhm cung cp cho cc em nhng kin thc mi, cp nht,p ng kp thi phng thc ra thi tuyn sinh ca cc trngi hc.

Phng php gi i bi tp Hp cht hu c c nhm chc c vitdnh cho hc sinh lp 1 1 v lp 1 2, sch gm 2 chng:

Chng 1: Cc dn xut ca hirocacbon (Hp cht c nhm chc)Chng 2: Hp cht hu c cha nit - Hp cht tp chc

Trong mi chng gm nhiu bi, mi bi c trnh by gm 2 phn chnh: phn mt: phng php gii bi tp; phn hai: gm bitp mu v bi tp t gii.

Vic bin son d t m cn thn n u cng kh trnh khinhng thiu st ngoi mun, rt mong nhn c nhng kin

xy dng qu bu ca ng nghip v cc em hc sinh.

Chc cc em n tp v thi t kt qu cao.

TC GI



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5/233

C h n g

CC DN XUT CA HIROCAGBON(HP CHT C NHOM CHC!

RU - PHENOL

ANEH1T - XETON

AXIT CACBOXYLIC

MU NATRI

ESTE - CHT BO



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6/233

i 1

Ru-PHENOL

V . . PHNG PHP G I I B I T P

NM VgMG CNO THC Vfi Tgw 0Q1 cc Ogl mj-_______________

1. Ru mch h

C n H 2n + 2 -2 a - x (O H ) x ( x n ) ; ( a : s n i 71 t r o n g g c R ) h a y R ( O H ) x .

2. Ru no m ch h

C n H 2n+2 - x (O H ) x h a y C n H 2n+ 20 x ( x < n ) .

3. Ru no n chc (ru no 1 ln ru)

CH2n+1OH hay CnH2n+20 (n > 1).

4. Ru cha no c 1 ni i n chc

CnHin-iOH hay CnH2nO (n > 3).

Nu nhm -O H nh vo c c ni 71 th ru km bn, chuyn th nh aneh it

CHs~CH=CH-OH ---- CH3-C H 2-CH O (anehit)

CH3-C(OH)=CH2 ----

> CH3-CO-CH 3 (xe ton)5. Ru b t k i

CxHy(OH)z hay CxHyOz (z > 1 )Ch : Mi nguyn t c ch c mang ti a mt nhm -O H, nn z < X.

, HO TNHCB RU______________ _____________________________________aHR

1. Xt ph n ng tc d ng vi kim o kim (Na, K)

R(OH)x + x N a ---- >R(ONa)x + x/2H2tT s mol H 2sy ra s" nhm -OH.

Ch ;Khi dung dch ru (c H20) tc dng vi kim loi kim th c 2 phn ng:

Na + H20 ---- > NaOH + 1/2H2f

R(OH)x + xNa ---- > R(ONa)x + x/2H2t

trng hp ny th tch H 2 c c H20 to nn. Nu ru tron g benzen hocexat th ch c ru phn ng vi Na.

Phn ng t i to ru:

, RONa + H20 ---- >ROH + NaOH (chng t ru khng phi l axit).



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7/233

2. Xt phn ng tch nc (loi H2O) t ru

a) Loi 1 phn t H20 t 1 phn t ru (to anken)

C J W O H .... H2s0 -ffc ^ CH2n + H20> 170 c

Ch : - Khi tch nc ca ru bc cao thng cho hn hp 2 anken, snphm chnh to ra theo quy tc Zai xp nhm -O H b tch cng H nguyn t cbc cao hn.

- Hn hp 3 ru tch H2O cho hn hp kh ch gm 2 anken: chng t c2 ru l ng phn.

- Sn phm tch H20 ca ru l anken (olefin) suy ra ru ban u l runo n chc (ankanol).

- Nu sau phn ng lm ngng t hi nc: n ru = ioien v V, T khng i suyra p khng i (Pi = P2).

- Nu 2 ru tch nc vi hiu sut nh nhau: c th dng cng thc tngng (trung bnh) gii. Nu 2 ru tch nc vi hiu sut khc nhau, hoc td kin cho, c th tm s" mol mi ru: nn t cng thc ring gii.

b) Loi 1 ph n t H20 t 2 phn t ru (to ete)

2ROH H > K-O -R + H20 ; n w , bi hoi = m,e+ m140 c h 2u

Ch : - T 1 ru cho 1 ete, t 2 ru cho 3 ete, t 3 ru cho 6 ete.

AOH v BOH I-g-) > AOA ; BOB ; AOB

n = T n = V n ,H2O ete cc ru

- Nu cc ete to ra c s moi bng nhau th suy ra cc ru tham gia phnng c s' mol bng nhau.

- C 3 ete c khi lng phn t bng nhau: th c 2 ru ng phn.

3. Xt phn ng 0 X1ho khng hon ton ru

a) Ru bc I oxi ho thnh anehit

R-CH2OH + V02 -U; t -> R-CHO + H20

Hoc R-CH2OH + CuO - *?> R-CHO + Cu + H20

^hn hp ru ~ Hhn hp anehit (l l6 U h ic u S U t 1 0 0 % )

b) Ru bc II oxi ho thnh xeton

R - C H O H - R ' + V2O2 R - C O - R + H 20

Hoc R-CH(OH)R' + CuO > R-CO-R + Cu + H20



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8/233

Ru bc III kh b oxi ho

'U: R-CH2OH + 0 2 M n 2+ ;-t0 >R-COOH + H20

i oxi ho khng hon ton ru bc I c th thu c axit, andehit tng

g s C) ru d v nc. Ho tnh ca sn phm ny rt phc tp, cn xtJ"ng trng hp.

i :Khi oxi ho khng hon ton ru metylic c th xy ra nh sau:H -COOH

CHqOH 2 Lxt, t

H - CHOCHLOHdh 2 0

ng hn hp sn phm c 4 cht, n s cho phn ng trng bc (ca HCHO,), phn ng vi baz (ca HCOOH) (s xt c th trong cc bi ton).

t phn n g ox ho hon ton (t chy)

CnH2n+1OH + 3n/202 > nC0 2+ (n+l)H20

CH2n+2 Ox + (3n + 1 ~^ Q 2 -t0 -> nC0 2 + (n+ l)H 20

:Nu sn phm chy c nH 0 > nc 0 th ru em t l ru no.

m O(trongru) + m o(trong khng kh) m o(tron gC 02) + m O (trong H20)

t phn ng tc dng vi axit (phn ng este ho)

r axit v c: R(OH)x + xHX H2sp-4?.c.- t0 ... R(X)x + xH20:Nu gc R khng no (c ni Ti) th s km theo phn ng cng vo gc R.

i axit hu c: ROOH + R'OH H2so4 c, t > RC0 0 R' + H20

i:Axit n chc + Ru a chc---- >Este a chc

2R - C O O H + 0 2H 4( 0 H ) 2 ; ........= R - C O O - C H 2 - C H 2 - O O C R + 2H 20

: n chc + Ru a chc ---- >Este a chc (vng) (s xt sau bi axit).

t phn ng t c dn g vi Cu(OH>2

c ru a chc c cc nhm -OH nh vo cc nguyn t cacbon k tipLOphn ng ho ta n Cu(OH)2 to thnh dung dch mu xanh lam.

CH2-OH CH2- 0 ^ ^ . - 0 -CH 2I I I

2C H - O H + C u ( O H ) 2 --------> C H - < P ^ ^ _ ^ H + 2 H 2

I H H Ic h 2- o h h2- o h h o - c h 2

a Cu(0H)2 s u y r a s m o 1 r ^ u -



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9/233

III. iu CM RU

1. Cc phng php ch ung

a) Xt ph n ng hrat ho anken CnH2n + H2O ----> CnH2n+iOH

Ch : Cng H2O vjo anken bt i xng, ta thu c 2 sn phm (theoMaccopnhicop).

lanken = nru (nu hiu sut 1 0 0 %).

b) Xt phn ng hro ho aneht hoc xeton

R(CHO)x + xH2 ^ - t0 > R(CH2OH)x (anehit---- >ru bc I)

R-CO-R' + H2 > R-CH(OH)-R' (xeton---- >ru bc II)

c) Xt phn ng thu phn dn xu t halogen trong dung dch kim

R(X)x+ xNaOH R(OH)x + xNaX

Lu : Nu c 2 hoc 3 nhm -OH nh vo nguyn t c, ru km bn, tchuyn thnh cc cht khc bn hn:

R-CH(OH ) 2 ---- >R-CHO + H20

R-C(OH)3 ---- >R-COOH + H20

R-C(OH)2-R' ---- > R-CO-R' + H20

2. Cc phng php r ing b it

a) iu ch ru metylicCO + 2H2 xtt0|P > CH3OH

b) iu ch rUu etylc

( C 6 H i o 0 6 )n + n H 20 ---------men- > n C 6H 120 6hocH

C 6H 120 6 - n r ^ u > 2 C 2 H 5 O H + 2C 0 2t

Lu : Bi ton tnh theo hiu sut phn ng, tnh khi lng dung dch ruto thnh (theo ru):

ru = ^ L.ch x 100 %V,

dung dch ru

c) iu ch ru a chc - glixerin

* Thu phn du m ng thc vt (lipit)

C3H5(OCOR) 3 + 3NaOH C3H5(OH)3 + 3RCOONa(lipit) (glixerin) (x phng)

* T c 3h 6 ---- >C1CH2-CH=CH2 ---- > CH2C1-CH(0H)-CH2C1---- > C3H5(OH)3.



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10/233

SiT Piictm VRU THM

henol l nhng hp cht hu c c nhm -OH nh trc tip vo vngV d:

CH3phen ol o-crezol m-crezol p-crezol

crezol l nhng ng ng ca phenol.

tu thm l nhng hp cht hu c c nhm -OH nh vo mch nhnh

ocacbon thm. V d:C H o O H C H 0 - C H 0 O H

rinh cht 'do nh m -OH gy nn

X t p h n n g t c d n g v i k im lo i k i m

C6H5OH + Na C6H5ONa + VH2t

CH3-C 6H4OH + Na ---- CH3C6H4ONa + VH2t

s" mol H2suy ra s' nhm -OH trong cht hu c.

Xt p hn ng tc dng vi dung dch kim

C6H5OH + NaOH---- >C6H5ONa + H20

(kh tan) (d tan)

r s" mol NaOH suy ra s nhm -OH nh trc tip vo nhn thm, natri phenolat ti to phenol:

CgHONa + H 2O + CO2---- ^C 6 5OH + NaHC03

enol th hin tnh axit yu < H 2 C O 3 . Phenol cn c gi l axit phenic.

Phenol cng cho phn ng este ho (tnh ch t ca ru) nhng r t kh

CH3-COOH + HO-C6H5 CH3-COO-C 6H5 + h 20e s te p h e n y l a x e t a t



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11/233

iu ch este phenyl, ta cho phenol tc dng vi anhirit tng ng.

(CH3C0)20 + C6H5OH ---- >CH3COO-C6Hr) + CH3COOH

Ch : Khc vi ru, phenol khng cho phn ng th nhm -OH khi tc

dng vi HC1, HBr, H2S0 4...2. Tnh ch t do nh&n benze n gy n n

a) Xt phn ng brom ho phenol

C 6 H 5 O H + 3B r 2 --------> C 6H 2B r 3( O H ) + 3H B r2, 4, 6 trbrom phenol

Nu brom ho khng hon ton:

C6H5OH + xBr2 ---->C6H5_ xBrx(OH) + xHBr (x < 3)

) ( )

C6H5OH + 3 HNO3 I.2^ 4lc_. > C6H2(N 02)3OH + 3H20

2, 4, 6 trinitro phenol ( cam)

(axit picric)

Nu nitro ho khng hon ton:

C6HOH + xHNOs H2S0 -4.c-..-> C6H5-x(N02)x0H h- xH20 (x < 3)

c) Xt phn ng cng H2

C 6 H 5 O H + 3H 2 -------C e H i i O H ( x ic lo h e x a no l)

VI. P llu CHPHCHOl____________________ ,11

C 6 H 6 ------- -+Br^ - -> C e H g B r - - +- N- ^ H > C 6H 5 O H .Fe ,tu tu , p

Nu hiu sut t 100% th s" mo cc cht tham gia v to thnh sau phnng bng nhau.

