Toan Bo Bai Tap Co Huong Dan Chuong Estelipit Ltdh Nam 2014

CHUYN ESTE LIPIT

A. TM TT L THUYT V ESTE

I. CTTQ MT S ESTE:

+ Este ca ru n chc vi axit n chc (este n chc): RCOOR ; CxHyO2

+ Este ca axit n chc vi ru a chc, c cng thc dng (RCOO)nR

+ Este ca axit a chc vi ru n chc, c cng thc dng R(COOR)n

+ Este ca axit a chc vi ru a chc, c cng thc dng Rn(COO)n.mRm

+ Este no n chc : CnH2nO2 + Este khng no c 1 ni i, n chc mch h: CnH2n - 2O2 (n 3)

+ Este no 2 chc mch h: CnH2n - 2O4 (n 2)

II. Danh php Tn Este = Tn gc hirocacbon ca ru + Tn axit ( trong ui oic i thnh at)

III. ng phn

- ng phn Axit

- ng phn este

- ng tp chc

- ng phn mch vng

Lu : CnH2nO2 c th c cc ng phn sau:- ng phn cu to:

+ ng phn este no n chc

+ ng phn axit no n chc

+ ng phn ru khng no c mt ni i hai chc

+ ng phn ete khng no c mt ni i hai chc

+ ng phn mch vng (ru hoc ete)

+ ng phn cc hp cht tp chc:

Cha 1 chc ru 1 chc anehit

Cha 1 chc ru 1 chc xeton

Cha 1 chc ete 1 chc anehit

Cha 1 chc ete 1 chc xeton

Mt ru khng no v mt ete no

Mt ete khng no v mt ru no

- ng phn cis tran (ng phn ru khng no c mt ni i hai chc - ng phn ete khng no c mt ni i hai chc - Mt ru khng no v mt ete no - Mt ete khng no v mt ru no)

- S ng phn este no n chc =2n-2 (1< n < 5)

- Cng thc tnh s triglixerit to bi glixerol vi n axit carboxylic bo =n2(n+1)*1/2

IV. T/c vt l

- Cc este l cht lng hoc cht rn trong iu kin thng,

- Cc este hu nh khng tan trong nc.

- C nhit si thp hn hn so vi cc axit hoc cc ancol c cng khi lng mol phn t hoc c cng s nguyn t cacbon. do gia cc phn t este khng to c lin kt hiro vi nhau v lin kt hiro gia cc phn t este vi nc rt km.Th d:CH3CH2CH2COOH

(M = 88) =163,50C

Tan nhiu trong ncCH3[CH2]3CH2OH

(M = 88), = 1320C

Tan t trong ncCH3COOC2H5(M = 88), = 770C

Khng tan trong nc

- Cc este thng c mi c trng

Iso amyl axetat c mi chui chn

Etyl butirat v etyl propionat c mi da

Geranyl axetat c mi hoa hngV. T/c ha hc

a) Thy phn trong mi trng kim(P x phng ha)

R-COO-R + Na-OH R COONa + ROH

b) Thy phn trong mi trng axit:R-COO-R + H-OH R COOH + ROH

* Nu Phng php P chuyn dich theo chiu thun

c) Ch :

- Este + NaOH 1Mui + 1 anehit

Este ny khi P vi dd NaOH to ra ru c nhm -OH lin kt trn cacbon mang ni i bc 1, khng bn ng phn ha to ra anehit.

VD: R-COOCH=CH2 + NaOH R-COONa + CH2=CH-OH

- Este + NaOH 1 Mui + 1 xeton

Este ny khi P to ru c nhm -OH lin kt trn cacbon mang ni i bc 2 khng bn ng phn ha to xeton.

+ NaOH R-COONa + CH2=CHOH-CH3

- Este + NaOH 2 Mui + H2O

Este ny c gc ru l phenol hoc ng ng phenol..

+ 2NaOH RCOONa + C6H5ONa + H2O

- Este + AgNO3/ NH3 ( P trng gng

HCOOR + 2AgNO3 + 3NH3 + H2O ( ROCOONH4 + 2Ag + 2NH4NO3

- Este no n chc khi chy thu c

d) P chy

VI. iu ch

a) P ca ancol vi axit cacboxylicRCOOH + ROH RCOOR + H2O

b) P ca ancol vi anhirit axit hoc anhirit clorua

+ u im: P xy ra nhanh hn v mt chiu

(CH3CO)2O + C2H5OH ( CH3COOC2H5 + CH3COOH

CH3COCl + C2H5OH ( CH3COOC2H5 + HCl

c) /c cc este ca phenol t P ca phenol vi anhirit axit hoc anhirit clorua(v phenol khng T/d vi axit cacboxylic)

(CH3CO)2O + C6H5OH ( CH3COOC6H5 + CH3COOH

CH3COCl + C6H5OH ( CH3COOC6H5 + HCl

d) P cng vo hirocacbon khng no ca axit cacboxylic

+ An Ken

CH3COOH + CH=CH CH3COOCH2 CH3

+ Ankin

CH3COOH + CH(CH CH3COOCH=CH2B. MT S BI TPV d - L thuyt

Cu 1: C cc nhn nh sau : (1) Este l sn phm ca P gia axit v ancol

(2) Este l hp cht hu c trong phn t c nhm - COO -

(3) Este no, n chc, mch h c CTPT l CnH2nO2, vi n 2

(4) Hp cht CH3COOC2H5 thuc loi este

(5) Sn phm ca P gia axit v ancol l este

Cc nhn nh ng l: A. (1), (2), (3), (4), (5). B. (1), (3), (4), (5) .C. (1), (2), (3), (4).D. (2), (3), (4), (5).Cu 2: Pht biu no sau y l ng?

A. phn bit benzen, toluen v stiren ( iu kin thng) bng phng php ha hc, ch cn dng thuc th l nc brom.

B. Tt c cc este u tan tt trong nc, khng c, c dng lm cht to hng trong cng nghip thc phm, m phm.

C. Phn ng gia axit axetic vi ancol benzylic ( iu kin thch hp), to thnh benzyl axetat c mi thm ca chui chn.

D. Trong phn ng este ha gia CH3COOH vi CH3OH, H2O to nn t -OH trong nhm

-COOH ca axit v H trong nhm -OH ca ancol.Cu 3: Metyl propionat l tn gi ca hp cht: A. CH3COOC2H5 B. CH3COOC3H7C. C3H7COOCH3D. C2H5COOCH3 Cu 4: Mt este n chc no mch h c 48,65 % C trong phn t th s ng phn este l: A. 1

B. 2

C. 3

D. 4

Cu 5: C3H6O2 c 2 ng phn T/d c vi NaOH, khng T/d c vi Na. CTCT ca 2 ng phn

A. CH3COOCH3 v HCOOC2H5

B.CH3CH2COOH v HCOOC2H5 C. CH3CH2COOH v CH3COOCH3

D. CH3CH(OH)CHO v CH3COCH2OH

Cu 6: S hp cht n chc c cng CTPT C4H8O2, u T/d vi dd NaOH

A.3

B.4

C.5

D.6

Cu 7: Cc ng phn ng vi CTPT C8H8O2 (u l n xut ca benzen) T/d vi NaOH to ra mui v Ancol l:

A. 2

B. 3

C. 4

D. 7

Hng Dn

C6H5COOCH3

HCOOCH2C6H5Cu 8: Thu phn este c CTPT: C4H8O2 ( xt H+ ), thu c 2 sn phm hu c X, Y.T X c th iu ch trc tip ra Y. Vy cht X l:

A. ancol metylic B. Etyl axetat C. axit fomic D. ancol etylic

Cu 9: Este c CTPT l C4H6O2, khi thu phn trong mi trng axit thu c hn hp cc cht u c kh nng trng gng. CTCT thu gn ca este l

A. HCOO-C(CH3) = CH2

B. HCOO-CH=CH-CH3.

C. CH3COO-CH=CH2.

D. CH2 =CH-COO-CH3.

Cu 10: Mt este c CTPT l C4H6O2 khi thy phn trong mi trng axit thu c imetyl xeton. CTCT thu gn ca C4H6O2 l cng thc no

A. HCOO-CH=CH-CH3 B. CH3COO-CH=CH2

C. HCOO-C(CH3)=CH2 D.CH2=CH-COOCH3Cu 11: Thu phn cht hu c X trong dd NaOH (d), un nng, c sn phm gm 2 mui v ancol etylic. Cht X l A. CH3COOCH2CH2Cl.

B. CH3COOCH2CH3.

C. ClCH2COOC2H5.

D. CH3COOCH(Cl)CH3. Cu 12: Cht hu c X c CTPT l C4H6O2Cl2. Khi cho X P vi dd NaOH thu c CH2(OH)COONa, etylenglicol v NaCl. CTCT ca X l:

A. CH2ClCOOCHClCH3.

B. CH3COOCHClCH2Cl.

C. CHCl2COOCH2CH3.

D. CH2ClCOOCH2CH2Cl.Cu 13: Hp cht hu c X cha mt loai nhom chc, co CTPT la C6H10O4. Khi thuy phn X trong NaOH thu c mt mui va 2 ancol ng ng lin tip nhau. X co CTCT la:

A. HOOC (CH2)2 COOH

B. CH3OOC CH2 COO C2H5 C. HOOC (CH 2)3 COO CH3

D. C2H5OOC CH2 - CH2 COOH

Cu 14: Cho s P : A (C3H6O3) + KOH ( Mui + Etylen glicol.

CTCT ca A l :

A. HOCH2COOCH3.

B. CH3COOCH2OH.

C. CH3CH(OH) COOH.

D. HCOOCH2CH2OH.Cu 15: Cho s : C4H8O2 ( X ( Y( Z( C2H6. CTCT ca X l

A. CH3CH2CH2COONa. B. CH3CH2OH.

C. CH2=C(CH3)-CHO. D. CH3CH2CH2OH.Cu 16: Cho cht X T/d vi mt lng va dd NaOH, sau c cn dd thu c cht rn Y v cht hu c Z. Cho Z T/d vi dd AgNO3/NH3 thu c cht hu c T. Cho T T/d vi dd NaOH li thu c cht Y. Cht X c th l:

A.HCOOCH=CH2

B.CH3COOCH=CH2

C.HCOOCH3 D.CH3COOCH=CH-CH3

Cu 17: Cho chui bin i sau: C2H2 X Y Z CH3COOC2H5 .X, Y, Z ln lt l

A. C2H4, CH3COOH, C2H5OH

B. CH3CHO, C2H4, C2H5OH

C. CH3CHO, CH3COOH, C2H5OH D. CH3CHO, C2H5OH, CH3COOHHng Dn

Cu 18: Thy phn hon ton este X bng dd NaOH. Sau khi P kt thc th s mol NaOH dng gp i s mol X. C

(1) X l este ca axit n chc v ancol hai chc

(2) X l este ca ancol n chc v axit hai chc

(3) X l este ca ancol n chc v axit n chc

(4) X l este c CTCT thu gn l RCOOC6H5(5) X l este ca ancol hai chc v axit hai chc

Cc pht biu ng l:

A. (1) (2) (3)

B. (3) (4) (5)

C. (1) (2) (3) (5)

D.(1) (2) (4) (5)Cu 19. Cht X T/d vi NaOH cho dd X1. C cn X1 c cht rn X2 v hn hp X3. Chng ct X3 thu c X4. Cho X4 trng gng thu c X5. Cho X5 T/d vi NaOH li thu c X2.Vy CTCT ca X l

A. HCOO- C(CH3)=CH2 B. HCOO-CH=CH-CH3

C. CH2=CH-CH2-OCOH D. CH2=CH-OCOCH3Cu 20: Dy cht no sau y c sp xp theo chiu nhit si ca cc cht tng dn

A. CH3COOH, CH3COOC2H5, CH3CH2CH2OH B. CH3COOH, CH3CH2CH2OH CH3COOC2H5C. CH3CH2CH2OH , CH3COOH, CH3COOC2H5 D. CH3COOC2H5 ,CH3CH2CH2OH , CH3COOHCu 21: Sp xp cc cht sau theo trt t tng dn nhit si: CH3COOH; CH3COOCH3; HCOOCH3; C2H5COOH; C3H7OH. Trng hp no sau y ng

A. HCOOCH3 < CH3COOCH3 < C3H7OH < CH3COOH < C2H5COOH.B. CH3COOCH3 < HCOOCH3 < C3H7OH < CH3COOH < C2H5COOH.

C. C2H5COOH< CH3COOH < C3H7OH < CH3COOCH3 < HCOOCH3

D. HCOOCH3< CH3COOCH3 < C3H7OH < C2H5COOH< CH3COOH

Cu 22: Cho s phn ng:

Este X (C4HnO2) Y Z C2H3O2Na.

CTCT ca X tha mn s cho l

A. CH2=CHCOOCH3.

B. CH3COOCH2CH3.

C. HCOOCH2CH2CH3.

D. CH3COOCH=CH2.V D - Bi TpCC CH KHI LM NHANH BI TP

- Nu cho bit s mol O2 phn ng ta nn p dng LBTKL tm cc i lng khc. nu bi cho este n chc ta c: neste + nO2(p) = nCO2 + 1/2nH2O- Nm chc l thuyt, cc phng trnh, cc gc hirocacbon thng gp khng phi nhp nhiu.

- t chy este no lun cho nCO2 = nH2O v ngc li. - Nu cho hay t p n suy ra este n chc th trong phn ng vi NaOH th s mol cc cht lun bng nhau.

- X phng ho este n chc cho 2 mui v nc => este ca phenol.

- Khi cho hh cht hu c tc dng vi NaOH:

+ to s mol ancol b hn s mol NaOH => hh ban u gm este v axit. Khi : nancol = neste; nmui = nNaOH(p) = nhh+ to s mol ancol ln hn s mol NaOH => hh ban u gm este v ancolDng 1: P chy

Cu 1: Khi t chy hon ton este no, n chc th P. Tn gi ca este l A. Metyl fomiat. B. Etyl axetat. C. Metyl axetat. D. n- Propyl axetat.

Hng DnGoi CT CnH2nO2

Ta c

Cu 2: t chy hon ton 7,4 gam hn hp hai este ng phn, thu c 6,72 lt CO2 (ktc) v 5,4 gam H2O. CTPT ca hai este l

A. C3H6O2B. C2H4O2C. C4H6O2D. C4H8O2Hng Dn

CTG ng thi cng l CTPT ca hai este l C3H6O2.

Cu 3: t chy hon ton hn hp 2 este, cho sn phm chy qua bnh P2O5d khi lng bnh tng ln 6,21 gam, sau cho qua dd Ca(OH)2 d c 34,5 gam kt ta. Cc este trn thuc loi:

A. Este no B. Este khng no C. Este no , n chc , mch h D. Este a chc

Hng Dn:

nn hai este l no n chc mch h.

