hóa nâng cao 12
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Hóa 12TRANSCRIPT
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Cu 1: Trung ha a mol axit hu c A cn 2a mol NaOH. t chy a mol A c 2a mol CO2. A l
A. CH3COOH B. HOOCCOOH.C. Axit n chc no. D. Axit n chc khng no.
Hng Dn
Gi CT axit l R(COO)k
T/d vi NaOH R(COO)k + k NaOH ( R(COONa)k + k H2O
a ( ka
ka = 2a k = 2
P chy a mol A c 2a mol CO2 (A c 2 cacbon v 2 nhn chc (B
Cu 2: Hn hp X gm hai axit cacboxylic no, mch h Y v Z (phn t khi ca Y nh hn ca Z). t chy hon ton a mol X, sau P thu c a mol H2O. Mt khc, nu a mol X T/d vi lng d dd NaHCO3, th thu c 1,6a mol CO2. Thnh phn % theo khi lng ca Y trong X l
A. 46,67%B. 40,00%
C. 25,41%
D. 74,59%
Hng dn:
a mol X ( a mol H2O ( ch s H = 2 vi a mol X ( 1,6 mol CO2 ( c 1 axit 2 chc ( x + y = 1 v x + 2y = 1,6 ( x = 0,4 v y = 0,6 %Y = = 25,41%
Cu 3: t chy hon ton 0,1 mol cht X l mui Na ca mt axit hu c thu c 0,15 mol CO2, hi H2O v Na2CO3. CTCT ca X l
A. C3H7COONa. B. CH3COONa. C. CH3CH2COONa D. HCOONa.
Hng dn:
Nhn v p n th y l mui ca axit hu c no n chc
2CnH2n+1 COONa + O2 ( (2n + 1)CO2 + (2n + 1) H2O + Na2CO3
0,1 0,05.(2n+1) mol
Vy 0,05(2n+1) = 0,15 n = 1 B
Cu 4: Hn hp X gm 0,01 mol HCOONa v a mol mui natri ca hai axit no n chc mch h l ng ng lin tip. t chy hn hp X v cho sn phm chy ln lt qua bnh 1 ng H2SO4 c bnh 2 ng KOH thy khi lng bnh 2 tng nhiu hn bnh mt l 3,51 gam. Phn cht rn Y cn li sau khi t l Na2CO3 cn nng 2,65 gam. CTPT ca hai mui natri l
A. C2H5COONav C3H7COONa.
B. C3H7COONav C4H9COONa.
C. CH3COONa v C2H5COONa.
D. CH3COONa v C3H7COONa.
Hng dn:
2H COONa + O2 CO2 + H2O + Na2CO3
0,01 0,005 0,005 0,005 mol
2CnH2n+1 COONa + O2 (2n + 1)CO2 + (2n + 1) H2O + Na2CO3
a 0,5.a (2n+1) 0,5.a (2n+1) 0,5.a mol
Cu 5: B sung dy chuyn ho sau:
C2H5OH A B C HCHO.
A. (A). CH3CHO ; (B) CH3COONa, (C) C2H4 B. (A) CH3CHO ; (B) CH3COONa , (C), CH4 C. (A). CH3CHO ; (B): HCOONa , (C) CH4 D. (A). CH3COOH; (B).C2H3COONa; (C)CH4Cu 6: Chia 23,6 gam hn hp gm HCHO v cht X l ng ng ca HCHO thnh 2 phn bng nhau. Phn 1, cho T/d vi H2 d (t0C, xc tc), sau P thu c 12,4 gam hn hp ancol. Phn 2, cho T/d vi dd AgNO3/NH3 d thu c 108 gam bc. CTPT ca X l
A. C2H3CHOB. (CHO)2C. CH3CHOD. C2H5CHO
Hng Dn
Cch 1: Loi A v B v X l ng ng ca HCHO, ti y c th th nghim chn c D.
Cch 2: HCHO ( x mol); RCHO ( y mol), c x + y = (12,4 23,6 : 2) : 2 = 0,3 mol
V 4x + 2y = 1 mol; 30x + My = 11,8; gii h c M =58 vC => RCHO l C2H5CHO
Cu 7: Cho cc pht biu:
(1) Ru bc 1 oxi ha khng hon ton ra Anehit (2) Tt c cc axit cacboxylic u khng tham gia P trng bc
(3) P thy phn este trong mi trng axit l P thun nghch(3) Tt c cc ancol no, a chc u ha tan c Cu(OH)2Pht biu ng l
A. (2) v (4)B. (3) v (4)C. (1) v (3) D. (1) v (2)
Cu 8: Chia hn hp X gm hai anehit no, n chc, mch h thnh hai phn bng nhau :- Phn 1 : em t chy hon ton thu c 5,4 gam H2O.