B . BI TP

1cc B&i Tp fflu _ _ _ _ _ _________ M||M|

B n

n nng 0,166g hn hp hai r i H2SO4 c a h c hn hp haiolefin l ng ng lin tip (hiu sut phn ng 100%). Trn hai olefin vi1,4336 lt khng khi (o ktc). Sau khi t chy ht olefin v lm ngng t hi nc th hn hp kh cn li () l 1,5 lt (o 27,3c v 0,9856 atm).

a) Tm cng th: phn t v khi lng cc ru .

b) Tnh khi lng hi nc ngng t.



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12/233

c) Tnh t khi ca hn hp () so vi khng kh. (Bit khng kh cha 20% oxiv 80% nit v th tch).

Hng dn

Phng php gii: Theo phng php s" c trung bnh ( n ), theo cc phngtrnh phn ng, lp h phng trnh, tnh c n , suy ra cng thc phn t 2 ruv khi lng cc ru. T n suy ra mH 0 . T Ma suy ra (A/kk-

a) Theogi thit, 2 olefin l ng ng lin tip nn 2 ru l ru no nchc ng dng lin tip. t cng thc phn t tng ng 2 ru lCsH2n-+1OH, cng thc phn t tng ng 2 olefin l C-H2_ ( l s nguyn t

c trung bnh ca 2 ru v 2 len), n < < m = n + l ; n l s nguyn t c cru I; m l s" nguyn t c ca ru II.

nkk = 1,4336 : 22,4 = 0,064 mol, trong

n 0 = 20%x0,064 = 0,0128 mol; nN = 0,064x80% = 0,0512 mol"2

0,9856x1,5n . . = ----- - = 0,06 moiA ( c o 2>N2, o2d) 0,082(273 + 27,3)

Theo cc phn ng:

C L H ^ O H H2s4 t ^ c H + H a 0 (1)

(moi)

n 2n+ l n 2n

C-H2_ + 1,5 n 0 2 >C 0 2 + n H 20 (2)

0,166 l,5n.0,166 .0,166

14n + 18 l n +18 14n +18Ta c: nA= nc + n + n = 0,06

0,166.n + 0 ) 0 5 1 2 + 0,0128- 1 5 n -0 1 6 614n + 18 V 14n +18

=> n = 2,667 suy ra n = 2 ; m = 3.Cng thc phn t ru I l C2H5OH (M = 46) ; cng thc phn t ruII l

C3H7OH (M = 60).

* Tnh m cc ru: Gi X, y ln lt l s mol 2 ru trong hn hp, tac:

X + y = 0,166 : (14 n + 18) = 0,166 : (14x2,667 + 18) = 0,003 (*)Theo cc phng trnh phn ng:

C2H 5 OH + 302---- >2C02 + 3H20X mol 2x mol

C 3H 7 O H + 9/ 2O 2 --------> 3C 0 2 + 4H 20y mol 3y mol

Ta c: 2x + 3y = r x0,003 = 2,667x0,003 = 0,008 (**)

12



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13/233

Gii h phng trnh (*) v (**) ta c X= 0,001, y = 0,002

=> 1C H QH = 0,001 X46 = 0,046g ; mc H 0H = 0,002 X 60 = 0,12g

b) nH G = nco = 0,008 mol => mH Q= 18 X0,008 = 0,144 g

c) M kk = 28x0,8 + 32x0,2 = 28,8

^ 28x0,0512 + 44x0,008 + 32x0,008Ma = ------- 1------------- --------------- 2----- = 34,02

0,06

Vy dA/kk = 34,02 : 28,8 = 1,18.

(li 2

C hn hp 2 ru n chc A, B (hn hp X). un nng hn hp X vi H2SO 4c 180c th thu c hn hp 2 olefin, cn 140c th thu c hn hp ete,trong c 1 ete c khi lng phn t bng khi lng phn t ca mt trong 2 ru. Trong mt bnh kn dung tch khng i 4,2 lt cha a gam hn hp X v2,88g oxi. Cho ru bay hi ht 136,5c th p sut trong bnh lc l 0,8 atm. Sau khi bt tia la in t chy ht ru v cho sn phm chy ln lt i qua ng 1 ng P2O5(d) v ng 2 ng 14ml dung dch KOH 32% (d = 1,3 g i m) thykhi lng bnh 2 tng l,408g.

a) Xc nh cng thc phn t, vit cng thc cu to v gi tn cc ru.

b) Tnh nng phn trm ca cc cht trong bnh KOH, nu khng cho snphm chy qua ng P2O5m cho hp th tt c vo bnh KOH.

Hrng dnPhng php gii: Theo phng php s c trung bnh (r), da vo cc

phng trnh phn ng v u bi cho, suy ra cng thc phn t 2 ru. Theophng php gii ton lng cht d, t nc 0 v KOH suy ra khi lng cc cht

trong dung dch v c%.

a) Theo bi, sn phm ca phn ng loi nc l 2 olefin nn 2 ru banu l ru no n chc c s c > 2. t cng thc phn t tng ng 2 ru:C-Hq_ ,OH (x mol).n 2n+l

Vi cng thc phn t ru I l CnH2n+iOH, ru II l CmH2m+iOH n < r < m.

Theo bi M =M , , nn m = 2nCmH2m+1OH (CnH2n+1)20

Theo cc phng trnh phn ng:

C-HL- OH HSo c c _n 2n+l 180c n 2n z

2 C-H _ OH ^ 4c- > (C-H,- 0 + H 20n 2n+l 140 c ' n 2n +l /2

13



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14/233

C_H25+1OH + p ^ ] o 2 ---- nC0 2 +(n + l )H 20

Xmo xn molx( n + 1) mol0 8 x 4 2

S" mol kh tron g bnh = --------- 1 --------- =0,1 mol0,082 X(273+ 136,5)Trong s' mol 0 2 = 2,88 : 32 = 0,09 mol

Suy ra s mol (X): nx = 0,1 - 0,09 = 0,01 mol = X

M nco = 1,408 : 44 = 0,032 mol = x n . Vy r = 0,032 : 0,01 = 3,2

(khi lng dung dch KOH tng 1,408gchnh l mco )

Xt iu kin 2 < n < 3,2 < m = 2n. Nghim thch hp n = 2 ; m = 4.

Cng thc 2 ru l: CH3-C H 2OH (ru etylic) v C4H9OH butanol c 4 dngln: CH3-C H 2-C H 2-C H 2OH (ru n-butylic), (CH3)2CH-CH2OH (isobutylic),

CH3-C H 2-CH(OH)-CH3(ru secbutylic), (CH3)3C-OH (ru tertbutylic).

b) Ta c nn = 0,032 mol ; iKOH = 0,104 mol2 100x56

Yi JiOH - ^9^ > 2 nn mui tao thnh l K2CO3nC 2 0,032

Theo phng trnh phn ng:

C02 + 2 KOH ---- > K2CO3 + H20

S" mol ban u: 0,032 0,104S mol phn ng: 0,Q32 0,064 0,032S" mol sau phn ng: 0 0,04 0,032

mdd = m ddK O H + m co + n H 0 = 14x1,3 + 0,032x44 + 0,01(3 ,2+ l)xl8 = 20,364g

Nng phn trm cc cht trong dung dch:

c% (K2C0 3) = Q , Q 3 2 x l 3 8 x l fi- - = 2 1 ,6 8 %20.364

c% (KOH d) = 0^04x56x100% = 11%

20.364

Bi 3

C 2 ru n chc X v Y, trong phn t mi ru cha khng qu 3 nguyn cacbon. un nng hn hp X, Y vi H2SO 4 c 140c ta thu c hn hp 3 ite vi s mol bng nhau.

Ly mt trong 3 ete cho vo bnh kn dung tch l V (lt). Thm vo bnh l lgn hp kh gm c o v O2 c khi lng phn t trung bnh bng 220/7. un

14



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15/233

nng bnh ete bay hi c hn hp kh B c khi lng phn t trung bnh l35. Bt tia la in t chy ht hn hp kh trong bnh sau a v 0c th p sut kh trong bnh bng 0,7 atm. Lng O2d bng 1 /6 lng 0 2ban u.

a) Tm cng thc phn t ca 2 ru.

b) Tnh khi lng ca mi ru ete ho.

c) Tnh dung tch V ca bnh.

Hng dn

Phng php gii: Bi ny b cho d d kin: Phn t cha khng qu 3nguyn t Q Nu cho iu kin ny th khng cn cc iu kin khc, ta suy rac c||||fllic phn t 2 ru (CH3OH v C 2 H 5 O H ). T Ma suy ra nco v nQ .

T Mbv nQ phn ng suy ra cng thc phn t ete => cng thc phn t 2 ru

v khi lng mi ru. Da vo v n d suy ra V.uu2 u2

a) Tnh nco v n 0 trong hn hp A. Gi a, b ln lt l s' mol c o v 0 2

trong 11 gam A. Ta c h phng trnh:

fa + b = l l : = 0,35


16/233

V s" mol 3 ete u bng nhau nn suy ra s mol ca ru ban u cng bngnhau bng 3 ln 's mol mi ete = 3x0,05 = 0,15 mol.

mCH3 0 H= >15x32 = 4,8g ; m c 2 H5 Q H = 0,15x46 = 6 ,9g

c) S moi cc kh trong bnh sau khi t chy:

**CC>2(cl0'tCO) ^C 02 (trong ete) d= 0,05 + 0,05x3 + (0,3 - 0 , 3 x ^ ) = 0,25 mol

nRT 0,25x0 ,082x273 _ _V = = ----------------------- = o lt.

p 0,7

0i 4

Chia hn hp 2 ru thnh hai phn bng nhau:

a) Ly phn th nht cho vo bnh kn dung' tch 0,9 lt, sau cho ru bayhi 136,5c (trong bnh khng c cht g khc, lgoi ru). Khi ru bay hi ht

th p. sut trong bnh l 851,2 mmHg. Tnh tng s mol ca 2 ru c trong bnh.b) un nng phn th hai vi H 2 S O 4 c 140c th thu c 0,742g hn hp

3 ete. Tch ly phn ru cha tham gia phn ng (gm 40% lng ru c khilng phn t nh v 60% lrig ru c khi ng phn t ln hn) v un nngvi H2SO 4 c 180c th thu c V lt hn hp 2 olefin, gi thit hiu sut cc

phn ng to th nh olefin l 1 0 0 %.

Xc rih cng thc phn t 2 ru, bit rng khi lng phn t ca chngkhc nhau 28 vC. Tnh V ( ktc).

Hng dnPhng p hp gii: T s mol hn hp 2 ru, khi lng ete, theo nh lut

bo ton khi lng lp quan h gia s nguyn t cacbon (n) vi s mol ca 2 rubin lun tm nghim thch hp. .

a) Tnh tng s mol 2 ru trong phn 1 (ni): . .p .v 851,2 0,9

ri! = z: = . -------------------- = 0,03 molR.T 760 0,082.(136,5 + 273)

b) Khi ehira t ho 2 ru ta .thu c 2 olefin, vy 2 ru l ru n n chc:

Cng thc phn t ru I: CnH2

n+iOH (hay ROH: Xmoi)Cngthc phn t ru II: CmH2m+iOH (hay ROH: y ml)

V khi lng ph n t ca chng khc nhau 28 dvC suy ra: m = n + 2 (1)

Theo cu 1, ta c: X + y = 0,03 (2)

Theo cc phn g trn h phn ng: 2ROH ------- R20 + H20

2R O H --------> R 20 + H 20

R O H + R O H --------> R O R ' + H 20

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17/233

Vi nh lut bo ton khi lng: ^ m ru phn ng = ^ m ete + ]>}m.h20

T thnh phn olefin suy ra c 60% (0,6) ru th nht tham gia phn ng v40% (0,4) ru th hai tham gia phn ng ete ho, ta c:

un nng 132,8g hn hp p gm ba ru no, ri chc AOH, BOH v ROH vi

H2SO 4c 140c ta thu c 111,2 gam hn hp 6 ete c s mo bng nhau. Mtkhc un nng 132,8g hn hp p vi H2SO 4c 180c th thu c hn hp khch gm 2olefin.

a) Xc nh cng thc phn t, vit cng thc cu to ca cc ru. Cho bithiu sut cc phn ng l 1 0 0 %.

b) Tnh phn trm khi lng ca mi ru trong hn hp p.

c) Tnh phn trm th tch ca mi olefin trong hn hp ca chng.