Cu 4: Hp cht X T/d c vi dd NaOH un nng v vi dd AgNO3/NH3.Th tch ca 3,7 gam hi cht X bng th tch ca 1,6 gam O2 (cng k v nhit v p sut). t chy hon ton 1 gam X th th tch CO2 thu c vt qu 0,7 lt ( ktc). CTCT ca X A. O=CH-CH2 CH2OH B. HOOC-CHO C. CH3COOCH3 D. HCOOC2H5Hng DnDo cng k v nhit v p sut

t chy hon ton 1 gam X th th tch CO2 thu c vt qu 0,7 lt ( ktc) DCu 5: t chy hon ton 11,6 gam este X thu c 13,44 lt CO2(ktc) v 10,8 gam H2O. Mt khc Cho 11,6 gam este T/d vi dd NaOH thu c 9,6 gam mui khan. CT ca X l:

A. C3H7COOC2H5 B. C2H5COOC2H5 C. C2H5COOC3H7 D. CH3COOC3H7Hng Dn

nn este l no n chc c CTTQ: CnH2nO2 CnH2nO2 nCO2

n= 6 C6H12O2

RCOOR + NaOH RCOONa + ROH

0,1 0,1 0,1

Ta c 0,1.(R+67)=9,6=> R=29: C2H5-

Vy CTCT ca este l C2H5COOC3H7Cu 6: Hn hp X gm hai este no, n chc, mch h. t chy hon ton X cn 3,976 lt O2 (ktc) c 6,38 gam CO2. Mt khc X T/d vi dd NaOH c mt mui v hai ancol l ng ng k tip. CTPT ca hai este trong X A. C2H4O2 v C5H10O2 B. C2H4O2 v C3H6O2

C. C3H4O2 v C4H6O2 D. C3H6O2 v C4H8O2Hng DnDo X l este no n chc v T/d vi dd NaOH, c mt mui v hai ancol l ng ng k tip Goi CTca hai este l

Ta c

Phn ng chy

Ta c

Cu 7: X l hn hp 2 este ca cng 1 ancol no, n chc v 2 axit no, n chc ng ng k tip. t chy han ton 0,1 mol X cn 6,16 lt O2(ktc). un nng 0,1 mol X vi 50 gam dd NaOH 20% n P han ton, ri c cn dd sau P c m gam cht rn. Gi tr ca m l:

A. 13,5 B. 7,5 C. 15 D. 37,5

Hng DnDo X l este ca cng 1 ancol no, n chc v 2 axit no, n chc ng ng k tip.

Goi CTca hai este l

HCOOCH3 V CH3COOCH3

Cu 8: t chy hon ton 6,8 gam mt este A no n chc cha vng benzen thu c CO2 v H2O. Hp th ton b sn phm ny vo bnh ng dd Ca(OH)2 ly d thy khi lng bnh tng 21,2 gam ng thi c 40 gam kt ta. Xc nh CTPT, CTCT c th c ca A

A. 2

B. 3

C. 4

D. 5

Hng DnTm CTG: d dng tm c CTPT C8H8O2

4 CTCT: phenyl axetat; 3 p: o, m, p -metyl phenyl fomat

Cu 9: Hn hp Z gm hai este X va Y tao bi cung mt ancol va hai axit cacboxylic k tip nhau trong day ng ng (MX < MY). t chay hoan toan m gam Z cn dung 6,16 lit O2 (ktc), thu c 5,6 lit CO2 (ktc) va 4,5 gam H2O. CT este X va gia tri cua m tng ng la

A. CH3COOCH3 va 6,7 B. HCOOC2H5 va 9,5

C. HCOOCH3 v 6,7 D. (HCOO)2C2H4 va 6,6

Hng Dn

X, Y l 2 este no n chc

p dng LBTKL : m = + 4,5 - = 6,7 (gam)

t cng thc ca X, Y :

n = 2 ; n = 3 X : C2H4O2 HCOOCH3 Y : C3H6O2 CH3COOCH3

Cu 10: t chy hon ton hn hp hai este X, Y, n chc, no, mch h cn 3,976 lt oxi(ktc) thu c 6,38 gam CO2. Cho lng este ny T/d va vi KOH thu c hn hp hai ancol k tip v 3,92 gam mui ca mt axit hu c. CTCT ca X, Y ln lt l

A. C2H5COOC2H5 v C2H5COOC3H7 B. C2H5COOCH3 v C2H5COOC2H5

C. CH3COOCH3 v CH3COOC2H5 D. HCOOC3H7 v HCOOC4H9

Hng Dnt CTTB ca 2 este X, Y l: CnH2n+1COO

V X, Y u l este n chc, no, mch h nn: = = 6,38/44 = 0,145 mol

meste + = 44. + 18. meste = 3,31 gam

Ta c : mO (trong este) = meste mC mH = 3,31 12.0,145 2.1.0,145 = 1,28 gam

nO = 1,28/16 = 0,08 mol neste = 0,04 mol

nmui = neste = 0,04 mol Mmui = 14n + 84 = 3,92/0,04 = 98 n = 1Mt khc: = 3,31/0,04 = 82,75 12.1 + 46 + 14 = 82,75 = 1,77

Vy: X l CH3COOCH3 v Y l CH3COOC2H5 ( p n C

Cu 11: Este X no, n chc, mch h, khng c P trng bc. t chy 0,1 mol X ri cho sn phm chy hp th hon ton vo dd nc vi trong c cha 0,22 mol Ca(OH)2 th vn thu c kt ta. Thu phn X bng dd NaOH thu c 2 cht hu c c s nguyn t cacbon trong phn t bng nhau. Phn trm khi lng ca oxi trong X l:

A. 43,24% B. 53,33% C. 37,21% D. 36,26% Hng Dn Cn nCO20,1 0,1n

CO2 + Ca(OH)2 CaCO3 + H2O (1)

0,22 0,22 0,22

CO2 + CaCO3 + H2O Ca(HCO3)2 (2)

0,22 0,22

Theo (1), (2): thu c kt ta th: nCO2 < 0,22+0,22 = 0,44

Hay: 0,1n < 0,44 n < 4,4

X + NaOH to 2 cht c C = nhau X c 2 hoc 4 C

X khng c P trng gng n = 4 C4H8O2

Cu 12: Hn hp X gm hai cht hu c no, n chc T/d va vi 100 mldd KOH 0,4M, thu c mt mui v 336 ml hi mt ancol (ktc). Nu t chy hon ton lng hn hp X trn, sau hp th ht sn phm chy vo bnh ng dd Ca(OH)2(d) th khi lng bnh tng 6,82 gam. CT ca hai hp cht hu c trong X l A. CH3COOH v CH3COOC2H5.

B. C2H5COOH v C2H5COOCH3. C. HCOOH v HCOOC2H5.

D. HCOOH v HCOOC3H7.

Hng Dn

Nhn vo p n cho thy hn hp X gm 1 axit v 1 este

Goi CT hn hp X l: CnH2n+1COOH x mol v CnH2n+1COOCmH2m+1 y mol

Tc dng KOH

P chy hp th ht sn phm chy vo bnh ng dd Ca(OH)2(d) th khi lng bnh tng 6,82 gam

Cu 13: t chy hon ton 1 mol axit cacboxylic n chc X cn 3,5 mol O2. Trn 7,4 gam X vi lng ancol no Y (bit t khi hi ca Y so vi O2 nh hn 2). un nng hn hp vi H2SO4 lm xc tc. P hon ton c 8,7 gam este Z(trong Z khng cn nhm chc khc). CTCT ca Z

A. C2H5COOCH2CH2OCOC2H5 B. C2H3COOCH2CH2OCOC2H3

C. CH3COOCH2CH2OCOCH3 D. HCOOCH2CH2OCOH

Bi gii:

Phn ng chy: CXHyO2 + (x + -1)O2 ( xCO2 + H2O (1)

Theo (1), ta c : x + -1= 3,5 ( x + = 4,5 ( ( X : C2H5COOH

Ancol no Y : CnH2n+2-m (OH)m (1 ( m ( n) ( este Z : (C2H5COO)mCnH2n+2-m

( Meste = 73m + 14n + 2 m = hay 14n + 2 = 15m (2)

Mt khc < 2 hay 14n + 2 + 16m < 64 ( 30m + 2 < 64 (v m ( n) ( m < 2,1

T (2) ( ( ancol Y : C2H4(OH)2

( Z : C2H5COOCH2CH2OCOC2H5 Cu 14: Hn hp X gm axit axetic, etyl axetat v metyl axetat. Cho m gam hn hp X T/d va vi 200 ml dd NaOH 1M. Mt khc, t chy hon ton m gam hn hp X cn V lt O2(ktc) sau cho ton b sn phm chy vo dd NaOH d thy khi lng dd tng 40,3 gam. Gi tr ca V l:

A. 17,36 ltB. 19,04 ltC. 19,60 ltD. 15,12 lt

Hng Dn

X c cng thc chung CnH2nO2 vi nX = 0,2 molm dd tng = mCO2 + mH2O = 0,2.n.44 + 0,2.n.18 = 40,3 n = 3,25

nO2 = (3n-2)/2 = (3.3,25-2)/2 V = 17,36

Cu 15: t chy hon ton 3,42 gam hn hp gm axit acrylic, vinyl axetat, metyl acrylat v axit oleic, ri hp th ton b sn phm chy vo dd Ca(OH)2 (d). Sau P thu c 18 gam kt ta v dd X. Khi lng X so vi khi lng dd Ca(OH)2 ban u thay i nh th no

A. Tng 2,70 gam. B. Gim 7,74 gam. C. Tng 7,92 gam. D. Gim 7,38 gam.Hng Dn

hn hp gm axit acrylic, vinyl axetat, metyl acrylat v axit oleic u c CT l:

. p dng LBT khi lng v nguyn t ta c:

Khi lng X so vi khi lng dd Ca(OH)2 ban u s gim l:

=> D ng.

Cu 16: t chy hon ton 10 ml hi mt este X cn va 45ml O2 thu c kh CO2 v hi nc c t l th tch l 4: 3. Ngng t sn phm chy th th tch gim i 30 ml . Bit cc th tch o cng iu kin. Cng thc X l:

A. C4H6O2 B. C4H6O4 C. C4H8O2 D. C8H6O4Hng Dn

Do cc th tch o cng iu kin nn th tch chnh l s mol

Gi CT este l CxHyOzCxHyOz + O2 xCO2 + H2O

10 10 10x 5y

Ta c 10 = 45 (1)

T l kh CO2 v hi nc: (2)

Ngng t sn phm chy th th tch gim i 30 ml : 10x + 5y 30 = 10xy = 6 (3)

T (1),(2),(3) Cu 17: Mt este A (khng cha chc no khc) mch h c to ra t 1 axit n chc v ru no. Ly 2,54 gam A t chy hon ton thu c 2,688 lt CO2 (ktc) v 1,26 gam H2O. 0,1 mol A P va vi 12 gam NaOH to ra mui v ru. t chy ton b lng ru ny c 6,72 lt CO2 (ktc). Xc nh CTPT, CTCT ca A

A. C3H5(OOCCH3)3.

B. C3H5(OOCC2H5)3. C. C2H4(OOCCH3)3.

D. C3H5(OOCCH = CH2)3.Hng Dn

nA:nNaOH = 1:3

(RCOO)3R + 3NaOH 3RCOONa + R(OH)3 0,1 0,1

S nguyn t cacbon ca ru

Khi t chy A => CTG: C6H7O3 . V este 3 chc => CTPT A: C12H14O6= 254

Ta c: 3(R1 + 44) + 41 = 254 R1= 27 CH2 CHVy A: (C2H3COO)3C3H5

Cu 18: em ha hi 6,7 gam hn hp X gm CH3COOH, CH3COOC2H5, CH3COOCH3 v HCOOC2H5 thu c 2,24 lt hi (ktc). t chy hon ton 6,7 gam X thu c khi lng nc

A. 4,5 gam.

B. 3,5 gam.

C. 5 gam.

D. 4 gam.

Hng dnGi cng thc chung ca X l CnH2nO2 ( MX = 14n + 32 = = 67 ( n = 2,5

S chy : CnH2nO2 ( nCO2 + nH2O

( n = 2,5.0,1 = 0,25 mol ( m = 0,25.18 = 4,5 gam

Dng 2: Xc nh CTPT da vo t khi hi

Cu 1: Este A iu ch t ancol metylic c t khi so vi oxi l 2,3125. CT ca A l:

A. C2H5COOC2H5. B. CH3COOCH3. C. CH3COOC2H5. D. C2H5COOCH3Hng Dn

Do Este A iu ch t ancol metylic

Cu 2: Este X khng no, mch h, c t khi hi so vi oxi bng 3,125 v tham gia P x phng ho to ra mt anehit v mt mui ca axit hu c. C bao nhiu CT ph hp vi X

A.2

B.3

C.4

D.5

Hng Dn

P x phng ho to ra mt anehit v mt mui ca axit hu c

Cu 3: X l mt este no n chc, c t khi hi so vi CH4 l 5,5. Nu em un 2,2 gam este X vi dd NaOH d, thu c 2,05 gam mui. CTCT ca X l:

A. HCOOCH2CH2CH3B. HCOOCH(CH3)2 C. C2H5COOCH3 D. CH3COOC2H5

Hng Dn

2,2 gam este X

+ NaOH

0,025 0,025 mol

Cu 4: Este n chc X c t khi hi so vi CH4 l 6,25.Cho 20 gam X T/d vi 300 ml dd KOH 1M (un nng). C cn dd sau P thu c 28 gam cht rn khan. CTCT ca X l

A.CH2=CH-CH2COOCH3

B.CH2=CH-COOCH2CH3

C.CH3COOCH=CH-CH3

D.CH3-CH2COOCH=CH2 Hng Dn

Cho 0,2 mol X T/d vi 0,3 mol KOH 28 gam cht rn khan gm mui v KOH d

+ KOH

0,2 0,2 0,2 mol

Cu 5: Mt este to bi axit n chac v Ancol n chc c t khi hi so vi CO2 bng 2. Khi un nng este ny vi NaOH to ra mui c khi lng ln hn este P. CTCT ca este l:

A. CH3COOCH3 B. HCOOC3H7 C. CH3COOC2H5 D. C2H5COOCH3.Hng Dn

+ NaOH

Ta c mui c khi lng ln hn este P

Cu 6: Este to bi axit n chc v Ancol n chc c t khi hi so vi CO2 bng 2. Khi un nng este ny vi dd NaOH to mui c khi lng bng 93,18% lng este P. CTCT ca este

A. CH3COOCH3 B. HCOOC3H7 C. CH3COOC2H5. D. C2H5COOCH3

Hng Dn

+ NaOH

Ta c mui c khi lng bng 93,18% lng este P

T (1) v (2)

Cu 7: Mt este ca ancol metylic T/d vi nc brom theo t l mol 1: 1 thu c sn phm trong brom chim 35,08% theo khi lng . Este l:

A. metyl propyonat B. metyl panmitat C. metyl oleat D. metyl acrylat

Hng Dn

Theo gi thit 1 mol este + 1 mol Br2 . Gi M l khi lng mol este ta c:

R l C17H33 . Vy este l: metyl oleat

Cu 8: X phng ha hon ton 20,4 gam cht hu c X n chc bng dd NaOH thu mui Y v Z .Cho Z T/d vi Na d thu c 2,24 lt H2 ( ktc) . Nung Y vi NaOH rn thu c mt kh R, dR/O2=0,5 , Z T/d vi CuO nung nng cho sn phm khng c P trng bc . Tn gi ca X l:

A. Etyl axetat B. Iso Propyl axetat C. Propyl propinoat D. Isopropyl fomat

Hng DnX l este n chc to bi acid c mui Y l R-COONa v ancol n chc Z , R- OH.

