Bồi dưỡng hóa học 12

^HCVNH
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Nam
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B6I
DI/ONG
HOA HOC
•
.
r. _-__B
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song
va nop
lu^u
chieu
quy I nam 2013.
&i
noi ddu
Nham giup
cac em hoc
sinh
c6
them
W
lieu
de
tiep

can,
nam
vilng
he
thong kien thufc
cof
ban, nang
cao va ren
luyen
ki
nang giai
bai tap
theo
yen cau doi
mdfi. Chung
toi
xin
tran
trgng gidi thieu
den cac em hoc
sinh
va cac ban
dong
nghiep
bo
sach
"BOI Dl/OfNG HOA HOC
12"
Noi
dung

sach
"BOI
DTJCfNG
HOA HOC
12"
difc/c
bien
soan
bam sat
chiicfng
trinh
sach
giao
khoa hien hanh
va
theo
tiing
chvfdng
ijfng
vdi
tiing
chu:dng trong
sach
giao
khoa.
Trong
moi
chifdng difgfc
trinh
bay gom 3

phan chlnh:
A.
KIEN
THL/C
CAN NHOf
B.
BAI
TAP
AP
DUNG
C. BAI TAP
NANG
CAO (CO
lofi
giai)
Cac
bai tap ap
dung
va
nang
cao
difcfc giai
chi
tiet,
Idi
gidi
phu hdp vdi moi do'i
tifdng
hoc
sinh, nham giup

cac em
hoc sinh
lam
quen
va
khac
sau
kien thiJc thong
qua cac bai
toan
va
phufdng phap giai
tifng
bai
toan.
Quy en
sach
la
tii lieu giiip
cac em hoc
sinh
Idp 12
nang
cao nang
l\ic t\i hoc d nha
cung nhu!
ren
luyen
de
tham

gia
vao
cac doi
tuyen
hoc
sinh gioi
cac cap. Va la
tai lieu tham
khao
cho cac
giao
vien tham
gia
giang
day d cac
tru:dng
pho
thong
va
giao
vien tham
gia boi
ditdng
hoc
sinh gioi.
Mac
du da co'
gang
trong
qua

trinh
bien
soan
song
khong
the
tranh
khoi nhGng thieu
sot
ngoai
y
muon.
Tac
giai
xin
chan thanh
cam dn cac y
kien
dong
gop, xay
diing
tijf
phia
ban doc de
Ian tai
ban sau
cuon
sach
cd
chat

liidng
tot
hdn.
Tdc
gid
CHl/dNG I.
ESTE
-
LIPIT
A.
KIEN
THLfC
CAN NH6
I.
ESTE.
1. Cong
thijtc
chung cua mot so este
-
Este
tgo hdi R-COOH vdi R'OH: R-COO-R'
Neu R va R' no thi este la CnHznOz (n >2)
-,
Este
tg o bdi R-COOH vdi R
'(OH),,:
(RCOO)„R'.
-
Este
tgo bdi

R(COOH),„
vdi R'OH: R(COOR')^.
-
Este
tgo bdi
R(COOH)n,
vdi R'iOH),,:
Rn(COO)„„,R'„,
2. Ten ggi
Ten
niia he thong cua este diigc ggi nhii sau: 'i '
Ten
este = Ten goc hidrocacbon cua ancol
+ ten goc axit (doi duoi ic at)
Vi du:
CH3COOC2H5
: etyl axetat
CHz^^CH-COO-CHj : metyl acrylat
CH3-OCO-[CH2]4-COO-CH3 : dimetyl adipat
C17H35COO—CH2
C17H35COO—CH
: glixerol tristearat
C17H35COO—CH2
3.
Tinh
chat vat li.
- Cdc este thiidng it tan trong niidc, nhe hdn niidc, di bay hdi.
- Co mill thdm dgc triing.
4.
Tinh

chat hoa hoc
a) Phan ihig thuy phan *
- Trong dung
dich
axit: •
RCOOR' + HOH
:^i===±RCOOH
+ R'OH
Phan
ling theo chieu tii trdi sang phdi la phdn ling thuy phdn este,
phdn
ling theo chieu tiiphdi sang trdi Id phdn ijtng este hoa. Phdn ving
thuy phdn este trong dung
dich
axit la phdn ling thugn nghich.
- Trong dung
dich
baza: dun ndng este trong dung
dich
natri hidroxit,
phdn
ling tgo
muoi
cua axit cacboxylic vd ancol. Phdn ling nay la
phdn
ling mot chieu vd con diiOc ggi phdn ting xd phdng hoa.
Bdi DI/ONG H(3A HCIC12 5
RCOOR'
+ NaOH > RCOONa + R'OH
CH3COOC2HS