Phng php gii: T khi lng hn hp p v khi lng 6 ete suy ra s" molH20 ri s" mol mi ete => s" mol mi ru. Lp quan h gia s c trong cc ruvi mp, bin lun tm nghim thch hp => cng thc phn t cc ru. T suyra %m v %v mi olefin.

Theo d bi, khi hirat ho hn hp 3 ru no p sn phm ch l 2 olefin,chng t 2 trong 3 ru l ng phn ca nhau, gi s l AOH v BOH cngthc phn t c dng ru no n chc: CnH2n+iOH v ROH c dng CmH2m+iOH.

0,6x(14n +18) + 0,4y(14m + 18) = 0,742 + 18(0,3x + 0,2y) (3)

Bin n: Vi iu kin 0 < X < 0,03 => 1,77 < n < 2,3.

M n < m = n + 2 . Nghim thch hp n = 2 ; m = 4

Cng thc phn t r.u I: C2H5OH(x = 0,01)

Cng thc phn t ru II: C4HgOH (y = 0,02)

* Theo cc phng trnh phn ng: C 2 H 5 O H -------->C 2 H 4 + H20(mol): 0,01x40% 0,004

C4H9OH ----->C4H8+ H20

(mol): 0,02x60% 0 , 0 1 2

Vy: V = (0,004 + 0,012-)22,4 = 0,3584 lt.

Bi 5

Hng d

2A-PPGBTHCHCCNC 17



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18/233

Theo cc phng trnh phn ng to ete:

2AOH A20 + H20 (1)

2B O H B 20 + H 20 ( 2)

2R O H r

20 + H

20 (

3)

A O H + B O H A O B + H 20 (4)

A O H + R O H A O R + H 20 ( 5)

B O H + R O H > B O R + H 20 ( 6 )

T a c : X n H O = ( 132,8 - 111,2) : 18 = 1,2 m o l

(V theo nh lut bo ton khi lng: m ru = m ete + X mH o ^

Suy ra nm5i ete = 1 , 2 : 6 = 0 , 2 mol

Theo cc phng trnh phn ng (1), (2), (3), (4), (5), (6 ) s" mol mi ru.ttAOH = n BOH= nROH= 4nmgi ete = 4x0,2 = 0,8 moi

Vy 2 ]m iu = 0,8x2(14n + 18) + 0,8(14m + 18) = 132,8 => 2n + m -- 8 .

Bin lun: OH v BOH l 2 ng phn th n > 3. Nghim thch hp n = 3v m = 2. Cng thc phn t AOH v BOH: C3H7OH ; cng thc cu to:

C H 3 - C H 2 - C H 2 O H ; C H 3- C H ( O H ) - C H 3

C ng t h c p h n t RO H : C2 H 5 O H .

b) Tnh %m cc ru: %I1A0H = %niBOH = xl00% = 36,14%132,8

%mROH= Q,8-4- x 100% = 27,72%.132,8

c) Tnh %v mi oen: Ta c %n = %v


C2H5OH t0xt > C2H4 + H20

C3H 7OH t0xt -> C3H6 + HsO(n- v iso)

V c ru n v isopropylic u to C3H6 nn s mol C3H6 gp 2 ln s moiC 2H 6 .

%Vn = xl00% = 33,3% ; %Vr = 100 - 33,3 = 66,7%.2 4 1 + 2 3 6

B6

Hn hp A gm ru metylic v ru propylic vi t l s mol tng ng l5 :1. Hn hp B gm 2 olefin khi iu kin thng. Chia A thnh 2 phn bng nhau:

18 2B-PPGBTHCHCCNC



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19/233

-Phn 1 cho tc dng ht vi Na ta thu c V lt H.2( ktc).

-Phn 2 un nng vi H2SO 4c 180c v cho hn hp sn phm kh (gm mt olefin v imetyl ete) li t t qua nc loi ht ete tan trong nc. Ly olefin cn li trn vi hn hp kh B ta c hn hp kh D c t khi so vi hiro l 21. Khi c mt Ni xc tc v un nng th D tc dng va ht vi V lt H2 trn.

a) Xc nh cng thc phn t ca cc olefin trong hn hp B.

b) Mun t chy B cn mt th tch O2gp bao nhiu th tch ca B ( cngiu kn nhit p sut). Gi thit cc phn ng xy ra hon ton.

Hng dn

Phng php gii: Chuyn bi ton hn hp thnh bi ton mt cht bngcch t cng thc phn t tng ng, da vo cc phng trnh phn ng v ccd kin trong d bi lp quan h gia n v s" mol cc cht, xc nh c r , suy

ra cng thc phn t 2 olefin v t tnh c Vn theo Vb.2

a) t cng thc phn t tng ng ca 2 olen l C_H2_ (olefin th 1 l

CnH2n, olefin th 2 l CmH2m) ; 2 < n < n < m < 4 .

Gi a l s" mol C3H7OH trong V2A suy ra s" mol CH3OH l 5a ; b l s" molC_H2_ trong B.


CH3OH + Na ---- >CH3ONa + VH2t (1)

5a moi 2,5a molC 3H 7 O H + N a --------> C 3 H 7 O N a + 2H2f ( 2)a mol 0,5a mol

2 CH3OH H2S 4c > (CH3)20 + H20 (3)

C3H7OH H 2S 4dc > C3H6 + H20 (4)a mol a mol

C 3H 6 + H 2 - N U > C 3H 8 ( 5 )a mol a mol

CLKL- + H2 > C-H_ (6 )n 2n 14 n 2n+2 v b mol b mol

Theo bi ra ta C: 2,5a + 0,5a = a + b=>b = 2a;

Md = 42a 1 145b = 42 (*)a + b

Thay b = 2a vo (*) ta c n - 3. Suy r a n = 2 < 3 < m = 4

Vy cng thc phn t 2 olefin l C2 4 v C4H8 .

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20/233

b) e-jig - + 1,5 i O2 ~ ^ n C0 2 + n H20

Rt ra: nQ = 1,5 nb = 1,5x 3iib hay VQ = 4,5VB.

Cho 47g hn hp hi ca 2 ru i qua AI2O3 nung nng (xc tc) ta c hnhp hi A gm ete, olefin, ru cn li v hi nc. Tch hi nc khi hn hp A ta c hn hp kh B. Ly nc tch ra trn cho tc dng ht vi kali thu c 4,704 lt H2 ( ktc). Lng olefin c trong B tc dng va vi 1,35 lt dungdch Br20,2 mol/l. Phn ete v ru c trong B chim th tch 16,128 lt 136,5c v 1 atm.

a) Tnh hiu sut ru b loi nc thnh olefin, bit rng hiu sut i vimi ru nh nhau v s mol cc ete bng nhau.

b) Xc nh cng thc phn t cc ru.Hng dn

Phng php gii: t cng thc tng ng ca hai ru l C5 H 2-+1OH .

Theo cc phng trnh phn ng tm s mol ru d v s mol ru ban u suy raH%. T suy ra cng thc phn t ru th iiht. Da vo s mol hn hp ru

ban u, s' mol ru d, khi lng hn hp ru, lp h phng tr nh, gii vbin lun tm nghim thch hp suy ra cng thc phn t ru 2 .

a) Khi loi nc khi hn hp ru ta c hn hp olefin, chng t 2 ru

ban du l ru no n chc. t cng thc phn t 2 ru l CnH2n+iOH,CmH2ni+1OH v cng thcphn t tng ng l C-H2- jOH .

2 < n < n< m ( l s' nguyn t c trung bnh ca 2 ru)


c l h 25+1o h -C-H 2 5 + h 20 (1)

2C A S+1 H Al2 3--t 0- > (CSH2, +i )2 + H * (2)

H20 + K ---- >KOH + y2H2T (3)

C h H 2 5 + B r 2 > C 5 H 2 S B r 2 ( 4 )

T (3) : nH 0 = 2 n H = (2x4,704) : 22,4 = 0,42 mol

T (4) : n r _ = n j = 1,35x0,2 = 0,27 molCnH2n Br2

^ A , .... ___ 16,128x1x273Tng s mol ete v ru d trong B = - _- = 0,48 moi

v 22,4(273 + 136,5)

T (1) : II = n olefin = 0,27 mol

20



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21/233

Ta c: n , , ,, = n ,, ,, - n , , v ,, = 0,42 - 0,27 = 0,15 molHUCKphn ng 2) H2 0(phnn g3) H2O (phn ng 1)

Mt khc: nrud = 0,48 - nete ; m: nete= nH 0(phnng2)

Nn: nrud = 0,48 - 0,15 = 0,33 mol

Vy tng s moi ru ban u = nru (1) + nru (2) + n ruud= 0,27 + 2x0,15 + 0,33 = 0,9 (mol)

0 27Hiu sut H = xl00% = 30%.

0,9

b) Khi lng phn t trung bnh ca 2 ru: M = 47 : 0,9 = 52,2

Ta c: 14 n + 18 = 52,2 r = 2,4Suy ra n = 2, cng thc phn t ru th nht l C2H5OH

Gi X, y ln lt l s' mol C2H5OH v CmH2m+iOH trong hn hp ru ban

u, ta c: fx + y = 0,9 _ 0,4|46x + (14m + 18)y =47 y m -2

Bin lun: Theo kt qu cu 1, s' mol ru CmH2m+iOH bin thnh olen bng30%xy = 0,3y.

S^ mol mi ete = 0,15 : 3 = 0,05 mol

V s" mol cc ete bng nhau nn s moi ru to ete nh nhau.

S mol CmH2m+iOH = 0,3 : 2 = 0,15 mol

Sauphnng ru CmH2m+iOH cnd, nh vy: y > 0,3y + 0,15 => y > 0,21Ta c0,21 < y < 0,9 o 0,21 < < 0,9 2,4< m


22/233

Hng dn

Phng ph p gii: T s mol hn hp u v s mol 2 suy ra s mol A, B, c.T mco , mH G, s" mol hn hp ru suy ra s" mol 0 2d v suy ra tng s" mol

hn hp sau phn ng :=> p. Dng phng php M, suy ra cng thc phn t 3ru A, B, c.

a) A, B, c l 3 ru n chc, c cng thc phn t tng ng l C-H-O ,

khi t chy:

C-H-O' + f x + - j o z > x C 0 2 + ^ H 20x y { 4 2 J 2

Ta bit H2SO4 c ht nc: nH = 3,78 : 18 = 0 , 2 1 moh2

KOH ht C02 theo phn ng: 2K0H + CO2 ---- K2 CO3 + H20

= 6,16 : 44 = 0,14 mol2. . . 16 x0,98x273

^ + n 0 2bl - 22,4(273 + 109,2) m 0

nOgban u = 1 3 4 4 : 3 2 = 4 2 m d => 2 > A+B,C = ' 5 - - 4 2 = 0 8 m o 1

Theo phng trnh phn ng, ta c:

n 0 2(trongru) + l 0 2(banu) ~ n 02(d) + n o 2(trongC02) + n 0 2(trongH20)

n O (d ) = 42 + (008 : 2 ) ~ 0)14 (0 21 : 2 ) = 21 5 mo1

Vy tng s" mol kh trong bnh sau khi chy bng._ n o i c ni / l 0,565x0,082(273 + 136,5)

ns = 0,215 + 0,14 + 0,21 = 0,565 ; p = --------- ------------- = 1,186 atm.16

b) Ta c: m(A+B+C) = Ittc + mH+ mo = 0,14x12 + 0,21x2 + 0,08x16 = 3,38g

Khi lng phn t trung bnh ca 3 ru M = 3,38 : 0,08 = 42,2

Nh vy c t nht 1 ru c khi lng phn t < 42,2 ; l CH3OH ng viA, v B v c c cng s nguyn t cacbon.

rn , 0,08x5 Ta co: nC30H - - 0,05 => mCHgOH- l , 6 g=> n(B+C) = 0,08 - 0,05 = 0,03 ; m(B+o = 3,38 - 1,6 = l,78g

Khi lng ph n t trung bnh M(B +C) = 1)78 : 0,03 = 59,3

Theo bi: B v c c cng s nguyn t cacbon, nn ta t cng thc phnt tng ng B, c l c H-,0 (' l s" nguyn t hiro trung bnh ca B v C).