S mol R-OH= s mol H =2,24: 11,2= 0,2 mol nn s mol X= 0,2 mol .

Kh R c khi lng mol = 32.0,5= 16: CH4 nn mui Y l CH3COONa.

Khi lng mol ca X = 20,4: 0,2 = 102g/mol

Ta c: CH3COOR = 59 + R= 102.

=> R= 43 nn R l C3H7 v este X l CH3-COOC3H7.

Cu 9: Thc hin P x phng ho cht hu c X n chc vi dd NaOH thu c mt mui Y v ancol Z. t chy hon ton 2,07 gam Z cn 3,024 lt O2 (ktc) thu c lng CO2 nhiu hn khi lng nc l 1,53 gam. Nung Y vi vi ti xt thu c kh T c t khi so vi khng kh bng 1,03. CTCT ca X l:

A. C2H5COOCH3 B. CH3COOC2H5 C. C2H5COOC3H7D. C2H5COOC2H5Hng Dn- Theo bi: X n chc, tc dng vi NaOH sinh ra mui v ancol

( X l este n chc: RCOOR.

Mt khc: mZ + = + ( 44.+ 18.= 2,07 + (3,024/22,4).32 = 6,39 gam

V 44.- 18.= 1,53 gam ( = 0,09 mol ; = 0,135 mol

> ( Z l ancol no, n chc, mch h c cng thc: CnH2n+1OH (n 1)

T phn ng t chy Z ( ==( n = 2.

Y c dng: CxHyCOONa ( T: CxHy+1 ( MT = 12x + y + 1 = 1,03.29

( ( C2H5COOC2H5 ( p n DDng 3: P x phng ha

TH1: Thy Phn Este n chc

Cu 1: Cho este X c CTPT l C4H8O2 T/d vi NaOH un nng c mui Y c phn t khi ln hn phn t khi ca X. Tn gi ca X l:

A. Metylpropionat B. Etyl axetat C. Propyl fomat D. Iso Propyl fomat

Hng Dn

C4H8O2 (X) =88 < C2H5ONa (Y) => CTCT l C2H5-COOCH3 Metylpropionat

Cu 2: Thu phn hon ton 8,8 gam mt este n chc, mch h X vi 100 ml dd KOH 1M (va ) thu c 4,6 gam mt ancol Y. Tn gi ca X l:

A. Etyl FomatB. Etyl PropionatC. Etyl Axetat D.Propyl Axetat

Hng Dn

Nhn vo p n nhn thy este X l no n chc, mch h

Gi CTCT este l CnH2n + 1COOCmH2m + 1

nru = nKOH = 0,1 mol

neste=nKOH =0,1 mol

Cu 3: Cho 12,9 gam mt este n chc, mch h T/d ht vi 150ml dd KOH 1M. Sau P thu c mt mui v anehit. S CTCT ca este tho mn tnh cht trn l:

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

Hng Dn:

HCOOCH=CH-CH3 v CH3COOCH=CH2

Cu 4: Hn hp M gm axit cacboxylic X, ancol Y (u n chc, s mol X gp hai ln s mol Y) va este Z c tao ra t X va Y. Cho hn hp M T/d va u vi dd cha 0,2 mol NaOH, tao ra 16,4 gam mui va 8,05 gam ancol. Cng thc cua X va Y la

A. HCOOH va CH3OH

B. CH3COOH va CH3OH

C. HCOOH va C3H7OH

D. CH3COOH va C2H5OHHng Dn:

Gi s mol: RCOOH a

ROH a

RCOOR b

Theo gi thit: ( nRCOONa = a + b = 0,2 mol. MRCOONa = 82 ( R = 15. (CH3). X l CH3COOH

Loi p n: A v C. (a + b) < nROH = a + b < a + b ( 0,1 < nROH < 0,2

40,25 < Mancol < 80,5. Loi p n B.

Cu 5: t chy hon ton 1 este n chc mch h X ( phn t c s lin kt < 3) c th tch CO2 bng 6/7 th tch O2 P ( cc th tch kh o cng iu kin) Cho m gam X T/d vi 200 ml dd KOH 0,7M c dd Y . C cn dd Y c 12,88 gam cht rn khan. Gi tr m l:

A. 8,88 B. 6,66 C. 10,56 D. 7,20

Hng Dn:

CTPT ca este l : CnH2n-2kO2 vi k R= 43 nn R l C3H7 v este X l CH3-COOC3H7.

Cu 8: Cho 27,6 gam hp cht thm X c CT C7H6O3 T/d vi 800 ml dd NaOH 1M c dd Y. Trung ha Y cn 100 ml dd H2SO4 1M c dd Z. Khi lng cht rn thu c khi c cn dd Z l

A. 31,1 gam.B. 56,9 gam.C. 58,6 gam.D. 62,2 gam.Hng Dn

= 0,2; = 0,8; = 0,2 0,6 mol NaOH phn ng vi C7H6O3.

HCOO-C6H4 OH + 3NaOH HCOONa + C6H4(ONa)2 + 3H2O

0,2 0,6 0,2 0,2

Khi lng cht rn = 0,2.68 + 0,2.154 + 0,1.142 = 58,6 gamCu 9: Cho axit salixylic (axit o-hiroxibenzoic) P vi anhirit axetic c axit axetylsalixylic (o-CH3COO-C6H4-COOH). P hon ton vi 43,2 gam axit axetylsalixylic cn va V lt dd KOH 1M. Gi tr ca V l

A. 0,72. B. 0,48. C. 0,96. D. 0,24.

Hng Dn

o-CH3COO-C6H4-COOH + 3KOH = CH3COOK +o-KO-C6H4-COOK+ 2H2O (1)

theo (1)

TH2: Thy phn hn hp Este n chc

Cu 1: X phng ha hon ton 1,99 gam hn hp hai este bng dd NaOH thu c 2,05 gam mui ca mt axit v 0,94 gam hn hp hai ancol l ng ng k tip nhau. CTCT ca hai este l:

A. HCOOCH3 v HCOOC2H5. B. C2H5COOCH3 v C2H5COOC2H5.

C. CH3COOC2H5 v CH3COOC3H7. D. CH3COOCH3 v CH3COOC2H5Hng Dn

Goi CTTB ca 2 Este l RCOO

RCOO + NaOH RCOONa + OHp dng LBTKL: meste + mNaOH = mmui + mru

1,99 + mNaOH = 2,05 + 0,94

Cu 2: X phng ha han ton 14,55 gam hn hp 2 este n chc X,Y cn 150 ml dd NaOH 1,5M. C cn dd thu c hn hp 2 ancol ng ng k tip v mt mui duy nht. CT 2 este l:

A. HCOOCH3, HCOOC2H5.

B. CH3COOCH3, CH3COOC2H5 C. C2H5COOCH3, C2H5COOCH3

D. C3H7COOCH3, C2H5COOCH3

Hng Dn

Goi CTTB ca 2 Este l RCOO

RCOO + NaOH RCOONa + OH 0,225 0,225 mol

Ta c

Cu 3: X l hn hp hai este ca cng mt ancol, no n chc v hai axit no, n chc, ng ng k tip. t chy hon ton 0,1 mol X cn 6,16 lt O2(ktc). un nng 0,1 mol X vi 50 gam dd NaOH 20% P hon ton, ri c cn dd sau P thu c m gam cht rn. Gi tr ca m l

A. 15 gam.B. 7,5 gamC. 37,5 gamD. 13,5 gamHng Dn

Do hai este ca cng mt ancol, no n chc v hai axit no, n chc, ng ng k tipGi CTTB ca hai este l n=2,5

P chy

Ta c

Cu 4: Hn hp X gm hai cht hu c. Cho hn hp X P va vi dd KOH th cn ht 100 ml dd KOH 5M. Sau P thu c hn hp hai mui ca hai axit no n chc v c mt ru no n chc Y. Cho ton b Y T/d vi Na c 3,36 lt H2 (ktc). Hai hp cht hu c thuc loi cht g

A. 1 axit v 1 este B. 1 este v 1 ancol C. 2 este D. 1 axit v 1 ancol Hng Dn

Ta c:

Ancol no n chc Y: CnH2n+1OH

CnH2n+1OH + Na CnH2n+1ONa + H2

0,3 mol 0,15 mol

Thu phn hai cht hu c thu c hn hp hai mui v mt ancol Y vi nY < nKOHVy hai cht hu c l: este v axit

Cu 5: Hn hp M gm hai hp cht hu c mch thng X v Y ch cha T/d va ht 8 gam NaOH c ru n chc v hai mui ca hai axit hu c n chc k tip nhau trong dy ng ng. Ru thu c cho T/d vi Na d c 2,24 lt H2 (ktc). X, Y thuc lai hp cht g A.1 axit v 1 este B.1 este v 1 ancol C.2 este D. 1 axit v 1 ancol Hng Dn

Thu phn hai X, Y v thu c nAncol = nNaOH. Vy X, Y l hai este.

Cu 6: Cho hn hp X gm ancol metylic v hai axit cacboxylic (no, n chc, k tip nhau trong dy ng ng) T/d ht vi Na c 6,72 lt H2 (ktc). Nu un nng hn hp X (c H2SO4 c lm xc tc) th cc cht trong hn hp P va vi nhau to thnh 25 gam este (gi thit P este ho t hiu sut 100%). Hai axit trong hn hp X l A. C3H7COOH v C4H9COOH. B. CH3COOH v C2H5COOH.

C. C2 H5COOH v C3H7COOH. D. HCOOH v CH3COOH.

Hng Dn

Gi CT hn hp X l

Do un nng hn hp X th cc cht P va vi nhau

T/d ht vi Na

COOH + CH3OH COOCH3 0,3 0,3 mol

Cu 7: Mt hn hp X gm 2 este n chc thy phn hon ton trong mi trng NaOH d cho hn hp Y gm 2 ru ng ng lin tip v hn hp mui Z

- t chy hn hp Y th thu c CO2 v hi H2O theo t l th tch 7:10

- Cho hn hp Z T/d vi lng va axit sunfuric c 2,08 gam hn hp A gm 2 axit hu c no. Hai axit ny va Pvi 1,59 gam natricacbonat

Xc nh CT ca 2 este bit rng cc este u c s nguyn t cacbon < 6 v khng tham gia phn ng vi AgNO3/NH3.

A. C2H5COOC2H5, CH3COOC3H7.

B. CH3COOCH3, CH3COOC2H5 C. C2H5COOCH3, C2H5COOCH3

D. C3H7COOCH3, C2H5COOCH3

Hng Dn

C: RCOOR RCOONa RCOOH + Na2CO3

0,03 0,015

t Y: nH2O > nCO2 => CH2 +1OH T ti l => = 2,33

=> 2 ru l: C2H5OH v C3H7OH (1)

axit = 2,08/0,03 = 69,3 => = 24,3 (2)

Do C < 6 v kt hp (1),(2) => C2H5COOC2H5, CH3COOC3H7 (khng c P vi AgNO3/NH3).

TH3: Thy phn Este ng phn ca nhau

Cu 1: Hn hp X gm hai este n chc l ng phn ca nhau. ung nng m gam X vi 300 ml dd NaOH 1M, kt thc cc P thu c dd Y v (m 8,4) gam hn hp hi gm hai anehit no, n chc, ng ng k tip c t khi hi so vi H2 l 26,2. C cn dd Y thu c (m 1,1) gam cht rn. Cng thc ca hai este l

A.CH3COOCH=CHCH3 v CH3COOC(CH3)=CH2 B. HCOOC(CH3)=CH2 v HCOOCH=CHCH3C. C2H5COOCH=CH2 v CH3COOCH=CHCH3. D. HCOOCH=CHCH3 v CH3COOCH=CH2.Hng Dn

= 52,4 CH3-CHO, C2H5-CHO loi p n A, B,

p dng BTKL ta c:

m + 0,3.40 = m 8,4 + 1,1 m = 21,5,

Cu 2: Hn hp A gm ba cht hu c X, Y, Z n chc ng phn ca nhau, u T/d c vi NaOH. un nng 13,875 gam hn hp A vi dd NaOH va thu c 15,375 gam hn hp mui v hn hp ancol c t khi hi so vi H2 bng 20,67. 136,50C, 1 atm th tch hi ca 4,625 gam X bng 2,1 lt. Phn trm khi lng ca X, Y, Z (theo th t KLPT gc axit tng dn) ln lt l:

A. 40%; 40%; 20% B. 40%; 20%; 40%

C. 25%; 50%; 25% D. 20%; 40%; 40%

Hng Dn

Ta c : ( MX =

Mt khc: X, Y, Z n chc, tc dng c vi NaOH ( X, Y, Z l axit hoc este

( CTPT dng: CxHyO2, d dng (

A ( ( ( p n B

Cu 3: t chy hon ton m gam hn hp X gm hai este ng phn cn dng 27,44 lt kh O2, c 23,52 lt CO2 v 18,9 gam H2O. Cho m gam X T/d ht vi 400 ml dd NaOH 1M, c cn dd sau P c 27,9 gam cht rn khan, trong c a mol mui Y v b mol mui Z (My < Mz). Cc th tch kh u o ktc. T l a : b l

A. 2 : 3 B. 4 : 3 C. 3 : 2 D. 3 : 5

Hng DnD dng c n CO2 = n H2O = 1,05 mol => Este no, n chc c cng thc chung CnH2nO2C nhhX = (3.1,05 2.1,225) : 2 = 0,35 mol (bo ton oxi) => n = 1,05 : 0,35 = 3

Hai este l HCOOC2H5 a mol; CH3COOCH3 b mol.

C a + b = 0,35 v 68a + 82b + 0,05.40 = 27,9

a = 0,2 mol; b = 0,15 mol => a: b = 4: 3

Cu 4: t chy hon ton m gam hn hp X gm hai este ng phn cn 6,272 lt O2(ktc), thu c 5,376 lt CO2(ktc) v 4,32 gam H2O. Thy phn hon ton m gam hn hp X bng lng va dd NaOH , Oxi ha hon ton ancol sinh ra ri cho sn phm to thnh T/d dd AgNO3/NH3 d thu c 23,76 gam Ag. Cc P xy ra hon ton. % khi lng hai este l

A. 62,5% v 37,5%

B. 60% v 40% C. 50% v 50% D. 70% v 30%

Hng Dn

p dng nh lut bo ton khi lng ( mX = 5,92 (g).