- Phn 2 : Cho T/d ht vi H2 d (Ni, to) thu c hn hp Y. t chy hon ton hn hp Y, thu c V lt CO2 (ktc). Gi tr ca V l
A. 3,36 ltB. 4,48 ltC. 1,12 ltD. 6,72 ltHng Dn
- Theo LBT nguyn t Cacbon
-
= 0,3.22,4 = 6,72 lt
Cu 9: Cho s chuyn ho sau: HCOONa A C2H5OH B D (COOH)2Cc cht A, B, D c th l
A. H2 ; C4H6 ; C2H4(OH)2
B. H2 ; C2H4 ; C2H4(OH)2C. CH4; C2H2; (CHO)2.
D. C2H6 ; C2H4(OH)2.
Hng Dn
Cu 10: Cho 19,8 gam mt anehit n chc A P hoan toan vi dd AgNO3/NH3 (d). Lng Ag sinh ra P ht vi dd HNO3 loang c 6,72 lit NO ktc. A co CTPT la
A. C2H4O.
B. C3H6O.
C. C3H4O.
D. C4H8O.
Hng Dn
Goi CT l RCHO:
RCHO + Ag2O RCOOH + 2Ag
T/d HNO3
3Ag + 4HNO3 3 AgNO3 + NO + 2H2O
Ta c
Cu 11: un nng 66,4 gam hn hp 3 ancol n chc vi H2SO4 c thu c 55,6 gam hn hp 6 ete vi s mol bng nhau. S mol ca mi ete l
A. 0,1 molB. 0,2 molC. 0,5 mol.D. 0,3 mol
Hng Dn p dung LBTKL:
Ta c
Cu 12: A c CTPT l C8H8, T/d vi dd KMnO4 nhit thng to ra ancol 2 chc. 1 mol A tc dng ti a vi:
A. 4 mol H2; 1 mol brom.
B. 3 mol H2; 1 mol brom.
C. 3 mol H2; 3 mol brom.
D. 4 mol H2; 4 mol brom.
Cu 13: Cho cc pht biu sau:
(a) Anehit va c tnh oxi ha, va c tnh kh.
(b) Phenol tham gia P th brom kh hn benzen.
(c) Anehit T/d vi H2 (d) c xc tc Ni, un nng, thu c ancol bc I.(d) Dd axit axetic T/d c vi Cu(OH)2.(e) Dd phenol trong nc lm qu tm ha .
(f) Trong cng nghip, axeton c sn xut t cumen.S pht biu ng trong s cc pht biu trn l:
A. 4 .B. 3.C. 5.D. 2.Cu 14: Cho 13,4 gam hn hp X gm hai axit no, n chc, mch h, k tip nhau trong cng dy ng ng T/d vi Na d, thu c 17,8 gam mui. Khi lng ca axit c s nguyn t cacbon t hn c trong X l
A. 3,0 gam.
B. 4,6 gam.
C. 7,4 gam.
D. 6,0 gam.Hng Dn
Gi CTTB ca 2 axit no, n chc, mch h, k tip trong cng dy ng ng
T/d Na
Cu 15: Cho 4 hp cht c CTPT l M : C3H6O ; N : C3H6O2 ; P : C3H4O ; Q : C3H4O2.
Bit : M v P cho P trng gng ; N v Q P c vi dd NaOH ; Q P vi H2 to thnh N ; oxi ha P thu c Q.
a) M v P theo th t l
A. C2H5COOH ; CH2=CHCOOH.
B. C2H5CHO ; CH2=CHCHO.C. CH2=CHCOOH ; C2H5COOH .
D. CH2=CHCHO ; C2H5CHO.
b) N v Q theo th t l
A. C2H5COOH ; CH2 = CHCOOH.
B. CH2=CHCOOH ; C2H5COOH.
C. C2H5CHO ; CH2=CHCHO.
D. CH2=CHCHO ; C2H5CHO.
B- L thuyt 1) CTCT - Ca este n chc : RCOOR- Ca este no n chc : CnH2nO2 2) Danh php Tn Este = Tn gc hirocacbon ca ru + Tn axit ( trong ui oic i thnh at)
3) 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
4) 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 chnEtyl butirat v etyl propionat c mi da
Geranyl axetat c mi hoa hng5) 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
6) 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 + HClc) /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 + HCld) P cng vo hirocacbon khng no ca axit cacboxylic
+ An Ken
CH3COOH + CH=CH CH3COOCH2 CH3
+ Ankin
CH3COOH + CH(CH CH3COOCH=CH2 C- Bi tp
V 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 esteCc 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 ngA. 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 TpDng 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 C3H6O2C. 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 CH3COOCH3Cu 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 C4H8O2Cu 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 molTc 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 DnX 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 + H2O10 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)3C3H5Cu 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=CH2Cu 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 Ete 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)
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)
(C15H31COO)3C3H5 + 3NaOH ( 3C16H33COONa+ C3H5(OH)3 (M=806) (834)
(C17H35COO)3C3H5 + 3NaOH ( 3C17H35COONa+ C3H5(OH)3 (M=890) (918)
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
CH3-CH=O
p ha
CH3-CO-CH3
p ha
H2O
H2O
EMBED Equation.DSMT4
EMBED Equation.DSMT4
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