Ta c:12X + ' + 16 = 59,3 => 12x + ' = 43,3

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23/233

Bin lun: y' < 2x + 2 ; X> 3. Nghim thch hp X= 3 ; y' = 6,3.

Vy ta chn cng thc phn t B v c l: C37OH v C3H5OH ; hoc C3H5OHv C3H7OH.

Hi 9

Cho hn hp hai ru no, n chc tc dng ht vi HBr ta thu c hn hphai ankyl bromua tng ng c khi lng gp i khi lng hai ru. Phn hu hai ankyl bromua chuyn brom thnh Br~ v cho tc dng vi AgN03 (d) ththu c 5,264g kt ta AgBr.

a) Tnh khi lng hai ru ban u.

b) Nu em t chy hn hp ru ban u v cho tt c sn phm hp th vo 280ml dung dch KOH 32% (d= l,3g/ml) th thu c dung dch A. Tnh nng

phn trm ca cc cht trong dung dch A.c) Cho bit tng s cacbon trong phn t hai ru bng 6, hy xc nh cng

thc phn t v s mol ca cc ru.

Hng dn

Phng php gii: Chuyn bi ton hn hp v bi ton mt cht, thng quacc phng trnh phn ng, gii bi ton theo phng php tng gim khi lng,suy ra khi lng 2 ru. Tnh n suy ra mco v mH 0 => c% dung dch A. T n

bin lun suy ra cng thc phn t v smol mi ru.

a) Tnh khi lng hai ru ban u: t cng thc phn t ru I lCnH2n+iOH, ru II l CmH2m+iOH

t cng thc phn t tng ng 2 ru: C-H2- 1OHvi: 1 < n < < m,

n l s" nguyn t cacbon trung bnh ca 2 ru.

Cc phng trnh phn ng:

C_H0_ .OH + H B r-----> C-HL- + H20 (1)n 2n+l n 2n+l

C-H2- jBr phflnh-l, > Br Agrc3 > AgBr (2)

Ta thy c 1 mol C_H2_ 1OH chuyn thnh C_H2_ 1Br khi lng tng

80 - 17 = 63g, to ra 188g AgBr.

Vy to ra 5,264g AgBr, s" mol ru phn ng l: 5,264 : 188 = 0,028 mol

Khi chuyn thnh C-H2- 1Brkhi lng tng: 0,028x63 = l,764g

m = 2 inxTheo bi v kt qu trn: < ^>mx= l,764g.

[mY- m x = 1,764

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24/233

''"'V.l'tf1

b) Tnh c% dung dch A:

C _ H 2 _ . J .O H + 1,55 0 2. -------- > n C 0 2 + ( n + l ) H 20 ( 3 )

(mol): 0,028 0,028 n 0,028(n + 1 )

Tnh n : M ru = = 1,764 : 0,028 = 14n + 18 => n = 3,21n .

Theo (3): n co = 0,028xn= 0,028x3,21 = 0,09 mol

nH 0 = 0,028( n +1) = 0,028(3,21 + 1) = 0,118 mol

_ 280x1,3x32 _ - nc _Theo bi: riKOH = ;------- = 2,08 mol

100x56Theo phng trnh phn ng:

C02 + 2KOH ---- > K2C0 3+H20

S" mol ban u: 0,09 2,08S" mol phn ng: 0,09 0,18 0,09S" mol sau phn ng: 0 1,9 0,09

Vy dung dch A c 0,09 mol K2CO3 ; 1,9 mol KOH d.

m ddA= mddKOH + ttVn + m w 280x1,3 + 0,09x44 + 0,118x18 = 370,084gL/Ug 2>

Vy c% (K2CO3) = (0,09x138x100%) : 370,084 = 3,35%

c% (KOH d) = (1,9x56x100%) : 370,084 = 28,75%.

c) Gi X, y ln lt l smoi CnH2n+iOH v CmH2Di+iOHTheo bi ta c h phng trnh:

n + m = 6nnoo _ 0,078 - 0,028nix + y = 0,028 o X = -------

J 6 - 2 nnx + my = 0,09

Bin lun: V n < m, nn n ch nhn 2 g tr thch hp bng 1 v 2 .

* Khi n = 1 th m = 5 ; X= 0,0125 ; y = 0,0155

Tc l: CH3OH (0,0125 mol) ; CgHnOH (0,0155 mol)

* Khi n = 2 th m = 4 ; X = 0,011 ; y = 0,017

Tc l: C2H5OH (0,011 mol) ; C4H9OH (0,017 mol).

Bi 10

Hn hp kh A gm etan v propan.

1 . t chy m t t hn hp A ta thu c kh C02 v hi nc theo t l thtch 11 : 15. Tnh phn trm v th tch mi cht trong hn hp A.

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25/233

2. un nng mt t hn hp A trong mt bnh kn khi c mt cht xc tcthch hp thc hin phn ng ehiro ho (tch mt phn t H2). Sau phn ngthu c hn hp kh (B) c t khi so vi hiro bng 13,5.

a) Tnh hiu sut phn ng ehiro ho, bit rng sn phm phn ng ch c

olefin v hiro; eian v propan b ehiro ho vi hiu sut nh nhau.

b) Tch hn hp olefin t hn hp B v hirat ho chng khi c mt axitH2SO 4 long thu c hn hp ru c. Ly m gam hn hp ru c cho tc dnght vi Na thy bay ra 448ml kh ( ktc). Oxi ho m gam hn hp ru c bng O2khng kh nhit cao v c Cu xc tc c hn hp sn phm D. Cho D tc dng vi Ag N 03trong N H 3 d thu c 2,806g bc kim loi.

Tnh phn trm s mol cc ru trong hn hp c. Gi thit cc phn nghirat ho olefin v phn ng oxi ho ru xy ra vi hiu sut 100%, D ch gmanehit v axeton.

Phng php gii: T t l V suy ra t l smoi, qua phn ng suy ra %Va-

Theo Mb ; %VAsuy ra H%. Theo cc phng trnh phn ng, lp h phngtrnh 3 n (s mol 3 ru) suy ra %n3ru.

1. Gi X, y ln lt l s mol C2H6v C38 trong A.

Theo cc phng trnh phn ng chy:

Hng dn

C2H6 + 3,502X mol

C38 + 5 O2 -

y molV

V. co9i ----- -

>2 > 2C02 + 3H202x mol 3x mol

(1)

> 3C02 + 4H203y mol 4y mol

(2)

Theo bi:Vh2 0

= 7 7 => y = 3x

2 . a) Theo kt qu 1, t l n : n = 1 : 3U2H6 3 8

Theo cc phng trnh phn ng v hiu sut phn ng h(%):

C2H6

aah

a - ah

c 2h 4 + h 2 (3)S" mol ban u:Smol phn ng:S' mol sau phn ng:

0 0ah ahah ah

25



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26/233

C 3H 8 xtr > C 3H 6 + H 2 (4)S mol ban u: 3aS" mol phn ng: 3ah 3ah 3ahS" mol sau phn ng: 3a - 3ah 3ah 3ah

Ta c: nA = a + 3a = 4a

nB = n C2H4 + n C3H6 + n H2 + n C2H6d + nC3Hgd = 4a + 4ah = 4a(l + h)

Theo nh lut bo ton khi lng th khi lng hn hp trc v sau phnng bng nhau:

rriA = fliB = 30a + 44x3a = 162a

Theo bi: Mb = = - 12~ - = 13,5x2 = 27 => h = 0,5 hay 50%.nB 4a(l + h)

b) Xt phn ng cng H2O vo olefin

CH 2=CH2+ H20 > CH3-CH2OH (5)

CH 3-CH=CH2 + H20 > CH3-CH(OH)-CH3 (6 )

CH 3-CH=CH2+ H20 CH3-C H 2-CH2OH (7)

Gi X, y, z ln lt l s mol ru C2H5OH, (CH3)2CHOH v C H 3 - C H 2 - C H 2 O Htrong m (g) c. Theo cc phng trnh phn ng:

C2H5OH + Na ---- > C2H5ONa + 0,5H2T

X moi 0,5x mol(CH3)2CHOH + N a ---- >(CH3)2CHONa + 0,5H2t

y mol 0,5y mol

C H 3- C H 2- C H 2O H + N a -------- C H 3- C H 2- C H 2O N a + 0,5H 2 t

z mol 0,5z mol

Theo bi ta c: 0,5(x + y + z) = 448 : 22400 = 0,02 mol (*)

Khi cho m (g) c tc dng vi 0 2/Cu, t:

C H 3- C H 2 O H + 0, 502 ^ - t0 - - - > C H 3 - C H 0 + H 20X m o X m ol

(CH3)2CHOH + 0,50 2 ( CH3)2C=0 + H20

CH3-CH 2-CH2OH + 0,502 - u--0 > CH3-CH2-CHO + H20z mol z mol

Khi cho D tc dng vi dung dch AgNC>3 trong NH3 d, anehit tham giaphn ng tr ng bc:

26



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27/233

CHgCHO + 2AgNOg + 3NH3+ H20X mol

C H 3 C O O N H 4 + 2NH 4NO3+ 2 Ag2 x mol

C2H5-CHO + 2AgN0 3+ 3NHg + H20 ---- >C2H5COONH4 + 2NH4NO3 + 2 Ag

Gii h phng trnh (*), (**), (***) ta c: X = 0,01 ; y = 0,027 ; z = 0,003

Vy trong (C) c %nc H 0H = (0,01 : 0,04)xl00% = 25%

T khi hi so vi khng kh (M-kk = 29) ca hn hp u (CO + H2) l 0,5 vca hn hp sau phn ng l 0,6 .

1. Tnh % th tch ca cc kh trong hn hp u v sau phn ng.

2. Nu thc hin phn ng trn trong bnh 10 lt 327c v p sut Pi; sau

phn ng gi nhi t khng i th p sut P2 - 110 atm. T nh Pj.3. Nu cho hi CH3OH thu c trn (phn 2) qua ng ng CuO nung

nng, thu c hn hp hi B gm HCHO v HCOOH. Ngng t B v thm H20 thnh 1 0 0ml ta c dung dch c.

Kh cho lOml dung dch c phn ng vi dung dch AgN03d trong amonac,thu c 75,6g Ag; mt khc lOml dung dch c trung ho c 50ml NaOH IM.

a) Tnh phn trm khi lng cc cht trong hn hp B.

b) Nu lng CH3OH ban u ch b oxi ho thnh HCHO vi hiu sut 75%

th lng CuO tham gia phn ng l bao nhiu?

Phng php gii:Theo dng ton lng cht d.

T t khi d => M => %n (hoc %V). Khi V, T khng i n t l vi p => Pi.

Tnh theo phng trnh ph n ng suy ra %mBv mCu0 -

1. Gi X, y ln lt l s" raol c o v H2trong hn hp X (hn hp du).

z mol

Ta c: 2x + 2z = 2,808 : 108 = 0,026Mt khc theo cu 1 th: y + z = 3x

2 z mol

CO + 2HZ > CH3OHt, p

Hng dn

27



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28/233

'

t l:

Theo d bi: Mx = ggg = 0,5x29 => y = l,08xx + y

Suy ra trong X c: %Vco = ---- ----- x l0 0 % = 48,08% ;J * uu 1 + 1,08

%v =1>- - xl00% = 51,92%2 1+ 1,08

Theo phng trnh phn ng: c o + 2H2 > CH3OHS" mol ban u: X yS' mo phn ng: Xi 2xi XiS" mol sau phn ng: (x-Xi) (y-2xi) Xi

* 28(x - X,) + 2(y - 2x,) + 32xn , ~ ,Theo bi: My= ---------- ------------ -----------= 0,6x29 (Y l hn hp sau)

(x -x 1) + ( y - 2 x1) + x1

10,6 x = 15,4y - 34,8xj m y = l,08x nn suy ra Xi = 0,173x. Vy trong Y c

Vco Vh2 VCH3OH = (x- x : : *1 = 4 7 8 : 4 2 4 1

Vy %VCo = (4,78x100%) : (4,78 + 4,24 + 1 ) = 47,7%%VH = (4,24x100%) : (10,02) = 42,32%

%VCH 0H = (1x100%) : (10,02) = 9,98%.