= = 0,24 (mol )( este no n chc, mch h (CnH2nO2).

Da vo phn ng t chy gii c n = 3, nX = 0,08 (mol).

CTPT: C3H6O2 ( CTCT ( x + y = 0,08 (*).

S hp thc: HCOOC2H5 ( C2H5OH ( CH3CHO ( 2Ag(

x 2x

CH3COOCH3 ( CH3OH ( HCHO ( 4Ag(

y 4y

(

2x + 4y = 0,22 (**).

Gii h (*) v (**), ta c: x = 0,05; y = 0,03.

hu c Z ; cn Y to ra CH2=CHCOONa v kh T. Cc cht Z v T ln lt l

TH4: Thy phn Este a chcCu 1: Hp cht hu c X cha C, H, O mch thng c phn t khi l 146. X khng T/d Na. Ly 14,6 gam X T/d 100ml dd NaOH 2M thu c 1 mui v 1 ru. CTCT X l:A. C2H4(COOCH3)2 B. (CH3COO)2C2H4 C. (C2H5COO)2 D. A v B ng Hng Dn

nX:nNaOH = 1:2 =>CT X: R(COOR)2 hoc (RCOO)2R

TH1: R + 2R = 58 => R = 28 (-C2H4) v R = 15 (-CH3)

TH2: 2R + R = 58 => R = 28 (-C2H4) v R = 15 (-CH3)

Cu 2: Thy phn hon ton 0,05 mol este ca 1 axit a chc vi 1 ancol n chc cn 5,6 gam KOH. Mt khc, khi thy phn 5,475 gam este th cn 4,2 gam KOH v thu c 6,225 gam mui. CTCT ca este l:

A. (COOC2H5)2B. (COOC3H7)2 C. (COOCH3)2D. CH2(COOCH3)2

Hng Dn

Thy phn 0,05 mol este ca 1 axit a chc vi 1 ancol n chc

P thy phn 5,475 gam

Cu 3: Este X c to thnh t etylen glycol v hai axit cacboxylic n chc. Trong phn t este, s nguyn t cacbon nhiu hn s nguyn t oxi l 1. Khi cho m gam X T/d vi dd NaOH (d) th lng NaOH P l 10 gam. Gi tr ca m l

A. 14,5.B. 17,5.C. 15,5.D. 16,5.Hng Dn

S nguyn t cacbon nhiu hn s nguyn t oxi l 1 nn c 4 nguyn t O th X c 5 C. Cng thc X l:

Cu 4: Cho 0,01 mol mt este X P va vi 100 ml dd NaOH 0,2 M, sn phm to thnh ch gm mt ancol Y v mt mui Z vi s mol bng nhau. Mt khc, khi x phng ho hon ton 1,29 gam este bng mt lng va l 60 ml dd KOH 0,25 M, sau khi P kt thc em c cn dd c 1,665 gam mui khan. CT ca este X l:

A. C2H4(COO)2C4H8 B. C4H8(COO)2C2H4 C. C2H4(COOC4H9)2 D. C4H8(COO C2H5)2

Hng Dn

Ta c: nZ = nY ( X ch cha chc este

S nhm chc este l: = = 2 ( CT ca X c dng: R(COO)2RR(COO)2R + 2KOH ( R(COOK)2 + R(OH)2T phn ng thy phn: naxit = nmui = nKOH = .0,06.0,25 = 0,0075 mol

( M mui = MR + 83.2 = = 222 ( MR = 56 ( R l: -C4H8-Meste = = 172 ( R + 2.44 + R = 172 ( R = 28 (-C2H4-)Vy X l: C4H8(COO)2C2H4 ( p n B.

Cu 5: Mt hp cht hu c X c CT C7H12O4. Bit X ch c 1 loi nhm chc, khi cho 16 gam X T/d va 200 gam dd NaOH 4% th thu c mt ancol Y v 17,8 gam hn hp 2 mui. Xc nh CTCT thu gn ca X.

A. CH3OOC-COOC2H5

B. CH3COO-( CH2)2-COOC2H5 C. CH3COO-(CH2)2-OCOC2H5

D. CH3OOC-COOCH3Hng Dn

p dng DDLBTKL tn khi lng Ancol

Cu 6: Cho 32,7 gam cht hu c X ch cha mt loi nhm chc T/d vi 1,5 lt dd NaOH 0,5M thu c 36,9 gam mui v 0,15 mol Ancol. Lng NaOH d c th trung ha ht 0,5 lt dd HCl 0,6M. CTCT ca X l

A. CH3COOC2H5

B. (CH3COO)2C2H4 C. (CH3COO)3C3H5

D. C3H5(COOCH3)3Hng Dn

PT T/d dd NaOH

Cu 7: un nng m gam hn hp X gm cc cht cng mt loi nhm chc vi 600 ml dd NaOH 1,15M c dd Y cha mui ca mt axit cacboxylic n chc v 15,4 gam hi Z gm cc ancol. Cho ton b Z T/d vi Na d, thu c 5,04 lt kh H2 (ktc). C cn dd Y, nung nng cht rn thu c vi CaO cho n khi P xy ra hon ton c 7,2 gam mt cht kh. Gi tr ca m l

A. 40,60 B. 22,60 C. 34,30 D. 34,51

Hng Dn

(R1COO)xR2 + x NaOH xR1COONa + R2(OH)x

0,45

0,45

0,45/x

R2(OH)x x/2 H2

0,45/x 0,225

RCOONa + NaOH Na2CO3 + RH

0,45 0,24 0,24

n ancol = 2n H2 = 0,45 mol

C n NaOH d = 0,6.1,15 0,45 = 0,24 mol

M kh = 7,2 : 0,24 = 30 => C2H6 => R1 = 29

Vy m = 0,45.96 + 15,4 0,45.40 = 40,6 gam

Chn A.

(RCOONa + NaOH => RH + Na2CO3)

Dng 4: Hiu sut P Este

Phn ng este ha:

RCOOH + ROH RCOOR + H2O

B/ a mol b mol

P/ x mol x mol x mol x mol

Sau p/ (a-x) mol (b-x) mol

1. Tnh hiu sut ca P este ha:

* Nu a b => H = xb . 100 => x = ; b =

* Nu a < b => H = xa . 100 => x = a =

2. Tnh hng s cn bng:

KC = Cu 1: un 12 gam axit axetic vi 1 lung d ancol etylic ( c H2SO4 c lm xc tc). n khi P dng li thu c 11 gam este. Hiu sut ca P este ho l bao nhiu

A. 70%B. 75%

C. 62,5%

D. 50%

Cu 2: Tnh khi lng este metyl metacrylat thu c khi un nng 215 gam axit metacrylat vi 100 gam ancol metylic. Gi thit P este ho t hiu sut 60%.

A. 125 gam

B. 175 gam

C. 150 gam

D. 200 gam

Cu 3: Khi un nng 25,8 gam hn hp ancol etylic v axit axetic c H2SO4 c lm xc tc thu c 14,08 gam este. Nu t chy hon ton lng hn hp thu c 23,4 ml H2O. Tm thnh phn trm mi cht trong hn hp u v hiu sut ca phn ng este ho.

A. 53,5% C2H5OH; 46,5% CH3COOH v hiu sut 80%

B. 55,3% C2H5OH; 44,7% CH3COOH v hiu sut 80%

C. 60,0% C2H5OH; 40,0% CH3COOH v hiu sut 75%

D. 45,0% C2H5OH; 55,0% CH3COOH v hiu sut 60%

Cu 4: un 12 gam axit axetic vi 13,8 gam etanol ( c H2SO4 c lm xc tc) n khi P t ti trng thi cn bng, thu c 11 gam este. Hiu sut ca P este ho l:

A. 55%

B. 50%

C. 62,5%

D. 75%

Cu 5: Khi thc hin P este ha 1 mol CH3COOH v 1 mol C2H5OH, lng este ln nht thu c l 2/3 mol. t hiu sut cc i l 90% (tnh theo axit). Khi tin hnh este ha 1 mol CH3COOH cn s mol C2H5OH l (bit cc P este ho thc hin cng nhit )

A. 2,115. B. 2,925.

C. 2,412. D. 0,456.

Hng Dn

CH3COOH + C2H5OH CH3COOC2H5 + H2O

Ban u 1 1 (mol)

P 2/3 2/3 2/3 2/3 (mol)

Cn bng 1/3 1/3 2/3 2/3

Ta c

Hiu sut cc i l 90% (tnh theo axit) (naxit p= 1.90%=0,9(mol)

CH3COOH + C2H5OH CH3COOC2H5 + H2O

Ban u 1 a (mol)

P 0,9 0,9 0,9 0,9 (mol)

Cn bng 0,1 (a 0,9) 0,9 0,9

Ta c

Cu 6: Hn hp X gm HCOOH v CH3COOH c s mol bng nhau. Ly 5,3 gam hn hp X cho T/d vi 5,75 gam C2H5OH (c H2SO4 c lm xc tc) thu c m gam hn hp este (hiu sut cc P este ha u bng 80%). Gi tr m l

A. 8,80B. 7,04C. 6,48D. 8,10

Hng Dn

RCOOH + C2H5OH RCOOC2H5 + H2O

B 0,1 0,125

Cu 7: Cho bit hng s cn bng ca phn ng este ho:

CH3COOH + C2H5OH CH3COOC2H5 + H2O KC = 4

Nu cho hn hp cng s mol axit v ancol T/d vi nhau th khi P t n trng thi cn bng th % ancol v axit b este ho l

A. 50%. B. 66,7%.

C. 33,3%. D. 65%.

Cu 8: Cho cn bng sau: CH3COOH + C2H5OH CH3COOC2H5 + H2O KC = 4

Khi cho 1 mol axit T/d vi 1,6 mol ancol, khi h t n trng thi cn bng th hiu sut ca P l

A. 66,67%. B. 33,33%.

C. 80%. D. 50%.

Cu 9: un nng hn hp X gm 1 mol ancol etylic v 1 mol axit axetic (c 0,1 mol H2SO4 c lm xt), khi P t n trng thi cn bng c hn hp Y trong c 0,667 mol etyl axetat. Hng s cn bng KC ca phn ng l

A. KC = 2. B. KC = 3.

C. KC = 4. D. KC = 5.

Cu 10: Bit rng P este ho CH3COOH + C2H5OH CH3COOC2H5 + H2O C hng s cn bng KC = 4, tnh % Ancol etylic b este ho nu bt u vi [C2H5OH] = 1M, [CH3COOH] = 2 M.

A. 80%

B. 68%

C. 75%

D. 84,5%

Cu 11: t chy hon ton 7,6 gam hn hp gm mt axit cacboxylic no, n chc, mch h v mt ancol n chc thu c 0,3 mol CO2 v 0,4 mol H2O. Thc hin P este ha 7,6 gam hn hp trn vi hiu sut 80% thu c m gam este. Gi tr ca m l

A. 4,08. B. 6,12. C. 8,16. D. 2,04.

Hng Dn

Do t axit no, n chc cho H2O = CO2 nn ancol cn tm l ancol no, n chc.

S mol ancol = 0,4 0,3 = 0,1 mol

S mol CO2 do ancol to ra s < 0,3 mol. Vy ancol A c mt hoc hai nguyn t C

TH1 Ancol c 1 nguyn t C vy ancol l CH3OH

S mol CO2 do axit to ra = 0,3 0,1 = 0,2 mol

Khi lng axit = 7,6 0,1.32 = 4,4 gam

CT axit : CnH2n+1COOH c s mol l x mol

Vy: (n+1).x = 0,2 v (14n+46)x = 4,4

Tm c: x = 0,05 v n = 3

Este: C3H7COOCH3 c s mol = 0,05.80% = 0,04 mol

Vy khi lng: 0,04.102 = 4,08 gam ( A.

TH2 Ancol c hai nguyn t C

Cu 12: Hn hp A gm axit axetic v etanol. Chia A thnh ba phn bng nhau.

+ Phn 1 T/d vi Kali d thy c 3,36 lt kh thot ra.

+ Phn 2 T/d vi Na2CO3 d thy c 1,12 lt kh CO2 thot ra. Cc th tch kh o ktc.

+ Phn 3 c thm vo vi git dd H2SO4, sau un si hn hp mt thi gian. Bit hiu sut ca P este ho bng 60%. Khi lng este to thnh l bao nhiu

A. 8,80 gam B. 5,20 gam C. 10,56 gam D. 5,28 gamHng Dn

Hn hp A ( (

V a < b (( hiu sut tnh theo axit) ( s mol este thc t thu c: n = 0,1.60% =

( Khi lng este thc t thu c: m = 0,06.88 = 5,28 gam ( p n D

Cu 13: Khi thu phn este A (khng T/d Na, c cu to mch thng di) trong mi trng axit v c c 2 cht hu c B v C. un 4,04 gam A vi dd cha 0,05 mol NaOH c 2 cht B v D. Cho bit MD = MC + 44. Lng NaOH d c trung ho bi 100 ml dd HCl 0,1M. un 3,68 gam B vi H2SO4 c, 170oC vi hiu sut 75% c 1,344 lit olfin (ktc). Tm CTCT A.

A. C3H5(OOCCH3)2.

B. C3H5(OOCC2H5)2.

C. C2H4(OOCCH3)2.

D. C4H8(COOC2H5)2Hng Dn

C: R(COOH)x ; D: R(COONa)x

67x 45x = 44 => x = 2

A: R(COOC2H5)2

R(COOC2H5)2 + 2 NaOH

0,02 0,04

MA = 202 => R = 56 (-C4H8)

A: C4H8(COOC2H5)2

Cu 14: Hn hp M gm ancol no, n chc X va axit cacboxylic n chc Y, u mach h v co cung s nguyn t C, tng s mol cua hai cht la 0,5 mol (s mol cua Y ln hn s mol cua X). Nu t chay hoan toan M thi thu c 33,6 lit khi CO2 (ktc) va 25,2 gam H2O. Mt khac, nu un nong M vi H2SO4 c thc hin P este hoa (hiu sut la 80%) thi s gam este thu c la

A. 34,20 B. 27,36 C. 22,80 D. 18,24Hng Dn

S C = nCO2/nhh = 3 vy ancol l C3H7OH 4H2O . V nNc < nCO2 nn axit khng no.

Axit c 3C c 2TH: CH2=CH-COOH 2H2O ; x + y = 0,5 v 4x + 2y = 1,4. Ta c x= 0,2 v y = 0,3 (nhn)

CHC-COOH 1H2O ; x + y = 0,5 v 4x + y = 1,4. Ta c x= 0,3 v y = 0,2 (loi nY < nX)

Este l CH2=CH-COOC3H7. Vi m CH2=CH-COOC3H7 = 0,2*0,8*114 = 18,24 (g)

C. L THUYT V BI TP LIPIT

LIPIT: Phn ln lipit l cc este phc tp, bao gm cht bo (triglixerit), sp, steroit v photpholipit,Trong chng trnh ta ch yu quan tm cht bo. CHT BO: Cht bo l trieste ca glixerol vi axit bo, gi chung l triglixerit hay l triaxylglixerol.

