4.2. Thy phn hon ton 0,1mol este X bng NaOH, thu c mui ca axit cacboxylic v 6,2 gam ancol Z. Mui thu c c phn ng trng bc, Z ha tan Cu(OH)2 cho dung dch mu xanh lam. Cng thc cu to ca X l:A. HCOOCH2CH2CH2OOCHB. HCOOCH2CH(CH3)OOCHC. HCOOCH2CH2OOCCH3D. CH3COOCH2CH2OOCCH334.2. Thy phn 0,01 mol este ca mt ancol a chc vi mt axit n chc cn 1,2 gam NaOH. Mt khc khi thy phn 4,36 gam este th cn 2,4 gam NaOH v thu c 4,92 gam mui. Cng thc ca este lA. (CH3COO)3C3H5B. (C2H3COO)3C3H5 C. C3H5(COOCH3)3D. C3H5(COOC2H3)3

4.3. Thy phn 37 gam hai este c cng cng thc C3H6O2 bng dung dch NaOH d. Chng ct dung dch sau phn ng thu c hn hp ancol Y v cht rn khan Z. un nng Y vi H2SO4 c 1400c thu c 15,7 gam hn hp cc ete. Bit cc phn ng xy ra hon ton. Khi lng mui trong Z l:A. 36,8 gamB. 32,3 gamC. 39,6 gamD. 38,4 gamCho cc dn xut halogen sau: (1) etyl clorua; (2) phenyl clorua; (3) benzyl clorua; (4) p- clotoluen; (5) 1,2-icloetan. Nhng dn xut b thy phn trong dung dch NaOH long, un nng l:A. (2) (3) (5) B. (1) (3) (5) C. (1) (2) (3) D. (2) (4) (5)30.5. Khi thu phn a gam mt este X thu c 0,92 gam glixerol, 3,02 gam natri linoleat(C17H31COONa) v m gam mui ca natri oleat (C17H33COONa). Gi tr ca a v m lA. 6,08 v 8,82.B. 3,94 v 7,88.C. 8,82 v 6,08.D. 6,08 v 9,2.37.5. Mt este X to bi axit n chc v ancol n chc c t khi hi so vi kh CO2 bng 2. Khi un nng X vi dung dch NaOH thu c ancol Y c khi lng bng 36,36% khi lng X phn ng. Cng thc ca X lA. HCOOC3H7.B. C2H5COOCH3.C. CH3COOC2H5D. CH3COOCH3.4.7. Cho cc pht biu sau: (a) Cht bo c gi chung l triglixerit hay triaxylglixerol; (b) Cht bo nh hn nc, khng tan trong nc nhng tan nhiu trong cc dung mi hu c; (c) Phn ng thy phn cht bo trong mi trng axit l phn ng thun nghch; (d) Tristearin, triolein c CT ln lt l (C17H33COO)3C3H5, (C17H35COO)3C3H5; (e) Triolein c kh nng tham gia phn ng cng hiro khi un nng c xc tc Ni; (f) Cht bo b thy phn khi un nng trong dd kim.S pht biu ng l:A. 3B. 5C. 4D. 620.7. Hn hp X gm 2 este ng phn ca nhau, t khi hi ca hn hp so vi khng kh bng 2,552. Cho 11,1 gam hn hp X tc dng vi dung dch NaOH d n khi phn ng xy ra hon ton thu c 11,6 gam hn hp mui. Tnh phn trm khi lng ca cc este trong hn hp X:A. 66,67% v 33,33%.B. 50% v 50%.C. 40% v 60%.D. 70% v 30%.22.7. Cho cc cht sau: CH3COOCH2CH2Cl, ClH3N-CH2COOH, C6H5Cl, HCOOC6H5, C6H5COOCH3, HO-C6H4-CH2OH, CH3CCl3, CH3COOCCl2-CH3. C bao nhiu cht khi tc dng vi NaOH c d, nhit v p sut cao cho sn phm c 2 mui:A. 7.B. 4.C. 5.D. 6.46.7. Thu phn triglixerit X trong NaOH thu c hn hp hai mui natrioleat v natristearat theo t l mol l 1 : 2. Khi t chy a mol X thu c b mol CO2 v c mol H2O. Mi lin h gia a, b, c lA. b = c aB. b c = 4aC. b c = 2a.D. b c = 3a23.8. Hn hp X gm 2 este ng phn. t chy hon ton m gam X cn dng 11,76 lit O2 (ktc) thu c 19,8 gam CO2 v 8,1 gam H2O. Cho m gam X tc dng ht vi 200ml dung dch NaOH 1M, c cn dung dch sau phn ng thu c 13,6 gam cht rn v hn hp ancol Y. t chy hon ton Y cn dng va V lit O2 (ktc). Gi tr ca V l:A. 6,72B. 3,36C. 5,60D. 4.4848.8. t chy hon ton m gam mt cht bo (triglixerit) cn 1,61 mol O2, sinh ra 1,14 mol CO2 v 1,06 mol H2O. Nu cho m gam cht bo ny tc dng va vi dung dch NaOH th khi lng mui to thnh l:A. 23,00 gamB. 18,28 gamC. 16,68 gamD. 20,28 gam24.9. X phng ha hon ton 66,6g hn hp hai este HCOOC2H5 v CH3COOCH3 bng dung dch NaOH, thu c hn hp X gm hai ancol. un nng hn hp X vi H2SO4 c 1400C sau khi phn ng xy ra hon ton thu c m gam nc. Gi tr ca m lA. 8,1 gam.B. 18 gam.C. 16,2 gam.D. 4,05 gam.