2. V, T khng i th p sut t l vi s" mol.

P. n x + y 2,08x ,- = - => Pi = llOx---------- = 110x = 132 atm.P2 n2 .x + y - 2 x1 l,734x

3. a) Theo kt qu cu 1:

1 1 0 x 1 0n C H O H = 9,98% X nB = 9,98%X

0,082(273 + 327)= 2,23 mol

Gi a, b ln lt l s" mol HCHO v HCOOH trong 1/10 hn hp B. Theo ccphng trnh phn ng:

CH3 OH + CuO > HCHO + Cu + H20a mol a mol a mol

CH3OH + 2 CuO > HCOOH + 2Cu + H20b mol b mol b mol

Khi cho (C) phn ng vi dung dch AgN3/NH3:

HCHO + 4AgN03+ 6NH3+ 2H20 ---- >4Ag + 4NH4NO3+ (NH4)2C03a mol 4a mol

28



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29/233

HCOOH + 2AgN03+ 4NH3+ H20 ---- * 2 Ag + 2NH4NO3+ (NH4)2C03b mol 2b mol

Khi cho (C) phn ng trung ho vi dung dch NaOH.

HCOOH + NaOH ---- > HCOONa + H20

b mol b mol

Theo cho: b = riNaOH = 0,05 X 1 = 0,05 mol

4a + 2b = nAg = 75,6 : 108 = 0,7 mol

Ta c: a = 0,15 mol v b = 0,05 mol.

Vy trong B c: mHcHO = 10x0,15x30 = 45g; mHcooH = 10x0,05x46 = 23g

m CH OHd = [ 2 2 3 - 10x(0>15 + 0,05)]x32 = 7,36g ;

raH 0 = 10x(0,15 + 0,05)xl8 = 36g

Khi lng hn hp B bng 45 + 23 + 7,36 + 36 = lll,3 6 g

Vy %HCHO = (45 : ll l,36)x l00% = 40,41%

%HCOOH = (23 : lll,36)xl00% = 20,65%

%CH3OH d = (7,36 : ll l,36)x l00% = 6,61%

%H20 = (100 - 40,41 - 20,65 - 6,61)% = 32,33%.

b) mCu 0 = 2,23x80x-^- = 133,8g.100

Bi 12

Chia hn hp A gm ru metylic v mt ru ng ng thnh 3 phn bngnhau: Cho phn th nht tc dng ht vi natri thy bay ra 336ml H2 ( ktc). Oxho phn th hai bng CuO thnh anehit (hiu sut 100%), sau cho tc dngvi AgN03 trong NH3 d th thu c 10,8g kim loi. Cho phn th ba bay hi vtrn vi mt lng d oxi th thu c 5,824 lt kh 136,5c v 0,75 atm. Sau khi bt tia la in t chy ht ru th thu c 5,376 t kh 136,5c v 1 atm.

a) Vit cc phng trnh phn ng xy ra.b) Xc nh cng thc phn t ca ru ng ng.

c) Nu khng bit ru th 2 l ng ng ca ru metylic m ch bit n l ru bc nht n chc th c th tm c cng thc ca ru th hai hai hay khng ?

Hng dn

Phng php gii:Theo cc phng trnh phn ng, lp h phng trnh tms mol 2 ru. Da vo phng trnh phn ng chy, t s mol hn hp kh => s

29



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30/233

mol C02, suy ra cng thc phn t ru. Theo phng trnh phn ng chy, tmc gii hn X > 4 (x l s nguyn t c trong cng thc phn t ru). Da von tm c X < 5. Suy ra cng thc phn t ru.u2

a) t cng thc phn t dng dng l CnH 2n+iCH2OH n > 1, nguyn.

Phn 1: CH3OH + N a ---- >CHgONa + VH2T (1)

CnH2n+1CH2OH + N a ---- >CnH2n+1CH2ONa + VH2t (2)

Phn 2: CH3OH + CuO HCHO + Cu + H20 (3)

C n H 2 n + i C H 2 O H + C u O > C nH 2n + 1 C H O + C u + H 20 ( 4 )

HCHO + 4AgN03 + 6NH3+ 2H20 -> (NH 4)2C03 + 4NH 4NO3+ 4A g (5)

CnH2n+1CHO + 2AgN03+ 3NH 3+ H20 ---- >CnH2n+1COONH4++ 2NH4NO3 +2A g (6)

Phn 3: CH3OH + 1,502 > C 0 2+ 2H20 (7)

CnH 2n+iCH2OH + l,5(n+l)0 2 ---- > (n+l)C02+ (n+2)H20 (8 )

b) Gi X, y ln lt l s' mol CH3OH v ru ng ng.

Theo (1), (2): X + y = 2n = 2x(0,336 : 22,4) = 0,03n 2

Theo (5), (6 ): 4x + 2y = nAg = (10,8 : 108) = 0,1

Gii h phng trnh ta c: X= 0,02 ; y = 0,01

Khi trn phn th 3 (ho hi) vi oxi, ta c:5,824x0,75 1 'n_ , ,, = -----------0,03 = 0,1 mol

o2 ban u 0,082(273 + 136,5)

Tng s' mol kh sau phn ng = n c 0 + nH 0 + nQ d =

= [0,02 + 0,01(n+l)] + [0,04 + 0,01(n+2)] + [0,1 - 0,03 - 0,015(n+l)]

= ------ 6j376xl----- _ 0 => n = 30,082(273 + 136,5)

V y h n h p r u l C H 3 O H v C 4 H 9 O H .

c) t cng thc phn t ca ru l xHyO (0,01 mol)

Theo phng trnh phn ng chy:

CxHyO + x + - - j 0 2 ---- > xC02+ | h 20

Ta c tng s' mol kh sau phn ng:

(0,02 + 0,01x) + (0,04 + 0 ,0 1 x ) + 0,1 - [0,03 + 0,0 1|x + - - | ] = 0,162 V 4 2)

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31/233

I"5*

Suy ra y = 10, m y < 2x + 2 => X > 4

V 2 d: 0,1 - [0,03 + 0,01 (x + - ) ] > 0, nn X < 5 => X = 4 4 2

Cng thc phn t ru l C4H9OH.

Bi 13Hn hp X gm 2 ru no mch thng A v B (trong A l ru n chc)

c t l khi lng l 1 : 1. Khi cho X tc dng ht vi Na th th tch kh H2 sinhra t B bng 16/17 th tch H2sinh ra t A (o cng iu kin nhit , p sut).

Mt khc, khi t chy ht 13f6g hn hp X thu c 10,36 lt CO2 (ktc).

a) Tm cng thc ca A v B, bit rng t khi h ca B so vi A l 4,25. Gitn ca A v B.

b) Oxi ho 19,2g A c xc tc thch hp thu c hn hp sn phm c. Chia c

thnh 3 phn bng nhau. Ly mt phn cho phn ng vi dung dch AgN03 trongamoniac d, c 64,8g Ag. trung ho phn th 2 cn 30ml dung dch KOH2M. Tnh hiu sut ca qu trnh oxi ho A.

Ly phn th 3 cho tc dng vi B kh i c cht xc tc thch hp thu ccht D ch cha mt loi nhm chc. Tnh khi lng cht B tham gia phn ng,bit hiu sut ph n ng l 1 0 0 %.

Hng dn

Phng php gii: Theo cc phng tr nh phn ng tc dng vi Na, da vot l th tch H2 do A, B gii phng v t l Mb/Ma = 4,25, suy ra s' nhm OHtrong B. Theo cc phng trnh phn ng chy ca A, B lp h phng trnh suy racng thc phn t ca A, B. T iAg v riKOH=> nAb oxi ho => H%.

a) t cng thc phn t ru A: CnH2n+iOH (a gam) ; cng thc phn t B l:CnH2n+2_t(OH)t (a gam)

Theo gi thit: MB= 4,25MA

CnH2n +iOH + Na ---- CnH2n+1ONa + VH2 (1)Ma (gam) 0,5 mola (gam) 0,5 a/MA

CnH2+2-t(OH)t + tN a ---- >CnH2+2-t(ONa)t + t/2H2t (2)4,25Ma (g) 0,5t mol

0 ,5ata (gam)

4,25MA16 0,5at 16 0,5a .

V v /0X= -V ,^nn - - = x-^- => t = 4h2 (2) 17 H2 (1) 4,25Ma 17 MA

Cng thc phn t B c dng CmH2m+2 0 4 .

31



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32/233

Xt phn ng t chy A: V t l khi lng 1 : 1 nn mA= me = 13,6 : 2= 6 ,8 g. Theo cc phng trnh phn ng:

CnH2n+20 + l ,5n0 2--> nC0 2 + (n+l)H 20

Ma (g) n (nol)o n 6 , 8 . n, ,,6 ,8 g (mol)

Ma

CmH2m+20 4 + l,5(m-l)0 2 ---- mC02 + (m+l)H20

4,25 Ma (g) m (mol)

6 ,8 g - % - t a o l ) * 4,25Ma

Ta c: + 10,36 : 22,4 = 0,4625

Ma 4,25Ma=> l , 6 m = 8,325 - 0,325n (*)

Mt khc Mb = 4,25Ma => 14m+66 = 4,25(14n+18) :=> m = 4,25n + 0,75 '(**).

Gii h (*), (**) ta c n = 1 (CH3OH metanol) ; v m = 5 (C5H12O4)

B c 3 ng phn mch thng (bn):

H C O O K + H 20

nHC00H = i c o h = 0 ,0 3 X2 = 0 ,0 6 m o l

HCOOH + 2AgN03+ 4NH3+ H20 ---- >(NH4)2C0 3+ 2 Ag + 2NH4NO30,06 mol 0 , 1 2 molVy s mol Ag do HCHO to ra = (64,8 : 108) - 0,12 = 0,48 mol

32



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33/233

HCHO + 4AgN03 + 6NH3 + 2H20 ---- >(NH4)2C0 3 + 4A g + 4NH 4NO30,12 mol 0,48 mol

Vy c 0,06 mol HCOOH v 0,12 mol HCHO to ra khi oxi ho 0,2 mol CH 3OH

Hiu sut phn ng oxi ho CH3 OH = ^ . 100% = 90%

* Tnh rriB: Theo phng tr n h phn ng:

C5H8(OH )4+ 4HCOOH---- >(HCOO)4 C5H8+ 4H20nB= V4nhcooh = V4 X0,06 = 0,015 mol ; mB = 0,015 X 136 = 2,04g.

Qi 14

Cho hn hp A gm mt ru no n chc v mt ru khng no (c 1 ni i)n chc. Chia A thnh 2 phn bng nhau, mi phn a (g). Ly phn 1 cho vobnh kn B dung tch 12 lt v cho bay hi 136,5c. Khi ru bay hoi ht th p

sut trong bnh l 0,14 atm. em este ho phn 2 vi 30g axit axetic; hiu sutphn ng este ho i vi mi ru u l h%.

a) Tnh tng khi lng este thu c theo a v h.

b) Bm 8 gam oxi vo bnh B, sau khi bt ta la in t chy ht ccru v a bnh v nhit ban u (136,5C) th p sut trong bnh l 0,98 atm. Cho sn phm chy hpth ht trong dung dchNaOH d, sau thmdung dch

BaC2 d vo th thyto thnh 23,64g kt ta. Xc nh cngthc ph n t, vitcng thc cu to gi tn cc ru.