Cc axit bo hay gp:C17H35COOH hay CH3[CH2]16COOH: axit stearic

C17H33COOH hay cis-CH3[CH2]7CH=CH[CH2]7COOH: axit oleic

C15H31COOH hay CH3[CH2]14COOH: axit panmitic

( Axit bo l nhng axit n chc c mch cacbon di, khng phn nhnh, c th no hoc khng no.

CTCT chung ca cht bo:

R1, R2, R3 l gc hirocacbon ca axit bo, c th ging hoc khc nhau.

Th d v cht bo :

(C17H35COO)3C3H5: tristearoylglixerol (tristearin) (rn m)

(C17H33COO)3C3H5: trioleoylglixerol (triolein) (lng du)

(C15H31COO)3C3H5: tripanmitoylglixerol (tripanmitin) (lng du) Phn ng thu phn trong mi trng axit: Phn ng x phng ho(thy phn trong mi trng baz): Phn ng cng hiro ca cht bo lng

S trieste c to thnh t glixerol v n phn t Axit bo l:

S trieste =

Ch s axit: S mg KOH dung trung ha lng axit t do trong 1 g cht bo.

Ch s axit = (khng i dn v ml)

Tnh cht git ra:

1) c im cu trc phn t mui natri ca axit bo:

Cu to phn t mui natri ca axit bo gm:

Mt u a nc, -COONa.

Mt ui k nc , nhm CXHY

2) C ch hot ng ca cht git ra natri stearat: ui a du m CH3[CH2]16- thm nhp vo vt du bn, cn nhm COONa a nc li c xu hng ko ra pha cc phn t nc.Kt qu: Vt du b phn chia thnh cc ht rt nh gi cht bi cc phn t natri stearat, ri b ra tri.

Dng 5: Tnh khi lng cht bo hoc khi lng x phng

Ta c PTTQ: (RCOO)3C3H5 + 3 NaOH ( 3RCOONa +C3H5(OH)3

( cht bo) (X phng) ( glixerol)

p dng LBT KL: mcht bo + mNaOH = mx phng + mglixerol => m ca cht cn tm

Cu 1: Cho 40,3 gam Trieste X ca Glyxerol vi Axit bo T/d va vi 6 gam NaOH. S gam mui thu c l:

A. 38,1 gam B. 41,7 gam C. 45,6 gam D. 45,9 gam

Hng Dn

S mol NaOH = 6 : 40= 0,15 mol :

C3H5(O-OC- )3 + 3NaOH C3H5(OH)3 + 3 COONa

0,05 0,15 0,05 0,15 mol

40,3 6 0,05.92 m gam

Theo nh lut BTKL ta c: Khi lng mui COONa =40,3+6-0,05.92=41,7 gam.

Cu 2: t chy hon ton m gam cht bo (triglixerit) cn 1,61 mol O2, sinh ra 1,14 mol CO2 v 1,06 mol H2O. Cng m gam cht bo ny T/d va vi dd NaOH th khi lng mui to thnh l

A. 23,00 gam.B. 20,28 gam.C. 18,28 gam.D. 16,68 gam.

Hng Dn

nO/cht bo = 1,06+1,14*2 - 1,61*2= 0,12 mol suy ra ncht bo=0,02mol

mmui=mcht bo+0,06*40-0,02*92 =18,28.

Cu 3: Khi thy phn mt Lipit X ta thu c cc axit bo l Axit oleic, Axit panmetic, Axit stearic. t chy hon ton 8,6 gam X cn th tch O2(ktc) A. 16,128 lt B. 20,16 lt C. 17,472 lt D. 15,68 lt

Hng Dn

X c CTCT l

Cu 4: un nng 44,5 gam cht bo l triglixerit ca 1 axit hu c no vi 70 ml dd NaOH 20% (d=1,2g/ml). trung ho lng kim d cn 22,5ml HCl 36,5%(d=1,2g/ml).CTCT ca cht bo.

A.(C17H29COO)3C3H5

B.(C17H31COO)3C3H5

C.(C17H35COO)3C3H5

D.(C15H29COO)3C3H5Cu 5: Thy phn hon ton 444 gam mt lipit thu c 46g glixerol v hai loi axit bo.Hai loi axit bo l

A.C17H31COOH v C17H33COOH B. C15H31COOH v C17H35COOH

C. C17H33COOH v C17H35COOH D. C15H31COOH v C17H33COOH

Hng Dn

n glixerol = 0,5.

Triglixerit + 3H2O 3 RCOOH + Glixerol

1,5 1,5 0,5

Theo nh lut bo ton khi lng: m axit = 444 +1,5.18 46 = 425g

Vy M tb axit = 425: 1,5 = 283,3. phi c 1 a xit < 283,3 c th l C17H33COOH (282)hoc C17H31COOH ( 280) hoc C15H31COOH (256)v 1 a xit > 283,3 l C17H35COOH (284)

Nhng th li ch c :0,5.282 + 0,5.2.284 = 425 l hp l. Chn C.

Cu 6: un nng 7,2 gam este X vi dd NaOH d. P kt thc thu c glixerol v 7,9 gam hn hp mui. Cho ton b hn hp mui T/d vi H2SO4 long thu c 3 axit hu c no, n chc, mch h Y, Z, T. Trong Z, T l ng phn ca nhau, Z l ng ng k tip ca Y. CTCT ca X

A. B.

C. D.A hoc B

Hng Dn

V Y, Z l ng ng k tip v Z, T l ng phn ca nhau

( c th t cng thc chung ca este X: C3H5(OCO)3

C3H5(OCO)3 + 3NaOH ( 3COONa + C3H5(OH)3 (1)

Theo (1), ta c : nmui = 3neste (

( ( CTCT cc cht: ( p n DCu 7: Mt loi cht bo c cha 25% triolein ,25% tripanmitin v 50% tristearin v khi lng. Cho m Kg cht bo trn P va vi dd NaOH un nng, thu c 1 tn x phng nguyn cht. Gi tr ca m l

A. 972,75

B. 1004,2

C. 1032,33

D. 968,68Hng Dn

(C17H33COO)3C3H5 + 3NaOH ( 3C17H33COONa+ C3H5(OH)3 (M=884) (912)



Cu 8: Mt loi m cha 40% triolein, 20% tripanmitin v 40% tristearin (v khi lng).. X phng ha hon ton m gam m trn thu c 138 gam glixerol. Gi tr ca m l:

A. 1,326 kg B. 1,335 kg C. 1,304 kg D. 1,209 kg

Hng Dn

Cu 9: A l mt este to bi 3 chc mch h. un nng 7,9 gam A vi NaOH d thu c 9,6 gam mui D v Ancol B. Tch nc t B c th thu c propenal. Cho D T/d dd H2SO4 thu c 3 axit no n chc mch h, trong 2 axit c phn t khi nh l ng phn ca nhau. CTPT ca axit c phn t khi nh l

A. C5H10O2

B. C7H14O2

C. C4H8O2

D. C6H12O2

Hng Dn

- Ancol B tch nc c th thu c propenal. Vy B l Glixerol

- T/d dd NaOH

2 axit c phn t khi nh l ng phn ca nhau l C3H7COOH

Cu 10: Mt este X pht xut t anol A v axit B n chc 0,01 mol X (mX = 8,90 gam) P va vi 0,3 lt dd NaOH 0,1M cho ra ancol B v mui C (mC = 9,18 gam). Xc nh CTCT ca X.

A. C3H5(OOCC15H31)3.

B. C3H5(OOCC17H35)3.C. C3H5(OOCC17H33)3.

D. C3H5(OOCC15H29)3.

Cu 11: Cho 2,4 gam este X bay hi trong mt bnh kn dung tch 0,6 lt. Khi este bay hi ht th p sut trong bnh 136,50C l 425,6 mmHg. thy phn 25,4 gam X cn 0,3 mol NaOH thu c 28,2 gam mt mui duy nht. Xc nh CTCT ca X, bit rng X pht xut t ancol a chc.

A. C3H5(OOCCH3)3.

B. C3H5(OOCC2H5)3.

C. C2H4(OOCCH3)3.

D. C3H5(OOCCH = CH2)3.Dng 6: Ch s axit v ch s x phng

Ch s axt ca cht bo: L s miligam KOH cn trung ho lng axit bo t do c trong 1 gam cht bo.

Ch s axt =

Ch s x phng ho ca cht bo: l tng s miligam KOH cn trung ho lng axit t do v x phng ho ht lng este trong 1 gam cht bo Ch s x phng =

Cu 1: X phng ho hon ton 2,5 gam cht bo cn 50 ml dd KOH 0,1 M. Ch s x phng ho ca cht bo l:

A. 280

B. 140

C. 112

D. 224

Hng Dn

Ch s x phng =

Cu 2: trung ho 14 gam mt cht bo cn 15 ml dd KOH 0,1 M. Ch s axit ca cht bo l

A. 6

B. 5

C. 7

D. 8

Hng Dn

Ch s axt =

Cu 3: x phng ho hon ton 2,52 gam mt lipt cn dng 90 ml dd NaOH 0,1M. Tnh ch s x phng ca lipit

A. 100

B. 200

C. 300

D. 400

Cu 4: trung ho 15 gam mt loi cht bo c ch s axit bng 7, cn dng dd cha a gam NaOH. Gi tr ca a l

A. 0,150 B. 0,280 C. 0,075D. 0,200

Hng dn:

Cu 5: Cho 200 gam mt loi cht bo c ch s axit bng 7 T/d va vi mt lng NaOH, thu c 207,55 gam hn hp mui khan. Khi lng NaOH tham gia phn ng l:

A. 31 gamB. 32,36 gamC. 30 gamD. 31,45 gam

Hng dn:

Gi s mol NaOH ban u l a

(RCOO)3C3H5 (Chtbo) + 3NaOH 3RCOONa + C3H5(OH)3 (1)

RCOOH (t do) + NaOH RCOONa + H2O (2) 0,025 0,025 mol

Bo + KOH mui (x phng) + C3H5(OH)3 + H2O (3)

( Ch s axit l 7 nn s mol KOH dng trung ha axit l:

200.7.10-3/56 = 0,025mol = s mol NaOH

( s mol H2O to ra: 0,025 mol

( s mol NaOH phn ng vi trieste l : a 0,025 ( s mol glixerol thu c:

LBTKL: mX + m NaOH = m mui + mglixerol +

200 + 40a = 207,55 + 92 + 18 . 0,025 ( a = 0,775 ( m NaOH = 31 gam

Cu 6: x phng ho hon ton 50 gam cht bo c ch s axit l 7 cn 0,16 mol NaOH. Tnh khi lng glixerol thu c?

A. 9,43gam B. 14,145gam

C. 4,715gam

D. 16,7 gam

Hng dn:

(RCOO)3C3H5 (Chtbo) + 3NaOH 3RCOONa + C3H5(OH)3 (1)RCOOH (t do) + NaOH RCOONa + H2O (2)Bo

+ NaOH mui (x phng) + C3H5(OH)3 + H2O (3)Cn nm r cc khi nim

1. Ch s axit: l smg KOH (2)cn trung ho ht axit t do c trong 1 gam cht bo2. Ch s este: l smg KOH (1)cn thu phn ht este bo c trong 1 gam cht bo3. Ch s x phng = ch s axit + ch s este

4. Khi lng x phng thu c khi x phng ha.

p dng nh lut bo ton khi lng cho phng trnh s (3)

mbo + mKOH = mx phng + mnc + mglixerol mx phng = mbo + mKOH - mnc - mglixerol

Cu 7: X phng ha 1 kg cht bo c ch s axit bng 7 ,ch s x phng ha 200.Tnh khi lng glixerol thu c A. 9,43gam B. 14,145gam

C. 4,715gam

D. 105,7 gamHng dn

Ch s este ha = 200 - 7 = 193S mol KOH este ha = 1000*0,193/56 = 3,446429==> mol Glyxerol = mol KOH/3 = 1,14881==> khi lng glyxerol = 92*1,14881 = 105,7Cu 8:X phng ho hon ton 12,5 gam cht bo c ch s x phng l 224, thu c 13,03 gam mui( Gi thit gc axit trong este v axit t do l nh nhau). Ly ton b lng glyxerol sinh ra em iu ch thuc n trinitro glyxerat. Ch s axit v khi lng thuc n thu c l: A.6,5 v 5,942g B. 5,6 v 4,125g C. 5,6 v 5,942g D. p n khc

Hng dn

(RCOO)3C3H5 (Chtbo) + 3NaOH 3RCOONa + C3H5(OH)3 (1) 3a a

RCOOH (t do) + NaOH RCOONa + H2O (2) b b

nNaOH=3.22412,5.1039 = 0,05mol.gi s mol glyxerol=a v s mol H2O=b 3a + b = 0,05BTKL: mchatbeo + mNaOH = mmuoi + mglyxerol + mH2O

12,5+0,05.40=13,03+92a+18ba=0,015 va`b=0,005Ch s axit:0,005.56.100012,5=22,4Thuc n l:C3H5(ONO2)3m=0,015.227=3,405 gamCu 9:Mt cht bo l trieste ca mt a xit v a xit t do cng l a xit cha trong cht bo.Ch s x phng ca cht bo l 208,77 v ch s a xit l 7.Cng thc Axit trong 1gam cht bo l

A. Stearic B.Oleic C. panmitic D. linoleic

Hng dn:

208,77mg = 0,20877g 0,003728mol KOH

7mg = 0,007g 0,000125mol KOH

(RCOO)3C3H5 + 3KOH 3RCOOK + C3H5(OH)3 0,001201 0,003603

RCOOH + KOH RCOOK + H2O

0,000125 0,000125

Bi cho: 0,001201.(3R +173) + 0,000125.(R+45) = 1

=>R= 211 vy a xit l 211+45 =256:panmitic =>Chn C.

Cu 10: x phng ha 100 kg cht bo c ch s axit l 7 cn dd cha 14,18 kg NaOH.Khi lng x phng cha 28% cht ph gia thu c l

A.143,7kg B. 14,37kg C. 413,7kg D.41,37kg

Hng dn:

nNaOH = 14180:40 = 354,5mol

n NaOH cn trung ha a xit bo d trong 100 kg l: 0,125.100 = 12,5 mol.

=> nNaOH x phng ha l: 354,5 -12,5= 342mol

(RCOO)3C3H5 + 3NaOH 3RCOONa + C3H5(OH)3 342 114

RCOOH + NaOH RCOONa + H2O

12,5 12,5

Theo nh lut bo ton khi lng c 2phng trnh:

mcht bo + mNaOH = mmui + m glixerol + m H2O

mmui = 100000 + 14180 114.92 12,5.18 =103467g

mx phng = 103467.100/72 = 143704,16g =143,7kg

Chn A.