3.10. X c cng thc phn t l C7H6O3. Cho 1 mol X tc dng va vi 2 mol NaOH thu chai cht hu c Y v Z. CTCT ca X lA. HCOOC6H4OH.B. HOOC-C6H4OH.C. (OH)2C6H3CHO.D. (OH)2C6H3CH2OH.29.11. Thy phn hon ton 0,1 mol este X cn 200 ml dung dch NaOH 1,5M. C cn dung dch sau phn ng thu c glixerol v 24,6 gam cht rn khan. Hy cho bit X c bao nhiu CTCT:A. 7B. 5C. 8D. 649.11. iu ch etyl axetat t tinh bt theo s : tinh bt glucoz ancol etylic axit axetic etyl axetat. Bit hiu sut ca mi giai on iu ch u t 50%. Khi lng tinh bt cn dng iu ch 1mol etyl axetat l:A. 1012 gamB. 1944 gamC. 405,0 gamD. 324 gam6.12. Khi cho 0,1mol este n chc, mch h X tc dng vi dung dch NaOH d sau khi phn ng kt thc th lng NaOH phn ng l 8g v tng khi lng sn phm hu c thu c l 21,0g. S ng phn cu to ca X tho mn cc tnh cht trn l:A. 5B. 3C. 4D. 2

Phn ng thy phn este no, n chc, mch h trong mi trng axit khng c c im no sau y:A. l phn ng thun nghchB. sn phm gm axit v ancolC. l phn ng mt chiuD. xy ra khi c xc tc axit, un nng

9.15. Cho 4,48 gam hn hp gm CH3COOC2H5 v CH3COOC6H5 (c t l mol 1:1) tc dng vi 800 ml dung dch NaOH 0,1M thu c dung dch X. C cn dung dch X th khi lng cht rn thu c l:A. 5,60 gamB. 4,88 gamC. 6,40 gamD. 3,28 gam