Hng dnPhng php gii: Dng cng thc phn t tng ng 2 ru ROH, theo

phng tr n h phn ng, lng me tnh theo s" mol hn hp ru v h. Da vo ccphng trn h phn ng lp h phng tr nh, bin lun suy ra cng thc phn t2 ru.

a) t cng thc phn t ru no CnH2n+iOH (hay ROH) n > 1; cng thc phnt ru khng no CmH2m_iOH (hay R'OH) m > 3.

Cng thc phn t tng ng: R OH ( M )

R O H + C H 3 C O O H - > C H 3 C O O R + H 20

0,14x12 _ .nru = ------ ----------------- = 0,05 mol

0,082(136,5 + 273)

naxit = 30 : 60 = 0,5 mol

V axit d nn khi lng este tnh theo ru

mE = 0,05[59 + (M -17)]x = 0,05 59 + 1---- 17100 to,05

X = ( a + 2 , l ) x 100 100

3A-PPGBTHCHCCNC 33



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34/233

b) Lp cng thc phn t, cng thc cu to, gi tn 2 ru:

Gi X, y ln lt l so" mol ru no v ru khng no trong mi phn, ta c:

X + y = 0,05 (1)

S" mol 0 2cho vo bnh B = 8 : 32 = 0,25 molTheo phng trnh phn ng t chy ru:

CnH2n+iOH + l,5n0 2 nC0 2 + (n+l)H20X l,5nx nx (n+l)x (mol)

C m H 2m- iO H + ( l , 5m - 0 ,5)0 2 --------> m C 0 2 + m H 20y (1,5m - 0,5)y my my (mol)

S mol hn hp C 0 2+ H20 + 0 2d: nh5n hp = - = 3 5 mou, Uo2 ( , 2 + 1 0 0 ,0 ,)

Qua NaOH d: C2

+2

NaO H----

>Na2

CO + H2

ONa2C0 3+ BaCla---- > BaC034 + 2NaCl

Suy ra nco = nB c 0 = 23,64 : 197 = 0,12 mol, haynx +my= 0,12 (2)

S' mol H20 + 0 2 d: (n+l)x + my + [0,25-l,5nx-(l,5m -0,5)y] =0,35 - 0 , 1 2

=> X + 0,5y - 0,(nx + my) = -0,02 (3)

Kt hp (1), (2), (3), gii ra: X = 0,03 ; y = 0,02

Thay vo (2) ta c: 3n + 2m = 12

iu kin CmH2m_iOH ru khng no m > 3 ; nghim thch hp m = 3 ; n = 2

Cng thc phn t v cng thc cu to 2 ru l:

C2H5OH c h 3- c h 2- o h

C 3 H 5 O H C H 2= C H - C H 2 O H .

fli 15

Hn hp X gm 3 ru AOH, BOH v B'OH, trong AOH v BOH cng dyng ng; BOH v B'OH c cng s nguyn t cacbon v mch cacbon thng.

un nng 20,3g hnhp X vi mt lng d CH3COOH khic mt H2SO 4cth thu c 41,3g hn hp 3 este (gi thit hiu sut ph n ng este ho l 100%).

Mt khc t chy 4,06g hn hp X th thu c 13,64g C02, cn nu cho 20,3ghn hp X tc dng vi nc brom thy c 40g Br2tham gia phn ng. Nu ly sn

phm cha brom em thu phn bng kim th thu c ru 3 ln ru.

a) T nh khi lng phn t trung bnh ca hn hp X v xc n h cng thcphn i cc ru trong hn hp X, bit rng c m t ru l ru metylc.

b) Tnh s mo mi ru trong 1 mol hn hp X.

34 3B-PPGBTHCHCCNC



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35/233

Hng d

Phng php gi i: t cng thc tng ng, theo phng php tng gimkhi lng suy ra Mx, bit AOH l CH3OH suy ra BOH l ru ng ngCnH2n+iOH. Mt khc X lm phai mu dung dch Br2, suy ra B'OH l CnH 2n-iOH.

Theo cc phng trnh phn ng, lp h phng trnh, bin lun tm nghim thchhp v s" moi cho mi trng hp.

a) t cng thc phn t tng ng 3 ru l ROH.

ROH + CH3COOH------->CH3COOR + H20 (1)

nROH = neste = 41l ~ 23 '3 = 5 mol ; Mx = 20,3 : 0,5 = 40,659-17

Suy ra t nht c 1 ru c khi lng phn t < 40,6 l CH3OH.V BOHv B'OH c cng s" nguyn t c (t 3 cacbon tr ln) nn AOH l CH3OH.

BOH cng dy ng ng vi CH3OH nn c cng thc dng CnH2n +lOH. Khiqua dung dch Br2 ch B'OH tc dng nn B'OH l ru cha no. Sn phm to rado B'OH tc dng vi Br2 khi thu phn cho ru c 3 nhm OH. Vy trong phnt sn phm c 2 nguyn t Br tc l B'OH l ru khng no c 1 n i:CnH2n_iOH.

CnHa^OH + Br2 ----------- >CnH2n - 1Br2OH (2 )

nB'OH = n = 40 : 160 = 0,25 mola r2

Gi X, y ln lt l s' mol ca CH3OH v CnH2n+OH trong 20,3g X. (hay

0,5 mol hn hp). Ta c:X + y + 0,25 = 0,5 (*)

CH3OH + 1,502 > C0 2 + 2H20

X m o l X m o l

CnH2n+1OH+ ^ 0 2 -> nC02 + (n+l)H20

y mol ny mol

C H 2 n - i O H + 0 2 --------> n C 0 2 + n H 200,25 mol 0,25n mol

Ta c: X+ ny + 0,25n = nco (t 20,3g X)

Khi t 4,06g X to ra 13,64 : 44 = 0,31 mol CO2

Vy khi t chy 20,3g X to ra (20,3 X0,31) : 4,06 = 1,55 mol C02

X+ ny + 0,25n = 1,55 (**)

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36/233

, , v ,, , l ,3-0 ,25nKt hp (*) v (**) ta rt ra: y = --------------

n - 1Bin lun: T khong xc nh 0 < y < 0,25. Suy ra khong xc nh 3,1 < n < 5,2

Nghim thch hp n = 4 v n = 5

- Khi n = 4 => X = 0,15 ; y = 0,1 ; cng thc phn t: CH3OH; C4H9OH;C4H7OH.

- Khi n = 5 => X =0,2375 ; y = 0,0125 ; cng thc phn t: CH3OH; C5H 11OH;C5H9OH.

b) Tnh si mol mi ru trong 1 mol hn hp X.

- Trng hp 1: CH3OH: 0,15 X2 = 0,3 mol

C4H9OH: 0,1 X2 = 0,2 mol

C4H7OH: 0,25 X2 = 0,5 mol

- Trng hp 2: CH3OH: 0,2375 X2 = 0,475 mol

C5H11OH: 0,0125 X2 = 0,025 mo

C5H9OH: 0,25 X2 = 0,5 mol

@i 1

Hn hp kh A ( ktc) gm 2 olefin l ng ng lin tip. t chy 8,96 lthn hp kh A J' cho n phm ch ln l i qa bnh ng P2O5 bnh 2 ng KOH rn thy khi lng binh 1 tng m gam, cn bnh 2 tng(m + 39) gam.

a) Xc nh cng thc phn t v tnh phn trm v th tchcamiolefintrong hn hp A.

b) Cho 8,96 lt hn hp A hp H2O (khi c mt axit xc tc) ta thu c hnhp 2 ru (gi s ch c 2 sn phm chnh ). Hiu sut phn ng hp nc ca mi olefin u l 50%. Ly hn hp 2 ru trn vi mtlng d axitfomic v axitaxetic ri un nng vi H2SO 4 c mt thi gian th thu cll,811g hnhp 4este. Bit rng c 60% ru c khi lng phn t nh hn v 55% ru c khilng phn t ln hn tham gia phn ng este ho. Tnh khi lng H2O tothnh v khi lng mi axit tham gia phn ng este ho.

Hng dnPhng php gi i: Dng cng thc tng ng, theo phng trnh phn ng

chy, d dng r t r a n = 3,75 suy ra cng thc phn t 2 olefin v %Va. Theo ccphng trnh phn ng, hiu sut cc phn ng hp H20 este ho suy ra s" moleste to ra t mi ru. Theo nh lut bo ton khi lng => mH 0 v mmi axit-

a) nh cng thc phn t v %VA:

S moi hn hp 2 olefin nA= 8,96 : 22,4 = 0,4 mol

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37/233

t cng thc phn t tng ng ca 2 olefin l: C-H2- (n l s' nguyn t

cacbon trung bnh ca 2 olefin)

C_H2_ + 1,5 n 0 2 ---- > n C 0 2 + nH 20

0,4 mol 0,4nmol 0,4nmolTheo bi: mco - mH 0 = (m + 39) - m = 39g

Hay 0,4nx44 - 0,4 n x l8 = 39 => n =3,75

Vy cng thc phn t 2 olefin k tip: C3H6v C4H8.

Gi X, y ln lt l s mol C3H6v C4H8 trong 0,4 mol A.f X+ y = 0,4

Ta c h: -T X= 0,1 ; y = 0,3[3x + 4y = 0,4x3 ,75

Vy %vr = x l0 0% = 25% ; %Vr = 100 - 25 = 75%.3 6 0,4 4 8

b) Tnh mnc to thnh v maxit tham gia phn ng este ho:

S' mol C3H6tc dng vi nc = 0,1x50% = 0,05 mol

S" mol C4H8tc dng vi nc = 0,3x50% = 0,15 mol

C3He + H20 > C3H7OH (sn phm chnh )0,05 moi 0,05 mol

C 4H 8 + H2O C4H9OH (sn phm ch n h)

0,15 mol 0,15 mol

C3H7OH + HCOOH -> HCOOC3H7+ H20C3H7OH 4- CH3COOH >CH3COOC3H7+ H20

C 4 H 9 O H + H C O O H -> H C 0 0 C 4H 9 + H 20

c 4h 9o h + CH3COOH ---- >CH3COOC4H9+ h 20

n C3H7OH phn ng esteho = n H20 = 0>05x60% = 0 ,0 3 m ol

nC4H9OHphn ngesteho = nH20 = 0>15x55% = 0,0825 moi

Vy mH 0 = (0,03 + 0,0825)xl8 = 2,025g

Khi lng axit tham gia phn ng:

maxit = meste + mH 0 - mru = 11,811 + 2,025 - 0,03x60 - 0,0825x74 = 5,931g

Xn a x i t P h n ns =z II ru phn ng (0,03 + 0,0825) molGi X, y ln lt l s mol HCOOH v CH3COOH tham gia phn ng:

x + y = 0,03 + 0,0825 x = 0,0585 HCOOH = 2,611g

(46x + 60y = 5,931 |y = 0,054 cHgCOOH = 3,24g

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38/233

Bi 17

t chy 5,8g cht A ta thu c 2,65g NCL2CO3, 2,25g H2O v 12,lg CO2.

a) Xc nh cng thc phn t ca A, bit rng mt phn t A ch cha mtnguyn t oxi.

b) Cho kh CO2 sc vo dung dch ca A thu c cht B l mt dn xut ca benzen. trung ho a gam hn hp gm B v mt ng ng tip theo (C) ca B

cn dng 200g dung dch NaOH nng %. T nh t l s mo ca B v c trong31

hn hp.

c ) C h o B c d n g i h n h p H N O 3 c d H 2 S O 4 c h c c h T .

Cho 18,32g T vo mt bnh chu p sut, dung tch khng i 560cm3, v lm ncht T 1911c. Tnh p sut trong bnh ti nhit , bit rng sn phm n l hn hp CO, CO2, N2, 2v p sut thc t nh hn p sut l thuyt 1 0 %.

Hng dn

Phng php gii: Lp cng thc phn t theo phng php khi lng.Theo 2 phng trn h ph n ng trung ho, lp h phng trnh tm t l n-B : ric.