Cu 11: x phng ho 35 kg triolein cn 4,939 kg NaOH thu c 36,207 kg x phng. Ch s axit ca mu cht bo trn l:

A. 7

B. 8 C. 9 D. 10

Hng Dn

Theo bi: nRCOONa (x phng) =

( nNaOH (dng x phng ho) = 119,102 mol

( nNaOH ( trung ho axit bo t do) =

( nKOH ( trung ho axit bo t do) = 4,375 mol

( mKOH (trong 1 g cht bo) =

( ch s axit = 7 ( p n A

Cu 12: Mt loi cht bo c ch s x phng ho l 188,72 cha axit stearic v tristearin. trung ho axit t do c trong 100 g mu cht bo trn th cn bao nhiu ml dung dch NaOH 0,05 M

A. 100 ml B. 675 ml C. 200 ml D. 125 ml

Hng Dn

axp = 188,72.10-3 ( phn ng vi 100 g cht bo cn mKOH = 188,72.10-3 .100 = 18,872 g

( nKOH = ( nNaOH = 0,337 mol

( (

Vy: Trong 100 g mu cht bo c 0,01 mol axit t do ( nNaOH (p) = 0,01 mol

( Vdd NaOH = 200 ml ( p n C

B- BI TP

(T d n kh)

Bi 1: Cho este C3H6O2 x phng ho bi NaOH thu c mui c khi lng bng 41/37 khi lng este. Tm CTCT ca este.

HD: RCOOR

Suy lun: Do este n chc m mmui > meste nn gc R < 23 nn CT este CH3COOCH3

Chi tit: Ta c: => (este n chc nn s mol cc cht bng nhau)

=> MRCOONa = = 82 => R = 15 => R = 15

CT: CH3COOCH3

Bi 2: Tm CTCT ca este C4H8O2 bit rng khi tc dng ht vi Ca(OH)2 thu c mui c khi lng ln hn khi lng ca este.

HD: 2RCOOR + Ca(OH)2 (RCOO)2Ca + 2R(OH)

a a/2

bi ra ta c: (2R + 88 +40)a/2 > (R + R + 44)a => R < 20 (-CH3)

CTCT: CH3CH2COOCH3Bi 3: Cho vo bnh kn (c V = 500 ml) 2,64 gam mt este A hai ln este ri em nung nng bnh n 273C cho n khi ton b este ha hi th p sut trong bnh lc ny l 1,792 atm. Xc nh CTPT

ca A

HD: => 12x+y = 68 => C5H8O4

Bi 4: un nng 0,1 mol cht hu c X vi mt lng va dung dch NaOH thu c 13,4 gam mui ca mt axit hu c Y v 9,2 gam mt ru. Cho ru bay hi 127C v 600 mmHg thu c mt th tch 8,32 lt. CTCT ca X l:

A. C2H5OOC COOC2H5

B. CH3OOC-COOC2H5

C. CH3OOC-CH2-COOC2H5

D. C2H5OOC CH2 COOC2H5

HD:

nru = 0,2 => Mru = 46 => C2H5OH

nru = 2 nX nn este phi l este ca axit hai chc v ru n chc c dng: R(COOC2H5)2 R(COOC2H )2 + 2NaOH 2C2H5OH + R(COONa)2

0,2 0,1

Mmui = 134 => R = 0 => A

Bi 5: Cho cc cht HCOOCH3; CH3COOH; CH3COOCH=CH2; HCOONH4; CH3COOC(CH3)=CH2; CH3COOC2H5; HCOOCH2-CH=CH2. Khi cho cc cht trn tc dng vi dd NaOH thu c sn phm c kh nng tc dng vi dd AgNO3/NH3. S cht tho mn iu kin trn l: A. 3 B. 4 C. 5 D. 6

HD: HCOOCH3; CH3COOCH=CH2; HCOONH4; HCOOCH2-CH=CH2

Bi 6: Cho 12,9g mt este n chc, mch h tc dng ht vi 150ml dd KOH 1M. Sau phn ng thu c mt mui v anehit. S CTCT ca este tho mn tnh cht trn l:

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

HD: HCOOCH=CH-CH3 v CH3COOCH=CH2

Bi 7: Hp cht hu c X cha C, H, O mch thng c phn t khi l 146. X khng tc dng Na. Ly 14,6g X tc dng 100ml dd NaOH 2M thu c 1 mui v 1 ru. CTCT X l:

A. C2H4(COOCH3)2 B. (CH3COO)2C2H4 C. (C2H5COO)2 D. A v B ng

HD nX:nNaOH = 1:2 =>CT X: R(COOR)2 hoc (RCOO)2R

TH1: R + 2R = 58 => R = 28 (-C2H4) v R = 15 (-CH3)

TH2: 2R + R = 58 => R = 28 (-C2H4) v R = 15 (-CH3)

Bi 8: Cho 21,8 gam cht hu c A ch cha 1 loi nhm chc tc dng vi 1 lt dung dch NaOH 0,5M thu c 24,6 gam mui v 0,1 mol ru. Lng NaOH d c th trung ha ht 0,5 lt dung dch HCl 0,4M. Cng thc cu to thu gn ca A l:

HD:Theo bi ra => (RCOO)3R

Theo pt => nmui = 0,3

Mmui = 24,6/0,3 = 82 MRCOONa = 82 =>R = 15 MA = 21,8/0,1= 218

3(15 + 44) + R = 218 R = 41 CT ca este l: (CH3COO)3C3H5

Bi 9: X l mt cht hu c n chc c M = 88. Nu em un 2,2 gam X vi dung dch NaOH d, thu c 2,75 gam mui. Cng thc cu to thu gn ca cht no sau y ph hp vi X:

A. HCOOCH2CH2CH3.

B. CH3CH2CH2COOH.

C. C2H5COOCH3.

D. HCOOCH(CH3)2.

* Nhn xt: Vi lp lun X l cht hu c no, n chc, phn ng vi dung dch NaOH nn X l axit hoc este (loi kh nng l phenol v Mphenol 94 > 88 ( = 94)).

Bi 10: un 20,4 gam mt hp cht hu c n chc A vi 300 ml dung dch NaOH 1M thu c mui B v hp cht hu c C. Khi cho C tc dng vi Na d cho 2,24 lt kh H 2 (ktc). Bit rng khi un nng mui B vi NaOH (xt CaO, t) thu c kh K c t khi i vi O2 bng 0,5. C l hp cht n chc khi b oxi ha bi CuO (t) cho sn phm D khng phn ng vi dung dch AgNO3 /NH3 da. CTCT ca A l:

A. CH3COOCH2CH2CH3

B. CH3COOCH(CH3)CH3

C. HCOOCH(CH3)CH3

D. CH3COOCH2CH3

b. Sau phn ng gia A v NaOH thu c dung dch F. C cn F c m(g) hn hp cht rn. Tnh m.

HD: a. Suy lun:

MK = 16 l CH4 nn axit to este l CH3COOH este c dng CH3COOR

D khng phn ng vi dung dch AgNO3 /NH3 d => D l xeton

=> cu B

Chi tit: este c dng CH3COOR

V este n chc: neste = nru = 2nH 2 = 0,2

=> 15+44+R = 102 => R = 43 ( -C3H7)

D khng phn ng vi dung dch AgNO3 /NH3 d => D l xeton

=> cu B

b. m = mCH3COONa + mNaOH d = 20,4

Bi 11: Hp cht hu c X c thnh phn C, H, O v ch cha 1 nhm chc trong phn t. un nng X vi NaOH th c X1 c thnh phn C, H, O, Na v X2 c thnh phn C, H, O. MX1 = 82%MX; X2 khng tc dng Na, khng cho phn ng trng gng. t 1 th tch X2 thu c 3 th tch CO2 cng iu kin. Tm CTCT X

HD: D dng nhn ra X l este.

Theo bi ra thy X2 l xeton v c 3C: CH3-CO-CH3

X: RCOO-C(CH3)=CH2 ; X1: RCOONa

C: R + 67 = 0,82(R + 85) => R = 15

Vy X: CH3-COO-C(CH3)=CH2Bi 12: Hn hp X c khi lng m(g) cha mt axit n chc no Y v mt ru n chc no Z cng s nguyn t cacbon vi Y. Chia hh X thnh 3 phn bng nhau.

Phn 1: Cho tc dng vi Na d thu c 2,8 lt H2 (ktc)

Phn 2: em t chy hon ton c 22g CO2 v 10,8g H2O

a. X CTPT ca Y v Z.

b. Tm m

c. un nng phn 3 vi H2SO4 c thu c 7,04g este. Tnh hiu sut phn ng este ho.

HD: nCO2 = 0,5; nH2O = 0,6

Do axit v ancol n chc nn: nX = 2nH2 = 0,25

s nguyn t C: = nCO2/nhh = 0,5/0,25 = 2

a. CH3COOH v C2H5OH

b. C: nru = nH2O nCO2 = 0,1 (do axit no th nCO2 = nH2O)

naxit = 0,15 => m = 13,6g

c. h = 80%

Bi 13: Thc hin phn ng x phng ho cht hu c X n chc vi dung dch NaOH thu c mt mui Y v ancol Z. t chy hon ton 2,07 gam Z cn 3,024 lt O2 (ktc) thu c lng CO2 nhiu hn khi lng nc l 1,53 gam. Nung Y vi vi ti xt thu c kh T c t khi so vi khng kh bng 1,03. CTCT ca X l:

A. C2H5COOCH3B. CH3COOC2H5C. C2H5COOC3H7D. C2H5COOC2H5

Gii :

- Theo bi: X n chc, tc dng vi NaOH sinh ra mui v ancol ( X l este n chc: RCOOR.

Mt khc: mX + = + ( 44.+ 18.= 2,07 + (3,024/22,4).32 = 6,39 gam

V 44.- 18.= 1,53 gam ( = 0,09 mol ; = 0,135 mol

> ( Z l ancol no, n chc, mch h c cng thc: CnH2n+1OH (n 1)

nZ = nH2O nCO2 => MZ = 46 (C2H5OH)

MT = 30 => C2H6 p n D

Bi 14: Hn hp X gm 2 cht A, B mch h, u cha cc nguyn t C, H, O v u khng tc dng Na. Cho 10,7g hh X tc dng va NaOH ri c cn sn phm thu c phn rn gm 2 mui natri ca 2 axt n chc no ng ng lin tip v phn hi bay ra ch c mt ru E duy nht. Cho E tc dng vi Na d thu c 1,12lt H2 (ktc). Oxi ho E bng CuO un nng v cho sn phm c th tham gia phn ng trng gng.

a. Tm CTCT ca E bit dE/KK = 2

b. Tm CTCT A, B bit MA < MB

HD: a. ME = 58 => E: C3H6O : CH2=CH-CH2OH (ru allylic)

b. Theo bi ra A, B l 2 este n chc, ng ng lin tip: COOC3H5

nX = nru = 2nH2 = 0,1 => MX = 107 => = 22

A: CH3COOCH2-CH=CH2

B: C2H5COOCH2-CH=CH2

Bi 15: Hn hp A gm 2 cht hu c X, Y u no, n chc v tc dng vi NaOH (MX > MY). T khi hi ca A so vi H2 l 35,6. Cho A td hon ton vi dd NaOH thy ht 4g NaOH, thu c 1 ru n chc v hh 2 mui ca 2 axit n chc. Cho ton b lng ru thu c td vi Na d c 672 ml H2 (ktc). Tm CTPT X, Y.

HD: nA = nNaOH = 0,1 ; nru = 2nH2 = 0,06

Ta thy X, Y n chc m nru < nNaOH nn hh A gm: X l axit (CxH2xO2) v Y l este (CyH2yO2)

nY = nru = 0,06 => nX = 0,1 0,06 = 0,04

mA = 71,2. 0,1 = 7,12 = (14x + 32)0,04 + (14y + 32)0,06

0,56x + 0,84y = 3,92

Vi x>y 2 => x = 4, y = 2

CTPT: C4H8O2 v C2H4O2

Bi 16: Khi thu phn este A (khng tc dng Na, c cu to mch thng di) trong mi trng axit v c c 2 cht hu c B v C. un 4,04g A vi dd cha 0,05 mol NaOH c 2 cht B v D. Cho bit MD = MC + 44. Lng NaOH cn d c trung ho bi 100ml dd HCl 0,1M. un 3,68g B vi H2SO4 c, 170oC vi hiu sut 75% c 1,344 lit olfin (ktc). Tm CTCT A.

HD: nNaOH d = 0,01 => nNaOH p A = 0,04

d dng tm c B: C2H5OH

Suy lun:C l axit; D l mui natri

mt khc MD = MC + 44 => axit 2 chc => nA = nNaOH = 0,02MA = 202 => R = 56 (-C4H8)

A: C4H8(COOC2H5)2

chi tit: C: R(COOH)x ; D: R(COONa)x

67x 45x = 44 => x = 2

A: R(COOC2H5)2

R(COOC2H5)2 + 2 NaOH

0,02 0,04

MA = 202 => R = 56 (-C4H8)

A: C4H8(COOC2H5)2

Bi 17: t chy hon ton 6,8 gam mt este A no n chc cha vng benzen thu c CO2 v H2O . Hp th ton b sn phm ny vo bnh ng dung dch Ca(OH)2 ly d thy khi lng bnh tng 21,2 gam ng thi c 40 gam kt ta. Xc nh CTPT, CTCT c th c ca A

A. 2

B. 3

C. 4

D. 5

HD: Tm CTG: D dng tm c CTPT C8H8O2

4CTCT: phenyl axetat; 3 p: o, m, p -metyl phenyl fomat

Bi 18: Hn hp X gm 1 ancol no, n chc v 1 axit no, n chc mch h. Chia X thnh 2 phn bng nhau.

- t chy hon ton phn 1 sn phm thu c cho qua bnh nc vi trong d thy c 30g kt ta.

- Phn 2 c este ho hon ton va thu c 1 este, t chy este ny thu c khi lng H2O l:

A. 1,8g B. 3,6g C. 5,4g D. 7,2g

HD:

Suy lun: Ta thy s C trong este bng tng C trong axit v ancol => Kh t este v hh (axit, ancol) th thu c CO2 nh nhau.