30.15. X phng ha hon ton m gam mt este n chc A bng 200ml dung dch KOH 1M. C cn dung dch sau phn ng c 19,3 gam cht rn khan v hi mt ancol B. Oxi ha B bng lng d CuO, sn phm thu c em trng bc hon ton thu c 64,8 gam Ag. Khi lng m v cng thc ca A l:A. 12,9 g v C2H3COOCH3B. 13,2 g v C2H5COOCH3C. 17,2 g v C2H3COOCH3D. 26,4 g v CH3COOC2H521.16. Hp cht hu c n chc X cha C, H, O. Trong phn trm khi lng ca C, H tng ng l 55,81 % v 6,98 %. Y l ng phn ca X v hu nh khng tan trong nc. C X v Y u c ng phn cis trans. Cng thc cu to ca X v Y l cng thc no sau yA. CH2=CHCH2COOH v CH3COOCH=CH2.B. HCOOCH=CHCH3 v CH3CH=CHCOOH.C. CH2=CHCOOH v HCOOCH=CH2.D. CH3CH=CHCOOH v HCOOCH=CHCH3.28.16. Cho este n chc E tc dng va vi NaOH thu c 12,3 gam mui v 0,15 mol ancol. t chy hon ton ancol ny ri hp th sn phm vo bnh cha 3 lt dung dch Ba(OH)2 0,125M thu c 59,1 gam kt ta. Cng thc cu to ca E c th c lA. E ch l CH3COOC2H5B. E ch l CH3COOC3H7.C. E l CH3COOC2H5 hay CH3COOC3H7D. E l CH3COOCH3 hay CH3COOC2H535.16. Cho cc pht biu:(a) Trong mt phn t tristearin c cha 3 lin kt pi.(b) Cht bo nh hn nc, khng tan trong nc(c) ht bo b thy phn hon ton trong mi trng axit(d) Cht bo l trieste ca etylen glicol vi cc axit bo S pht biu ng l:A. 4B. 1C. 2D. 33.17. Thu phn hon ton 8,8g este n chc, mch h X vi 100ml dung dch KOH 1M (va ) thu c 4,6g mt ancol Y. Tn gi ca X l:A. Propyl axetat.B. Etyl fomiat.C. Etyl propionat.D. Etyl axetat.18.17. Thu phn este X (C4H6O2) trong mi trng axit ta thu c mt hn hp cc cht u c phn ng trng gng. Cng thc cu to ca X l:A. HCOO CH2 CH = CH2.B. CH2 = CH COO CH3.C. CH3 CH = CH OCOH.D. CH2 = CH OCO CH3.Cu 1. MH_2015: X phng ha hon ton m gam mt este no, n chc mch h E bng 26 gam dung dch MOH 28% (M l kim loi kim). C cn hn hp sau phn ng thu c 24. 72 gam cht lng X v 10,08 gam cht rn khan Y. t chy hon ton Y, thu c sn phm gm CO2, H2O v 8,97 gam mui cacbonat khan. Mt khc, cho X tc dng vi Na d, thu c 12,768 lt kh H2 (ktc). Phn trm khi lng mui trong Y c gi tr gn nht viA. 67,5. B. 85,0. C. 80. D. 97,5. p n B.CnH2n+1COOCmH2m+1 + MOH CnH2n+1COOM + CmH2m+1OH

Cht lng (X):

Rn khan (Y):

Kali

Mui: HCOOK: 0,1 (mol) mHCOOK = 0,1.84 = 8,4g

%HCOOK = . Dng bi hay v kh v phn ng thy phn este Chng 1 Lp 12.

Cu 2. VST_ln 9: C bao nhiu ng phn este mch h, khng phn nhnh c cng thc phn t C6H10O4 khi cho tc dng vi NaOH to ra 1 ancol v 1 mui?A. 3.B. 5.C. 4.D. 2.p n CV C6H10O4 khi tc dng vi NaOH thu c 1 ancol v 1 mui nn C6H10O4 c th l este 2 chc ca ancol 2 chc vi axit n chc hoc este 2 chc ca ancol n chc vi axit n chc. Cc ng phn tha mn gm: CH3COOCH2CH2OOCCH3; HCOOCH2CH2CH2CH2OOCH; CH3OOCCH2CH2COOCH3; C2H5OOCCOOC2H5. Cu 3. Cho 0,01 mol mt este X phn ng va ht vi 100 ml dung dch NaOH 0,2M, sn phm to ra ch gm mt mui v mt ancol, u c s mol bng s mol este. Mt khc x phng ha hon ton mt lng este X bng dung dch KOH va , th va ht 200 ml dung dch KOH 0,15M v thu c 3,33 gam mui v 0,93 gam ancol. Phn trm khi lng ca cacbon trong phn t X lA. 55,81%.B. 60,24%.C. 54,32%.D. 50,73%.p n A

C X l ieste hoc este n chc ca phenol; v nmui X l este 2 chc mch vng c cng thc cu to dng:

C = = nKOH = 0,015 (mol)

= = 222 R = 56 R l gc C4H8 axit to ra X l C4H8(COOH)2.

Mancol = = 62 ancol: CH2OH CH2OH X l C4H8(COO)2C2H4 %(khi lng ) C 55,81%.Bi ton v phn ng x phng ha este Chng 1 Lp 12.Cu 4. VST_ln 9: C bao nhiu ng phn este mch h, khng phn nhnh c cng thc phn t C6H10O4 khi cho tc dng vi NaOH to ra 1 ancol v 1 mui?A. 3.B. 5.C. 4.D. 2.p n CV C6H10O4 khi tc dng vi NaOH thu c 1 ancol v 1 mui nn C6H10O4 c th l este 2 chc ca ancol 2 chc vi axit n chc hoc este 2 chc ca ancol n chc vi axit n chc. Cc ng phn tha mn gm: CH3COOCH2CH2OOCCH3; HCOOCH2CH2CH2CH2OOCH; CH3OOCCH2CH2COOCH3; C2H5OOCCOOC2H5. 1. HA2009: X phng ha mt hp cht c cng thc phn t C10H14O6 trong dung dch NaOH (d), thu c glixerol v hn hp gm ba mui (khng c ng phn hnh hc). Cng thc ca ba mui l:A. CH2=CH-COONa, HCOONa v CHC-COONa.B. CH3-COONa, HCOONa v CH3-CH=CH-COONa.C. HCOONa, CHC-COONa v CH3-CH2-COONa.D. CH2=CH-COONa, CH3-CH2-COONa v HCOONa.p n DCch 1: CTTQ ca este: CnH2n+ 2 2k 2xO2x( k l s lin kt + v trong gc hirocacbon v x l s nhm chc)Ta c CTPT: C10H14O6