Theo phng rn h ph n ng n, riT => ng mo k h => p.

a) Tm cng thc phn t A:

_ _ 2 , 6 5 ^ 0 12,1 1 0 0 _ _ 2,25 _ n ocmc =-rp- xl2 + xl2 = 3,6g ;mH=- L x2 = 0,25g

106 44 18

mNa = 0,025x46 = l,15g ; m0 = 5,8 - (3,6 + 0,25 + 1,15) = 0,8g

t cng thc phn t,ca A l CxHyOzNat, ta c:

X: y \ z : t i l $ . 6 : 5 : 1 : !12 1 16 23

V phn t A c 1 nguyn t oxi nn cng thc phn t A l C 6H5ONa.

b) Tnh t l s" mol B v c trong hn hp:

C6H5ONa + H20 + C02 ---- >C6H5OH (B) + NaHCOa

c l ng ng k tip B nn c l CH3C6H4OH

Gi X, y ln lt l s" mol B, c trong a gam hn hp ta c:

94x + 108y = a (1 )

C6 H5OH + NaOH ---- > C6H5ONa + H20

C H 3 C 6H 4O H + N aO H > C H 3 C 6H 4ON a + H 20

y moi y mol

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39/233

m , 200 X 6 a 0,3a ,Ta c: X + y = ------------ = (2)

100x31x40 31Gii h phng trnh (1 ), (2): y = 2x. Vy nB : nc = 1 : 2.

c) Tnh p sut ca bnh: B + H N O 3 ---- > T

C 6H 5O H + 3 H N O 3 --------> C 6H 2( N 0 2 ) 30 H + 3H 20Theo bi ra iT = 18,32 : 229 = 0,08 mol

Theo phn ng n: C6H^37N3 ---- 1,5N2+ 1,5H2+ 5CO + C0 2

Tng smol kh = 0,08x(l,5 + 1,5 + 5 + 1) = 0,72 mol0 _ 0,72.0,082.(273 + 1911) _ OQn OK ,Pi thuyt = --------1 7:----------------- = 2 3 0 25 a t m0,56

Pthc t = Pi thuyt = ^ x230,25 = 207,22 atm.

1. Mt ru n chc A tc dng vi HBr cho hp cht B cha c, H, v58,4%Br. Nu un nng A vi H2SO4 c 170c th thu c 2 anken. Vit cngthccu to A, B v cc anken.

2. Khi phn tch cht hu c A (cha c, H, O) th c mc + mn = 3,5mo

a) Tm cng thc n gin ca A.

b) Ly 2 ru n chc X, Y em un nng vi H2SO4 c t th ch hp ththu c A. Xc nh cng thc cu to (mch h) A, X, Y. Bit A l ete.

3. t chy hon ton 0,05 mol ru noX (h) cn 5,6gO2 to ra 6 ,6 g CO2 .

a) Lp cng thc cu to ca X.

b) Vit phn ng iu ch X t butan v t este c sn trong t nhin.

4. Vit cng thc cu to v gi tn tt c cc hp cht ch cha c, H, o cM = 60 vC. Nhng cht no trong s' cc cht c th chuyn ho theo s :CxHyOz---- >CxHy_2>A i ---- >B i---- >Glixerin. Vit phng trnh phn ng.

5*. Ru A c mt loi nhm chc. t hon ton 10,4g cn ht 15,68 lt O2(ktc) v thu c t l : n = 5 : 6 . Xc inh cng thc n gin nh t v

2 2 *cng thc phn t ca A. Ly 5,2g A cho tc dng va vi 4gCuO (t) thu ccht hu c B c kh nng trng bc. Lp cng thc cu to ca A.

6 . t chy hp cht Y (n chc) thu c nco : nR 0 ; nQ tiu tn = 4nY.Xc

nh cng thc phn t, cng thc cu to Y (mch h) bit Y phn ng vi dungdch Br2 , cng H2 to ru n chc, vit cc phng trnh phn ng.

39



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40/233

7. Ru A bc I, mch h, c th no hay c 1 lin kt i, cng thc phn tCxHioO. Ly 0,02 mol CH3OH v 0,01 mol A trn vi 0,1 mol 0 2 ri t chy honton 2 ru. Sau phn ng thy c 0 2d. Lp cng thc cu to A.

8 . t hon ton 7,4g ru A ch thu c 8,96 lt CO2 (27,3c v 1,1 atm) v

9g H2O. Tin hnh phn ng loi nc ca A thu c hn hp 2 anken. Tm cngthc phn t, cng thc cu to ca A.

9. Mt ru no a chc, mch h X c n nguyn y y j v m nhm OH. Cho 7,6gru trn tc dg Na d c 2,24 lt kh (ktc).

a) Lp biu thc lin h gia n v m.

b) Cho n = m + 1 , tm cng thc phn t, cng thc cu to ca X.

1 0 . M cha (C, H, ) mch h ch cha 1 loi nhm chc. t chy M thu cs mol H20 = 2 ln s mol C 02. Cn nu cng ly lng M trn tc dng vi Na

cho s mol hiro bng V2s mol M phn ng. Tm cng thc cu to ca M.

l. t hon ton m (g) hn hp X gm 2 ru A v B thuc cng dy ngng thu c 6,72 lt C0 2 v 7,65g H20. Mt khc m (g) X tc dng ht vi Nac 2 , 8 lt kh H2.

a) Xc nh cng thc cu to A, B, Bit t khi hi ca mi cht trong X sovi H2u nh hn 46.

b) Tnh phn trm v khi lng mi cht trong X (cc kh o ktc).

12. Ho hi hon ton 6,42g hn hp X gm 2 ru no A v B 81,9c v

1,3 atm c mt hn hp hi ca 2 ru c th tch 2,352 lt. Cho cng mt lnghn hp ru X ny tc dng vi K d thu c 1,848 lt kh H 2 (ktc). Xc nhcng thc phn t v khi lng mi ru, bit rng s" nhm chc trong B nhiuhn trong A mt n v.

13. Hn hp 2 ru no n chc c sc < 4. Khi cho chng tc dng vi Nau thu c 5,6 lt H2 (ktc), cn khi t chy hon ton cc hn hp ru ucn 47,04 lt 0 2 (ktc). Hy xc nh thnh phn cc hn hp ru (gm nhngru g v s mol ca chng).

14. un nng hn hp 3 ru X, Y, z vi H2SO4 c 180c thu c hn hp2 olefin l ng ng lin tip. Ly 2 trong s" 3 ru trn un vi H2SO4 c 140c c l,32g hn hp 3 ete. Mt khc lm bay hi l,32g ete ny c th tchng, bng th tch ca 0,48g 0 2(do cng diu kin). Xc nh cng thc cu to. 3ru X, Y, z.

15. t chy hon ton hn hp A gm 2 ru n chc, mch h, k tiptrong dy ng ng thu c kh CO2v hi H2o c t l v th tch l 5 : 7.

a) Xc nh cng thc phn t ca 2 ru.

40



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41/233

b) Tnh phn trm khi lng trong hn hp ru.

. 16. Cho 9,4g hn h p A gm hi ca 2 ru i qua cht xc tc thch hp chn hp B gm 3 ete (s mol cc ete bng nhau), 2 anken, hai ru v hi nc.Cho hn hp B i qua P 2O5 th to ra 5,488g axit H3PO4 , Hn hp B c th lm

mt mu va ht lng nc brom c cha 8,64 gam Br2 . Nu tch ring hn hpru v ete t hn hp B v cho bay hi th c 4,3008 lt 91c v 2/3 atm. Lpcng thc phn t ca 2 ru ban u. Bit rng hiu sut phn ng to anken ca2 ru bng nha.

17. Chia hn hp 2 ru no n chc, mch h, lin tip trong dy ng ngthnh 2 phn bng nhau. Phn 1 cho tc dng vi Na d c 0 , 2 mol H2. Phn 2 un nng vi H2SO4c c 7,704g hn hp 3 ete. Tham gia phn ng ete ho c50% lng ru c khi lng.phn t nh v 40% lng ru c khi lng phnt ln. Lp cng thc phn t 2 ru.

18. a) Trnh by cc tnh cht ho hc chnh ca C2H5OH v C6H5OH.

b) M t c im cu to gii thch v so snh nh hng qua li canhm chc OH v phn gc hirocacbon ca hai phn t C2H5OH v C6H5OH.

19. a) Bng phn ng ho hc chng t rn g trong phn t phenol nh n .benzen c nh hng n tnh cht ca nhm OH v nhm OH c nh hngn tnh cht ca nhn benzen.

b) Mt hp ch t hu c A thuc loi hp cht-thm , c cng thc phn tC6H 7 0N, c th phn ng vi NaOH v HC1. Tim*cng thc cu to ca A. So

snh tnh axit ca A (ng phn para) vi phenol v gii thch ngn gn.20. Vit cng thc cu to v gi tn cc hp cht thm c cng cng thc

phn t C78O,

21. Ln lt cho hp cht c cng thc cu to sau: H O - ^ ^ - C H 2OH

tc dng vi: a) Na ; b) NaOH ; c) Br2; d) HC1 (H2S04).

Vit cc phng trnh phn ng xy ra.

22. Phn ng gia phenol, toluen, glixerin v xenlulo vi HNO3 c c mtH2SO4 c c g ging, nhau v khc nhau? Cc sn phm phn ng to thnh cthuc cng loi hp chtkhng? Ti sao? c tn cc sn phm phn ng.

23. Vit cc phng trnh phn ng khi cho axit HNO3 c, d (c mt xctc) tc dng vi toluen, glixerin, phenol v xenluloz.

24. Cho 3 cht A, B, c u l hp cht thm c cng thc phn t l C78O.Khi cho mi cht trn ln lt tc dng vi Na v vi NaOH th thy: A phn ngvi c hai; B ch phn ng vi Na; c khng phn ng. Hy vit cng thc cu toca A, B, c v cc phng trnh phn ng.

41



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42/233

25.Hy cho bit nhng tnh cht ho hc ch yu ca phenol v ru benzyliekhc nhau th no ? Gii thch s khc nhau gia hai loi cht .

26. C bao nhiu ru n chc v bao nhiu phenol n chc tng ngvi mi cht toluen v metylxiclohexan? i vi mi trng hp (touen vmetylxicohexan) hy nu hai v d in hnh bng cch vit cng thc cu to,gi tn v ch r bc ru (nu c).

27.Vit cc phng trnh phn ng:

a) T phenol iu ch: axit picric, nha phenolfomandehit.

b) Ru benzylic tc dng vi: Na, CuO (nung nng) to th nh an ehi t vCH3COOH.

28. Phenol c th iu ch c t clobenzen hoc t cumen; cn t phenol c

th iu ch c 1,3-xiclohexaien. Vit cc phng trnh phn ng.29. a) T toluen, vit cc phng trnh phn ng ( dng cng thc cu to)

iu ch v gi tn cc hp cht thm c cng thc phn t C 7 8O. Bit rng cccht ny tc dng c vi Na.

b) Vit cc phng tr n h phn ng chuyn ho propanol - 1 thnh propanol'2.

30. T benzen v cc cht v c thch hp hy vit cc phng trnh phnng ru ch 2,4,6-triamino phenol.

1. ROH* + HBr ---- > RBr + H20R l gc hirocacbon ho tr I, cng thc cu to c dng CxHy (y < 2x+l)

rr . 8 0 _ 584%Ta c: ------------ = = 0,5841 2 x + y + 80 1 0 0 %12x + y = 57. Nghim thch hp X = 4 ; y = 9

Cng thc phn t ca A: C4H9OH ; Cng thc phn t ca B: C49Brrra__ i.1 -A-, A H2S 0 4c,180C ,Theo gi thit: A ----- ----------- > 2 anken.

Chng t A l ru bc II, cng thc cu to: C H 3 - C H 2 - C H - - C H 3

OH2. a) t cng thc ca A l CxHyOz, ta c: 1 2 x + y = 3,5xl6z = 56z

Khi z = 1 => 12x + y = 56. Nghim thch hp ~ 4 ; y = 8

Cng thc n gin ca A l C4H80.

b) Cng thc phn t ca A: C4H80, v A l ete to ra t ru n chc chcha 1 nguyn t o.