Mt khc t este no, n chc c nH2O = nCO2 = 0,3

Chi tit:

CnH2n+1OH nCO2

CmH2m+1COOH (m+1)CO2

CmH2m+1COOCnH2n+1(n+m+1) H2O

phn ng va => nax = nancol = x => nCO2 = (n+m+1)x = 0,3

t este: nH2O = (n+m+1)x = 0,3 => C

Bi 19: Thu phn hon ton m gam este X n chc bng NaOH thu c mui hu c A v ancol B. Cho B vo bnh Na d thy khi lng bnh tng 3,1g v c 1,12 lt kh (ktc) thot ra. Mt khc cng cho m gam este X phn ng va 16g brom thu c sn phm cha 35,1% brom theo khi lng. CTCT ca X:

A. C15H33COOCH3

B. C17H33COOCH3

C. C17H31COOCH3

D. C17H33COOC2H5

HD: Ta c: mB = 3,1 + = 3,2

neste = nru = 2nH2 = 0,1 => R = 15 (-CH3)

LBTKL: mg X + 16g Br2 (m + 16)g SP

Ta c: => m = 29,6 => Meste = 296 => R = 237 (-C17H33)

Bi 20: Mt este n chc E c dE/O2 = 2,685. Khi cho 17,2g E tc dng vi 150ml dd NaOH 2M sau c cn c 17,6g cht rn khan v 1 ancol. Tn gi ca E l:

A. Vinyl axetat B. anlyl axetat C. Vinyl fomiat D. Anlyl fomiatHD: nNaOH p = nE = 0,2

=> mmui = 17,6 40(0,3-0,2) = 13,6 => R = 1 => R = 41

Bi 21: Mt hn hp X gm 2 este n chc thy phn hon ton trong mi trng NaOH d cho hn hp Y gm 2 ru ng ng lin tip v hn hp mui Z

- t chy hn hp Y th thu c CO2 v hi H2O theo t l th tch 7:10

- Cho hn hp Z tc dng vi lng va axit sunfuric c 2,08 gam hn hp A gm 2 axit hu c no. Hai axit ny va phn ng vi 1,59 gam natricacbonat

Xc nh CT ca 2 este bit rng cc este u c s nguyn t cacbon < 6 v khng tham gia phn ng vi AgNO3/NH3.

HD: C: RCOOR RCOONa RCOOH + Na2CO3

0,03 0,015

t Y: nH2O > nCO2 => CH2 +1OH T ti l => = 2,33

=> 2 ru l: C2H5OH v C3H7OH (1)

axit = 2,08/0,03 = 69,3 => = 24,3 (2)

Do C < 6 v kt hp (1),(2) => C2H5COOC2H5 v CH3COOC3H7 (khng c phn ng vi AgNO3/NH3).

Bi 22: Mt este A (khng cha chc no khc) mch h c to ra t 1 axit n chc v ru no. Ly 2,54 gam A t chy hon ton thu c 2,688 lt CO2 (ktc) v 1,26 gam H2O . 0,1 mol A phn ng va vi 12 gam NaOH to ra mui v ru. t chy ton b lng ru ny c 6,72 lt CO2 (ktc). Xc nh CTPT, CTCT ca A

HD: nA:nNaOH = 1:3

(RCOO)3R + 3NaOH 3RCOONa + R(OH)3 0,1 0,1

s nguyn t cacbon ca ru: n = 0,3/0,1 = 3 => C3H5 (OH)3Khi t chy A => CTG: C6H7O3 . V este 3 chc => CTPT A: C12H14O6 = 254

Ta c: 3(R1 + 44) + 41 = 254 R1= 27 CH2 CHVy A: (C2H3COO)3C3H5

Bi 23: un nng 0,1 mol este no, n chc mch h X vi 30 ml dung dch 20% (D = 1,2 g/ml) ca mt hiroxit kim loi kim A. Sau khi kt thc phn ng x phng ho, c cn dung dch th thu c cht rn Y v 4,6 gam ancol Z, bit rng Z b oxi ho bi CuO thnh sn phm c kh nng phn ng trng bc. t chy cht rn Y th thu c 9,54 gam mui cacbonat, 8,26 gam hn hp CO2 v hi nc. Cng thc cu to ca X l:

A. CH3COOCH3

B. CH3COOC2H5

C. HCOOCH3

D. C2H5COOCH3

Gii :

X l este no, n chc, mch h : CnH2n+1COOCmH2m+1 ( 0 ( n; 1 ( m)

Ta c: nX = nAOH (p) = nZ = 0,1 mol ( MZ = 14m + 18 = = 46 ( m = 2

Mt khc:

nA = = 2.( MA = 23 ( A l Na ( nNaOH (ban u) =

Y

EMBED Equation.3

EMBED Equation.3Vy: mY + =

Hay 0,1(14n+68) + 0,08.40 + = 9,54 + 8,26 ( n = 1 ( X : CH3COOCH3 ( p n A

Bi 24: Mt hn hp A gm 2 este n chc X, Y (MX < My). un nng 12,5 gam hn hp A vi mt lng dung dch NaOH va thu c 7,6 gam hn hp ancol no B, n chc c khi lng phn t hn km nhau 14 vC v hn hp hai mui Z. t chy 7,6 gam B thu c 7,84 lt kh CO2 (ktc) v 9 gam H2O. Phn trm khi lng ca X, Y trong hn hp A ln lt l:

A. 59,2%; 40,8%B. 50%; 50%C. 40,8%; 59,2%C. 66,67%; 33,33%

Bi gii :

T bi ( A cha 2 este ca 2 ancol ng ng k tip

t cng thc chung ca ancol l

= 7,84/22,4 = 0,35 mol; = 9/18 = 0,5 mol ( nB = -= 0,5 0,35 = 0,15 mol

( = = 2,33. Vy B

t cng thc chung ca hai este l ( neste = nNaOH = nmui = nY = 0,15 mol

( mZ = 12,5 + 0,15.40 7,6 = 10,9 g ( = + 67 = =72,67 ( = 5,67

Nh vy trong hai mui c mt mui l HCOONa

Hai este X, Y c th l:

(I) hoc (II)

- trng hp (I) (

- trng hp (II) ( 12x + y = 8 ( loi)

Vy A( n n A

Bi 25: un nng 7,2 gam este X vi dung dch NaOH d. Phn ng kt thc thu c glixerol v 7,9 gam hn hp mui. Cho ton b hn hp mui tc dng vi H2SO4 long thu c 3 axit hu c no, n chc, mch h Y, Z, T. Trong Z, T l ng phn ca nhau, Z l ng ng k tip ca Y. Cng thc cu to ca X l:

A. B.

C. D.A hoc B

Gii :

V Y, Z l ng ng k tip v Z, T l ng phn ca nhau

( c th t cng thc chung ca este X: C3H5(OCO)3

(1) C3H5(OCO)3 + 3NaOH ( 3COONa + C3H5(OH)3Theo (1), ta c : nmui = 3neste (

( ( CTCT cc cht: ( p n DBi 26: Cho 0,01 mol mt este X ca axit hu c phn ng va vi 100 ml dung dch NaOH 0,2 M, sn phm to thnh ch gm mt ancol Y v mt mui Z vi s mol bng nhau. Mt khc, khi x phng ho hon ton 1,29 gam este bng mt lng va l 60 ml dung dch KOH 0,25 M, sau khi phn ng kt thc em c cn dung dch c 1,665 gam mui khan. Cng thc ca este X l:

A. C2H4(COO)2C4H8B. C4H8(COO)2C2H4C. C2H4(COOC4H9)2D. C4H8(COO C2H5)2

Gii:

Ta c: nZ = nY ( X ch cha chc este

S nhm chc este l: = = 2 ( CT ca X c dng: R(COO)2RT phn ng thy phn: naxit = nmui = nKOH = .0,06.0,25 = 0,0075 mol

( M mui = MR + 83.2 = = 222 ( MR = 56 ( R l: -C4H8-Meste = = 172 ( R + 2.44 + R = 172 ( R = 28 (-C2H4-)Vy X l: C4H8(COO)2C2H4 ( p n B.

Bi 27: Hn hp A gm axit axetic v etanol. Chia A thnh ba phn bng nhau.

+ Phn 1 tc dng vi Kali d thy c 3,36 lt kh thot ra.

+ Phn 2 tc dng vi Na2CO3 d thy c 1,12 lt kh CO2 thot ra. Cc th tch kh o ktc.

+ Phn 3 c thm vo vi git dung dch H2SO4, sau un si hn hp mt thi gian. Bit hiu sut ca phn ng este ho bng 60%. Khi lng este to thnh l bao nhiu?

A. 8,80 gamB. 5,20 gamC. 10,56 gamD. 5,28 gam

Bi gii:

Hn hp A ( (

V a < b (( hiu sut tnh theo axit) ( s mol este thc t thu c: n = 0,1.60% =

( Khi lng este thc t thu c: m = 0,06.88 = 5,28 gam ( p n D

Bi 28: t chy hon ton 1 mol axit cacboxylic n chc X cn 3,5 mol O2. Trn 7,4 gam X vi lng ancol no Y (bit t khi hi ca Y so vi O2 nh hn 2). un nng hn hp vi H2SO4 lm xc tc. Sau khi phn ng hon ton thu c 8,7 gam este Z (trong Z khng cn nhm chc no khc). Cng thc cu to ca Z l:

A. C2H5COOCH2CH2OCOC2H5B. C2H3COOCH2CH2OCOC2H3

C. CH3COOCH2CH2OCOCH3D. HCOOCH2CH2OCOH

Bi gii:

Phn ng chy: CXHyO2 + (x + -1)O2 ( xCO2 + H2O (1)

Theo (1), ta c : x + -1= 3,5 ( x + = 4,5 ( ( X : C2H5COOH

Ancol no Y : CnH2n+2-m (OH)m (1 ( m ( n) ( este Z : (C2H5COO)mCnH2n+2-m

( Meste = 73m + 14n + 2 m = hay 14n + 2 = 15m (2)

Mt khc < 2 hay 14n + 2 + 16m < 64 ( 30m + 2 < 64 (v m ( n) ( m < 2,1

T (2) ( ( ancol Y : C2H4(OH)2

( Z : C2H5COOCH2CH2OCOC2H5 ( p n A.

Bi 29: t chy hon ton mt lng hn hp hai este X, Y, n chc, no, mch h cn 3,976 lt oxi (ktc) thu c 6,38 gam CO2. Cho lng este ny tc dng va vi KOH thu c hn hp hai ancol k tip v 3,92 gam mui ca mt axit hu c. Cng thc cu to ca X, Y ln lt l

A. C2H5COOC2H5 v C2H5COOC3H7B. C2H5COOCH3 v C2H5COOC2H5

C. CH3COOCH3 v CH3COOC2H5D. HCOOC3H7 v HCOOC4H9

Bi gii :

t cng thc trung bnh ca 2 este X, Y l: CnH2n+1COO

V X, Y u l este n chc, no, mch h nn: = = 6,38/44 = 0,145 mol

( meste + = 44. + 18. ( meste = 3,31 gam

neste = nCO2 + 1/2nH2O nO2 = 0,04 mol

( nmui = neste = 0,04 mol ( Mmui = 14n + 84 = 3,92/0,04 = 98 ( n = 1

Mt khc: = 3,31/0,04 = 82,75 ( 12.1 + 46 + 14 = 82,75 ( = 1,77

Vy: X l CH3COOCH3 v Y l CH3COOC2H5 ( p n C

Bi 30: t chy 0,8 gam mt este X n chc c 1,76 gam CO2 v 0,576 gam H2O. Cho 5 gam X tc dng vi lng NaOH va , c cn dung dch sau phn ng c 7 gam mui khan Y. Cho Y tc dng vi dung dch axit long thu c Z khng phn nhnh. Cng thc cu to ca X l:

A. B. C. D. CH2=CH-COOC2H5Bi gii :

Cng thc X: CxHyO2 ( 2 ( x; y ( 2x )

Theo bi: mc = gam; mH = gam ( mO (X) = 0,256 gam

( x : y : 2 = 0,04 : 0,064 : 0,016 = 5 : 8 : 2

( Cng thc ca X: C5H8O2

V X l este n chc (X khng th l este n chc ca phenol) ( nX = nY = nz = nNaOH = 0,05 mol

Ta c : mX + mNaOH (p) = 5 + 0,05.40 = 7 gam = mmui Y

( E l este mch vng ( p n C

Bi 31: Hn hp A gm ba cht hu c X, Y, Z n chc ng phn ca nhau, u tc dng c vi NaOH. un nng 13,875 gam hn hp A vi dung dch NaOH va thu c 15,375 gam hn hp mui v hn hp ancol c t khi hi so vi H2 bng 20,67. 136,50C, 1 atm th tch hi ca 4,625 gam X bng 2,1 lt. Phn trm khi lng ca X, Y, Z (theo th t KLPT gc axit tng dn) ln lt l:

A. 40%; 40%; 20%B. 40%; 20%; 40%C. 25%; 50%; 25%D. 20%; 40%; 40%

Bi gii :

Ta c : ( MX =

Mt khc: X, Y, Z n chc, tc dng c vi NaOH ( X, Y, Z l axit hoc este

( CTPT dng: CxHyO2, d dng (

Vy A ( ( ( p n BBi 32: trung ho 14g cht bo X cn 15ml dd KOH 0,1M. Ch s axit ca cht bo l?

HD: mKOH = 0,015.0,1.56 = 0,084g = 84mg KOH

14g cht bo...............84mg KOH

Vy 1g cht bo.................6 mg KOH => ch s axit l 6

Bi 33: trung ho 10g cht bo c ch s axit l 5,6 th khi lng NaOH cn dng l bao nhiu?

HD: 0,04g

Bi 34: x phng ho hon ton 2,52g mt lipit cn dng 90ml dd 0,1M. Tnh ch s x phng ho ca lipit?

HD: 200

Bi 35: thu phn hon ton 8,58Kg mt loi cht bo cn va 1,2Kg NaOH, thu c 0,368kg glixerol v hh mui ca axit bo. Bit mui ca cc axit bo chim 60% khi lng x phng. Khi lng x phng c th thu c l:

HD: 15,69kg

Bi 36: trung ho 14g cht bo X cn 15ml dd KOH 0,1M. Ch s axit ca cht bo l?

HD: mKOH = 0,015.0,1.56 = 0,084g = 84mg KOH

14g cht bo...............84mg KOH

Vy 1g cht bo.................6 mg KOH => ch s axit l 6

Bi 37: x phng ho 35 kg triolein cn 4,939 kg NaOH thu c 36,207 kg x phng. Ch s axit ca mu cht bo trn l:

A. 7

B. 8

C. 9

D. 10

Bi gii :

Theo bi: nRCOONa (x phng) = ( nNaOH (dng x phng ho) = 119,102 mol

( nNaOH ( trung ho axit bo t do) =

( nKOH ( trung ho axit bo t do) = 4,375 mol

( mKOH (trong 1 g cht bo) =

( ch s axit = 7 ( p n A

Bi 38: Mt loi cht bo c ch s x phng ho l 188,72 cha axit stearic v tristearin. trung ho axit t do c trong 100 g mu cht bo trn th cn bao nhiu ml dung dch NaOH 0,05 M

A. 100 ml

B. 675 ml

C. 200 ml

D. 125 ml

Bi gii :

axp = 188,72.10-3 ( phn ng vi 100 g cht bo cn mKOH = 188,72.10-3 .100 = 18,872 g

( nKOH = ( nNaOH = 0,337 mol

( (

Vy: Trong 100 g mu cht bo c 0,01 mol axit t do ( nNaOH (p) = 0,01 mol

( Vdd NaOH = 200 ml ( p n CBI TP TRONG THI

Bi 1: X l mt este no n chc, c t khi hi i vi CH4 l 5,5. Nu em un 2,2 gam este X vi dung dch NaOH (d), thu c 2,05 gam mui. Cng thc cu to thu gn ca X l: (khi B 2007)

A. C2H5COOCH3.B. HCOOCH2CH2CH3.C. CH3COOC2H5D. HCOOCH(CH3)2.

Gii:

Meste = 5,5.16 = 88 neste = 2,2/88 = 0,025 mol

( nEste = nmui = 0,025 mol ( Mmui = 2,05/0,025 = 82

( R=82 67 = 15 ( R l CH3- ( p n C ng

* Ch : Ta c th dng phng php loi tr tm p n:

T bi: meste > mmui ( X khng th l este ca ancol CH3OH ( p n A loi.