2n + 2 2k 2x = 14 2.10 + 2 2k 6 = 14 k =11 lk gc nn chn phng n B hoc DB CH3CH=CHCOONa c ng phn hnh hc nn chn DCch 2: Theo cng thc tnh bt bo ha ta c CxHyOzNtClv

k = (2x +2 - y + t -v)/2 = 4. C 3 lk ti gc COO,1lk gc nn chn phng n D. 1. HB.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.p n Daxit acrylic : C3H4O2, vinyl axetat : C4H6O2,metyl acrylat : C4H6O2,axit oleic : C17H34O2D thy cc hp cht hu c trn u c dng : CnH2n-2O2

Phng trnh phn ng chy : CnH2n-2O2 + O2 nCO2 + (n-1)H2O

D thy :.

1. p dng nh lut bo ton khi lng

m dung dch gim = = 18-10,62 = 7,38 gam.1. VST_ln 11: Hn hp X gm 2 este n chc. Cho 0,05 mol hn hp X tc dng va vi dung dch NaOH, thu c dung dch Y. C cn dung dch Y, thu c cht rn khan Z. t chy hon ton Z th thu c 5,28 gam CO2; 3,18 gam Na2CO3 v m gam H2O. Gi tr ca m lA. 0,9.B. 1,8.C. 2,7.D. 1,98.p n AnCO2 = 0,12 mol; nNa2CO3 = 0,03 mol.nNa+ = 2nNa2CO3 = 2.0,03 = 0,06 (mol)

Ta c nNaOH : nX = 1,2 hn hp c 1 este ca hp cht phenol.

- Bo ton C = 0,15 mol = 0,15/0,05 = 3 C este c s C < 3. Hn hp c HCOOCH3 (a mol) v CxHyO2 (b mol) vi CxHyO2 l este ca hp cht phenol nn tc dng vi NaOH theo t l mol 1:2.

Ta c:

Bo ton C 0,04.2 + 0,01.x = 0,15 x = 7 este cn li l HCOOC6H5HCOOCH3 + NaOH HCOONa + CH3OH 0,04 mol 0,04 molHCOOC6H5 + 2NaOH HCOONa + C6H5ONa 0,01 mol0,01 mol 0,01 molnH/Z = 0,04 + 0,01 + 0,01.5 = 0,1 (mol) nH2O sinh ra khi t chy = 0,05 mol. m = 0,9 (g).Bi ton hay v este Chng 1 Lp 12.1. VST_ln 11: t chy hon ton 34,0 gam hn hp X gm 1 ancol n chc v 1 este no, n chc, mch h, thu c 48,4 gam CO2 v 28,8 gam H2O. Mt khc, nu un 34,0 gam hn hp X vi 200 ml dung dch KOH 1,2 M ri c cn dung dch sau phn ng th khi lng cht rn khan thu c lA. 19,92 gam.B. 23,76 gam.C. 16,32 gam.D. 20,16 gam.p n D- Gi CTPT este no, n chc, mch h l CnH2nO2. nCO2 = 1,1 mol; nH2O = 1,6 mol.Khi t CnH2nO2, sinh ra nCO2 = nH2O t ancol sinh ra nCO2< nH2O Ancol n chc l no, mch h v nancol = 1,6 1,1 = 0,5 (mol).Bo ton khi lng mO2 = 48,4 + 28,8 34 = 43,2 (g) nO2 = 1,35 (mol) nO/X = 2.1,1 + 1,6 2.1,35 = 1,1 (mol) neste = (1,1 0,5)/2 = 0,3 (mol)

Ancol l CH3OH Meste = (34 32.0,5)/0,3 = 60 (g/mol) Este: HCOOCH3HCOOCH3 + KOH HCOOK + CH3OHnHCOOCH3 = 0,3 mol > nKOH = 0,24 mol nHCOOK = 0,24 mol.Vy khi lng cht rn khan = 84.0,24 = 20,16 (g)Bi ton hay v este Chng 1 Lp 12 v ancol Chng 8 Lp 10.