X + Y ---- > C4H80 (ete)



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43/233

Cng thc cu to A, X, Y c th l:

CHg-OH + CH2=CH-CH2-OH

(X) (Y)3. a) CnH2n+2Ox + - g-y ~-x 0 2

3n +1 - X

c h 3- o - c h 2- c h = c h 2 + h 20

(A)

0,05

25,632

nC02+ (n+l)H20

6 , 6

44

(mol)

Rt ra n = 3 ;x = 3=> C3 8O3 (glixerin)

b) C4H 10 ac-k--g --> CH3-CH=CH 2

-> CH2-CH~CH2

Cl OH Cl

-> C1-CH2-CH=CH 2 +HOC1

+NaOH c h 2- c h - c h 2I I IOH OH OH

(RCOO)3C3H5 n^ c 3h 5(OH ) 3

4. T cng thc phn t CxHyOz suy ra: 12x + y + 16z = 60 (y < 2x + 2 )

Khi z = 1 th 12x + y = 44 => x = 3 ; y = 8 C3H80 (c 3 ng phn)

Khi z = 2 th 1 2 x + y = 28=>x = 2 ; y = 4 C2 4O2 (c 4 ng phn)

CH3-CO OH ; H-COOCHa ; HO -CH 2~CHO

OHC-O-CHg ; HO - CH - o

c h 2

CH3-C H 2-CH2OH (hay CH3-CH(OH)-CH3) ---- > CH2=CH-CH3 ---- >

CH2 =CH-CH 2C1---- > H2-CH-CH 2 ---- > C3H5(OH )3 X = 5 ; y = 12 ; z = 2

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44/233

Cng thc n gin nht ca A: C 5 H 1 2 O 2 => Cng thc phn t ca A l(C5H i202)n. A l ru nn c nhm O H , cng thc c dng C5nHi0n(OH)2n

Ta c: lOn < 2x5n + 2 - 2n = > n < l

Cng thc phn t A: C5H 10(OH)2 (M = 104)nA= 5,2 : 104 = 0,05 moi ; nCuo = 0,05 mol

R(CI2 0 H)x + xCuO ---- R(CHO)x +xCu + 2xH20

1 mol X mol (B)

0,05 mol 0,05 mol

Suy ra X = 1 v A ch c 1 nhm -CH 2OH, nguyn t o cn li rubc III.

6. C xHyO z + x + 1 - 1j 02 > xC 02 + I H 20

Ta c:9

=> =3 ; y = 6 ; z = 1 ; cng thc phn t: C3I 6Oy z .

X + - - = 44 2

Cng thc cu to: CH2=CH-CH 2OH.

7. CH 3 OH + 1,502 ---- C0 2 + 2H200,02 mol 0,03 mol

CxH 10O + ---- > xC02 + 5H20

0,01 mol 0,005(2x + 4)

Ta c: 0,03 + 0,005(2x + 4) < 0,1 (O2cn d)

=> X < 5 ; vi iu k i n 10 < 2x + 2 => X > 4.

Vy A l C 4 1 0 O (c 2 cng thc cu to ru bc I)

1 , 1 x 8 , 9 6

0,082(273 + 27,3)

Ta thy nH 0 > nco => A l ru no. Khi loi nc t A thu c hn hp 2

anken, vy A l ru no n chc bc cao. Cng thc phn t c dng CnH2n+iOH.

CnH2n+1OH + 0 2 * nC0 2+ (n+l)H20

(14n + 18)g n mol

7,4g 0,4 mol

Rt ra n = 4: C 4 9 O H (secbutyic)

8. n = ------- I-11---- = 0,4 mol ; n , = 9 : 18 = 0,5 moiCOo n 0 7 q Ho0

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9. Cng thc phn t: CnH2n+2-ni(OH)miu kin: n > m ; n, m nguyn dng.

C n H 2n +2 - m ( O H ) m + mNa --------> CnH2n+2-m(ONa)m + ^ H 2z

(14n + 16m + 2)g mol

na. 2,24 .7,6g - mol22,4

Rt ra: 7n + 1 = llm.

10. C x H y O z + j 2 xC02 + ^ H 202

yTa c: = 2x => y = 4x. Cng thc phn t ca M c dng: (CH4)xOz.

- Khi z = 1 => (CH4)xO ; y < 2 x + 2 = > x < l . Cng thc phn t : CH4O

CH3OH + Na -------> CH3ONa + 1/2H2

Ta thy:nCH 0H = 2nH (ph hp gi thit) . Vy M l C H 3 O H .

Khi z = 2 => CH4O2: Cng thc cu to: HOCH2OH km bn v c 2 Hlinh ng.

- Cc trng hp z = 3 ; z = 4 ... cng loi v khng thch hp.

1 1 . nco = 6,72 : 22,4 = 0,3 moi ; nH 0 = 7,65 : 18 = 0,425 moi

n = 2,8 : 22,4 = 0,125 mol =>nH= 2x0,125 = 0,25 mol2

nco < n 0 => A, B thuc dy ng ng ru no.

a) t cng thc phn t tng ng ca A, B l C_H2~ 20

3n +1 - X"2

_ , n n + 1 _ _ 0 .Ta c: - = => n = 2,4

0,3 0,425+ xNa >


46/233

C2 H4(0H ) 2(Ma - 62) rc 2 h 4(0 H ) 2(Ma = 62)Cp nghim thch hp |c 2 ( O H ) 2 (Mb = 76) ; = 90)

b) Gi a, b l n lt l s" mol A v B.

- Vi cp C 2H4(OH ) 2v C3H6(OH) 2 ta c: - = 2,4 a : b = 3 : 2a + b

%mA= ----------------- xl00% = 55% => %mB = 45%.3x62 + 2x7 6

- Vi cp C2H4(OH ) 2v C4H8(OH)2 ta c: %I1A= 73,37% v %mB= 26,63%.

12. nx = ---- 1 3 x ^ 3 5 2---- = n = 2 2 _ 0 Q025 mol0,082(273 + 81,9) H2

CSH2S+2_ ,(0 H ), + XK _ > C5 H2H+2_,(OK)s + | h 2

1 mol mol2

0,105 mol 0,0825 mo

Rt ra X = 1,57

Vy ru cha 1 nhm OH, ru B cha 2 nhm OH

CnH2n+1OH + K ---- CnH2n+1OK + VH2ta mol a/2 mol

C m H 2m( O H ) 2 + 2K --------> C m H 2m ( O K ) 2 + H 2b mol b mo

a + b = 0,105in => a = 0,045 mol ; b = 0,06 molj - + b = 0,0825

(14n + 18)x0,045 + (14m + 34)x0,06 = 6,42 => 3n + 4m = 17

Nghim thch hp: n = 3 ; m = 2 . Cng thc phn t ca A: C3H7OH ; Cngthc phn t ca B: C24(OH)2 .

13. a) nH = 5,6 : 22,4 = 0,25 ml ; nQ = 47,04 : 22,4 = 2,1 mol

C-H0- .OH + Na ---- > C_h " ,ONa + *H2tn 2n+l n 2n+l ^1 mol 0,5 mol

0,5 mol 0,25 mol

C5H2n+2 + -170 2 ---- > n C 0 2 + (n + l ) H20

1 mol 1,5 n mol0,5 mol 2,1 mol

Suy ra n = 2 ,8 . C 4 cp nghim:

A C



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47/233

b) Gi a, b ln lt l s' mol ru c s c < 2,8 v ru c s c > 2,8

Vi cp I ta c:

14. Trong s 3 ru X, Y, z c 2 ru l ng phn ca nhau chng u l runo n chc, cng thc tng ng l C-H2_ jOH vi 2 < n ; Cng thc chung

ca 3 ete l (C5 H2 S t 2 ) 2 o .

Ta c: n3ete = nn = 0,48 : 32 = 0,015 mol2

Mete = 1,32 : 0,015 = 8 8 2 8 n + 18 = 8 8 = 2,5

Cng thc 3 ru l: C2H5OH ; CH3-C H 2-CH2OH ; (CH3)2CH-OH.

nBr = 8,64 : 160 = 0,054 mol

Phn ng tch H20 ra khi ru to anken, iu ny chng t 2 ru l runo n chc.

15. 1 < n => ru no C-H_ o0C02 H20 y n 2n+2

C n H 2n+2 + 2 2 * n C 0 2 + ( r + 1) H 20

Ta c: n : ( n + 1) = 5 : 7 => n = 2,5

Cng thc 2 ru C2H5OH (a mol) v C3H7OH (b mol)

Ta c: = 2,5 - = 1 : 1a + b b

%mn = c 2H5OH 46 + 60

xlOO% = 43,4% ; %mc 3h 7o h = 100 - 43,4 = 56,6%.

16. 3H20 + P 2 O 5----

> 2 H3PO4 (1)_ 3 3 M 8 8 _ An0/) _ ,

nHn = 7 : x nw pn = - x - ~ r - = 0,084 mol2 2 3 4 2 98

0,096 mol

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48/233

C_H9_ o h C-H- + H20n 2n+l n c H 1n 2n+l (2)

(4)

(3)

Ta thy nete = nH2G(3) = 0,084 - d(2) = 0,084 - 0,054 = 0,03 mol

tthn hp ru = n ru (2) + lru (3) + ttru d ~ 0 ,0 5 4 + 2x0,03 + (0,09 60,03 ) 0 ,18

M ru = 9,4 : 0,18 = 52,2 => = 2,4. Cng thc ru th nht l C2H5OH.

- Tm cng thc phn t ru th 2: Theo gi th it cho s mol cc ete bngnhau suy ra s" mol ru cng bng nhau = 0,18 : 2 = 0,09 mol

Ta c: 0,09x46 + (14m + 18)x0,09 = 9,4 => m = 2,89 3.Cng thc ru th 2 l C 3 H 7 O H .

17. C-HL- .OH + Na ---- > C-H- ,ONa + VHa? (1)n 2n+l n 2n+l *

2CmH2m+1OH ---- > (CmH2m+1)20 + H200,4b mol 0,2b moi

0,25ax(28n + 18) + 0,2bx(28m + 18) = 7,704

Thay m = n + l v a = 0,4 - b vo phng trnh trn ta c:

5,904 - 4,7bn 2,8 - l ,4b

Vi 0 < b < 0,4 => 1,8 < n < 2 ,1 . Cng thc 2 ru C 2 H 5 O H v C 3 H 7 O H .

(2)

0 < a < 0,4 => 2,108 < n < 2,796

Cng thc 2 ru: C2H5OH v C3H7OH.

Cch 2 : 2CnH2n+1OH0,5a mol

> (CnH2n+1)20 + H200,25a mol

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49/233

ANEHIT - XETON

A e PHNG PHP GII BI TP

t lim VjjfNO wg THC m TN Gi fiWPCHrr

1. Anehit no n chc (ankanal) CnH2n+1CHO (n > 0 )

hay CxH2xO (x > 1 )

2. Anehit no nh chc (ankanial) CnH2n(CHO) 2 (n > 0 )

3. Anehit n chc c 1 ni i(ankenal)

CH2n_iCHO (n > 2 )

4. Anehit n chc CxHyCHO

5. Anehit no CnH2n+2_z(CHO)z (n > 0)

hay C nH 2n+22kOk (n > 1 ; k > 1)

6 . Anehit mch h CxHy(CHO)z hay CnH2n+2_2a_z(CHO)z

a l s lin kt n (a > 1 )

1L HO tnh fi ftNOH!T

Khc vi ru, phenol, trong andehit khng cha H linh ng nn khng tcdng vi kim loi kim v dung dch kim.

1. Xt phn ng cng Hg to thnh ru bc I

R(CHO)z + zH2 Ni|t R(CH2OH)z

Lu : Nu gc R l gc khng no th c th cho phn ng cng vo R.CH2=CH-CHO +2H 2 Ni>t-..> CH3-C H 2-CH2OH

CH2n+1_2aCHO + (a+l)H2 Nit0- - > CnH2n+1CH2OH

CnH2n+2_2a_z(CHO)z + (a+z)H2 N -- > CnH2n+2_z(CH2O

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PHƯƠNG PHÁP GIẢI BÀI TẬP HỢP CHẤT HỮU CƠ CÓ NHÓM CHỨC - PHẠM ĐỨC BÌNH