T phn ng thy phn ta ch xc nh c CTPT ca cc gc R v R m khng th xc nh c cu to ca cc gc do B v D khng th ng thi ng do ta loi tr tip B v D.

Vy ch c p n C ph hp

Bi 2: C=10: Hn hp Z gm hai este X va Y tao bi cung mt ancol va hai axit cacboxylic k tip nhau trong day ng ng (MX < MY). t chay hoan toan m gam Z cn dung 6,16 lit khi O2 (ktc), thu c 5,6 lit khi CO2 (ktc) va 4,5 gam H2O. Cng thc este X va gia tri cua m tng ng la

A. CH3COOCH3 va 6,7 B. HCOOC2H5 va 9,5

C. HCOOCH3 va 6,7 D. (HCOO)2C2H4 va 6,6

HD: Gii : => X, Y l 2 este no n chc

p dng LBTKL : m = + 4,5 - = 6,7 (gam)

t cng thc ca X, Y : => =>

=> => n = 2 ; n = 3 X : C2H4O2 HCOOCH3 Y : C3H6O2 CH3COOCH3

Ch : gp bi ton hu c m khi t chy th cn bao nhiu lt oxi hoc cn tnh th tch oxi th chng ta nn ngh ngay n pp LBTKL hoc LBTNT ty thuc vo d kin bi ton cho.

Bi 3: C10: Thuy phn cht hu c X trong dung dich NaOH (d), un nong, thu c san phm gm 2 mui va ancol etylic. Cht X la

A. CH3COOCH2CH3 B. CH3COOCH2CH2Cl

C. ClCH2COOC2H5 D. CH3COOCH(Cl)CH3

HD: ClCH2COOC2H5 + 2NaOH NaCl + HO-CH2COONa + C2H5OHBi 4: C10: t chay hoan toan 2,76 gam hn hp X gm CxHyCOOH, CxHyCOOCH3, CH3OH thu c 2,688 lit CO2 (ktc) va 1,8 gam H2O. Mt khac, cho 2,76 gam X phan ng va u vi 30 ml dung dich NaOH 1M, thu c 0,96 gam CH3OH. Cng thc cua CxHyCOOH la

A. C2H5COOH B.CH3COOH C. C2H3COOH D. C3H5COOH

HD: + Axit v este u khng no (CH3OH chy cho s mol H2O ln hn s mol CO2) loi A, B

+ = amol

Cch 1: neste = 0,03-a. Ta c: nCO2 = (x+1)a + (x+2)(0,03-a) + a = 0,12 => x = 2

Cch2: Coi X gm CxHyCOOCH3 v H2O Vi neste mi = 0,03 mol

p n C

Bi 5: C10: Cho 45 gam axit axetic phn ng vi 69 gam ancol etylic (xc tc H2SO4 c), un nng, thu c 41,25 gam etyl axetat. Hiu sut ca phn ng este ho l

A. 62,50% B. 50,00%C. 40,00% D. 31,25%

HD: %

Bi 6: C10: trung ho 15 gam mt loi cht bo c ch s axit bng 7, cn dng dung dch cha a gam NaOH. Gi tr ca a l

A. 0,150 B. 0,280 C. 0,075 D. 0,200

HD:

Cch khc:

thi C H ch c 1 cu v ch s axit v cn nh cng thc tnh th bi ton tr nn nh nhng hn nhiu,

Bi 7: HB-2011: Cho 200 gam mt loi cht bo c ch s axit bng 7 tc dng va vi mt lng NaOH, thu c 207,55 gam hn hp mui khan. Khi lng NaOH tham gia phn ng l:


Hng dn: cu ny tng t thi C 2010 chc cc bn lm thun thc ri


ADLBTKL

Bi 8: C11: phn ng ht vi mt lng hn hp gm hai cht hu c n chc X v Y (Mx < MY) cn va 300 ml dung dch NaOH 1M. Sau khi phn ng xy ra hon ton thu c 24,6 gam mui ca mt axit hu c v m gam mt ancol. t chy hon ton lng ancol trn thu c 4,48 lt CO2 (ktc) v 5,4 gam H2O. Cng thc ca Y l :

A. CH3COOC2H5 B. CH3COOCH3 C. CH2=CHCOOCH3 D. C2H5COOC2H5HD: nancol =nH2O-nCO2 => C = nCO2/nancol => C2H5OH

V X, Y n chc m nNaOH > nancol => Y l este; X l axit (do Mx < MY v c cng gc axit)Mmui = 24,6/0,3 = 82 => R =-CH3

=> A

Bi 9: C11: Cho m gam cht hu c n chc X tc dng va vi 50 gam dung dch NaOH 8%, sau khi phn ng hon ton thu c 9,6 gam mui ca mt axit hu c v 3,2 gam mt ancol. Cng thc ca X l:

A. CH3COOC2H5B. C2H5COOCH3C. CH2=CHCOOCH3 D. CH3COOCH=CH2Bi 10: C11: x phng ho hon ton 52,8 gam hn hp hai este no, n chc , mch h l ng phn ca nhau cn va 600 ml dung dch KOH 1M. Bit c hai este ny u khng tham gia phn ng trng bc. Cng thc ca hai este l

A. CH3COOC2H5 v HCOOC3H7B. C2H5COOC2H5 v C3H7COOCH3

C. HCOOC4H9 v CH3COOC3H7D. C2H5COOCH3 v CH3COOC2H5Bi 11: C11: Hp cht hu c X c cng thc phn t l C4H8O3. X c kh nng tham gia phn ng vi Na, vi dung dch NaOH v phn ng trng bc. Sn phm thu phn ca X trong mi trng kim c kh nng ho tan Cu(OH)2 to thnh dung dch mu xanh lam. Cng thc cu to ca X c th l:

A. CH3CH(OH)CH(OH)CHOB. HCOOCH2CH(OH)CH3

C. CH3COOCH2CH2OH.D. HCOOCH2CH2CH2OH

HD: - X c phn ng vi dung dch NaOH, trng bc loi A, C

- Sn phm thu phn ca X trong mi trng kim c kh nng ho tan Cu(OH)2 to thnh dung dch mu xanh lam (tnh cht ru a chc c 2 nhm OH k nhau) => B

Bi 12: C11: Este X no, n chc, mch h, khng c phn ng trng bc. t chy 0,1 mol X ri cho sn phm chy hp th hon ton vo dung dch nc vi trong c cha 0,22 mol Ca(OH)2 th vn thu c kt ta. Thu phn X bng dung dch NaOH thu c 2 cht hu c c s nguyn t cacbon trong phn t bng nhau. Phn trm khi lng ca oxi trong X l:

A. 43,24%B. 53,33%C. 37,21%D. 36,26%

HD: Cn nCO2

0,1 0,1n

CO2 + Ca(OH)2 CaCO3 + H2O (1)

0,22 0,22 0,22

CO2 + CaCO3 + H2O Ca(HCO3)2 (2)

0,22 0,22

Theo (1), (2): thu c kt ta th: nCO2 < 0,22+0,22 = 0,44

Hay: 0,1n < 0,44 => n < 4,4

X + NaOH to 2 cht c C = nhau => X c 2 hoc 4 C

X khng c p trng gng => n = 4 C4H8O2

HA -2011

Bi 13: HA -2011 : t chy hon ton 3,42 gam hn hp gm axit acrylic, vinyl axetat, metyl acrylat v axit oleic, ri hp th ton b sn phm chy vo dung dch Ca(OH)2 (d). Sau phn ng thu c 18 gam kt ta v dung dch X. Khi lng X so vi khi lng dung dch Ca(OH)2 ban u thay i nh th no?

A. Tng 2,70 gam.B. Gim 7,74 gam.C. Tng 7,92 gam.D. Gim 7,38 gam.

Gii:

Cch 1: Sau phn ng thu c 18 gam kt ta th Khi lng X so vi khi lng dung dch Ca(OH)2 ban u s gim ri, vn l gim 7,74 hay 7,38 gamCng thc chung ca cc cht trn l CnH2n-2O2 do nu gi x l mol CO2, y l mol H2O

BTKL : 3,42 + 3/2y.32 = 44x + 18y . mt khc x = 0,18 ----> y = 0,18 ---> tng (CO2+H2O) =10,62< 18 gam kt ta nn dd gim 7,38gam => D ng.

Cch 2: hn hp gm axit acrylic, vinyl axetat, metyl acrylat v axit oleic u c ctc l:

. p dng lbt khi lng v nguyn t ta c:

Khi lng X so vi khi lng dung dch Ca(OH)2 ban u s gim l:

=> D ng.

Ch : ta lun c mCO2 + mH2O = mCaCO3 m(dd gim)

Bi 14: HA -2011: Cho axit salixylic (axit o-hiroxibenzoic) phn ng vi anhirit axetic, thu c axit axetylsalixylic (o-CH3COO-C6H4-COOH) dng lm thuc cm (aspirin). phn ng hon ton vi 43,2 gam axit axetylsalixylic cn va V lt dung dch KOH 1M. Gi tr ca V l

A. 0,72.B. 0,48.C. 0,96.D. 0,24.

Gii: 1mol axit axetylsalixylic (o-CH3COO-C6H4-COOH) th cn 3 mol KOH, nn d dng suy ra

=> A ng.

Nu cha hiu th theo cch gii sau: ptpu xy ra:

o-CH3COO-C6H4-COOH + 3KOH = CH3COOK +o-KO-C6H4-COOK+ 2H2O (1)

theo (1) => A ng.

Phn tch: cu ny nu khng cho sn phm v ctct ca axit axetylsalixylic th mc s kh hn nhiu, nhng cho ctct th nhn vo s tnh ra ngay. nu khng cn thn th s chn p n B: 0,48 lt.

Bi 15: HA -2011: Este X c to thnh t etylen glicol v hai axit cacboxylic n chc. Trong phn t este, s nguyn t cacbon nhiu hn s nguyn t oxi l 1. Khi cho m gam X tc dng vi dung dch NaOH (d) th lng NaOH phn ng l 10 gam. Gi tr ca m l

A. 14,5.B. 17,5.C. 15,5.D. 16,5.

Gii:

Cch 1. Cu ny bn phi tnh to th d dng suy ra cng thc ESTE l C5H8O4 (132)

Nu vn kh hiu th xem hng dn sau.

Cch 2. S nguyn t cacbon nhiu hn s nguyn t oxi l 1 nn c 4 nguyn t O th X c 5 C. cng thc X l:

Cch 3. (-COO)2C2H4 ( = 1 ( HCOOH v CH3COOH ( ME = 132

nNaOH = 0,25 ( nX = 0,125 ( m = 132.0,125 = 16,5 gam

Bi 16: HA -2011 : Cho dy cc cht: phenylamoni clorua, benzyl clorua, isopropyl clorua, m-crezol, ancol benzylic, natri phenolat, anlyl clorua. S cht trong dy tc dng c vi dung dch NaOH long, un nng l

A. 4.B. 3. C. 6.D. 5.

Gii: phenylamoni clorua, benzyl clorua, isopropyl clorua, m-crezol, anlyl clorua

Bi 17: HA -2011: t chy hon ton 0,11 gam mt este X ( to nn t mt axit cacboxylic n chc v mt ancol n chc) thu c 0,22 gam CO2 v 0,09 gam H2O. S este ng phn ca X l:

A. 2B. 5C. 6D.4

Gii:

Cch 1: theo quy lut ng phn ca este l: 1-2-4-9. nh vy ch c A hoc D ng m thi. m cho 0,11 gam nn D ng. v C2H4O2 (60) c 1 p este.

C3H6O2 (74) c 2 p este.

C4H8O2 (88) c 4 p este.

C5H10O2 (102) c 9 p este.

Ch : ly 0,11 chia cho 60, 74, 88... p n c s mol p th ta chon thi.

Cch 2: = 0,005 = ( Este no, n chc CnH2nO2 ( M = 14n + 32

n = 0,005 ( n = 4 ( S p este CnH2nO2 = 2n-2 => D ng.Bi 18: HB -2011: Cho dy cc cht: phenyl axetat, anlyl axetat, metyl axetat, etyl fomat, tripanmitin. S cht trong dy khi thy phn trong dung dch NaOH (d), un nng sinh ra ancol l:

A. 4B. 2C. 5D. 3Hng dn: bn phi thuc tt c cc cht hu => Ch c CH3COOC6H5 thy phn to 2 mui

Bi 19: HB -2011: Cho 200 gam mt loi cht bo c ch s axit bng 7 tc dng va vi mt lng NaOH, thu c 207,55 gam hn hp mui khan. Khi lng NaOH tham gia phn ng l:


Hng dn: cu ny tng t thi C 2010 chc cc bn lm thun thc ri


ADLBTKL

thi C H ch c 1 cu v ch s axit v cn nh cng thc tnh th bi ton tr nn nh nhng hn nhiuBi 20: HB -2011: Khi cho 0,15 mol este n chc X tc dng vi dung dch NaOH (d), sau khi phn ng kt thc th lng NaOH phn ng l 12 gam v tng khi lng sn phm hu c thu c l 29,7 gam. S ng phn cu to ca X tha mn cc tnh cht trn l:

A. 4B. 5C. 6D. 2

Hng dn: k thut bm my tnh: =>A ng.C1. p dng nh lut BTKL => C 4 ng phn. => A ng.

Nu khng hiu lm th xem cch sau.

Cch 2: nNaOH:nEste = 2:1 l este to bi axit v gc ancol dng phenol RCOOR + 2NaOH RCOONa + RONa + H2O 0,15 0,3 0,15

mEste = 29,7 + 0,15.18 12 = 20,4 gam MX = 136 = R + 44 R = 92 ( C7H8

( CTPT C8H8O2 ( ng phn ca X: CH3-COO-C6H5; HCOOC6H4 CH3 (c 3 p )Bi 21: HB -2011: Triolein khng tc dng vi cht (hoc dung dch) no sau y? A. H2O (xc tc H2SO4 long, un nng)B. Cu(OH)2 ( iu kin thng)

C. Dung

Toan Bo Bai Tap Co Huong Dan Chuong Estelipit Ltdh Nam 2014

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