+ NaOH —> CHjCOONa +
CzHsOH
Tuy nhien, mot so triidng hap ngoai muoi cua axit cacboxylic khong
tgo ra ancol ma tgo thanh andehit, xeton, muoi cua phenol, hogc chi
tgo thanh mot sdn phdm duy nhdt.
CH,COOCH=CH2 + NaOH —> CHjCOONa + CH,-CH=0
CHsCOOCiCHshCH^
+ NaOH —> CHjCOONa + (CHjJzC =0
CH3COO—y
+ 2NaOH —> CHsCOONa + ^ ^^ONa
ONa
+ NaOH > C
^OH
b)
Phdn
thig
khvT
R-COO-R'
—> R-CH2OH + R'OH
c)
Phan
ihig
cgng
d goc
hidrocacbon:
Phdn ling cgng vd trtmg hap
khi
goc R chiia lien ket n.
.0
COOCH3

n
H2C=:C
-^-^"-P )
CH,
COOCH3
CH,—C
CH2=C(CH3)COOCH3 + H2 ——>
CHs-CHiCHaJCOOCHs
'
d)
Phdn
iftig
dot
chay:
Dot
chdy
este
no
dofn
chv^c:
CnH2n02 +
(^^^)
O2 > /jCO^ + /iH^O
2
Chu
y: Ngoai cdc phdn ring tren, rieng este cua axit fomic, con c6
phdn iJtng trdng giiang giong nhii andehit.
5.
Dieu
che

a)
Este cua ancol
-
Thiic hien phdn ling este hda:
RCOOH
+ R'OH < "^^"^'^ z> RCOOR' + HOH
-
W muoi vd ddn xudt halogen cua hidrocacbon
RCOOAg
+ R'Cl > RCOOR' + AgCU
BO'I
oadNG
HdA
HOC
12
b)
Este cua phenol
Tif
halogenua axit va phenolat:
RCOCl
+
NaOCoHs
>
RCOOCeH^
+
NaCl
-
Tir anhidrit axit vd phenol:
(CH3CO)20
+

HOCeHs
>
CHsCOOCgHs
+ CH3COOH
11.
LIPIT.
1.
Khdi
niem
va cd'u tgo
-
Chat beo (nguon goc dong vat, thitc vat] la este cua glixerol vdi axit
beo (axit hHu ca mot idn axit mgch thdng, khdi liigng phdn tit Idn).
Cdc chat beo nay diigc goi chung Id triglixerit.
Cong thdc tong qudt cua chat beo: CH2—OCOR1
CH—OCOR2
CH2—OCOR3
Trong
do: Ru R2, Rj cd the gio'ng nhau hogc khdc nhau.
+) Mot so' axit beo thitdng gap:
Axit
panmitic :
C15H31COOH
Axit stearic :
C^HssCOOH
Axit
oleic : CiyH33COOH Axit linoleic :
CMiCOOH
+] Thitdng gap cdc triglixerit pha tgp (R^ ^ R^ ^R^)
-

Trong chat beo, ngoai este cua glixerol vdi axit beo con cd mot
liigng nhd axit d
dgng
tiJ do diigc dgc tritng hdi chi so axit.
-
Chi so
axit
cua mot chat beo la so miligam KOH can thiet de trung
hda axit tit do cd trong 1 gam chat bdo.
Vi
du: Mot chat beo cd chi so axit
bd-ng
9. Nghia la de trung hda
1 gam chat beo can 9 mg KOH.
-
Chi so
este
cua mot chat beo la so miligam KOH can thiet de thuy
phdn hoan toan litgng este cd trong 1 gam chat beo.
-
Chi so xd
phong
hda cua mot chat beo Id so' miligam KOH can thiet
de trung hda axit tii do vd thuy phdn hoan todn litgng este cd trong
1 gam chat beo.
2.
Tinh
chat
hda hoc
a)

Phdn
vtng
thuy
phdn
-
Trong mdi trifdng nitdc hogc axit:
Chat beo khong bi thuy phdn bdi nitdc, khi cd xuc tdc axit hogc enzim
chat beo bi thuy phdn thugn nghich cho glixerol vd axit beo:
CH2—OCOR1
CH2-OH
R1COOH
CH—OCOR2
+ 3H2O < ^
CH-OH
+ R2COOH
CH2—OCOR3
CH2-OH
R3COOH
B(5i
DI/SNG
HOA
HQC
12 7
-
Trong
moi triidng
kiem
(phan
ling xa
phdng

hoa) pMn
itng
thuy
phan
xay ra
hoan
toan:
CH2—OCOR1
CH2-OH
RiCOONa
CH—OCOR2
+
3H2O
<===^==±
CH-OH
+
R2C00Na
CH2—OCOR3
CH2-OH
RgCOONa
triglixerit
ghxerol
xd
phdng
b)
Phan
vtng
cong
hidro:
Bien

glixerit
chifa
no
thanh
glixerit
no
(CM^COOJ^CsH,
+ 3H2 '° > (C.yHssCOOhCsHs
chat
long
chat
ran
[II.
XA PHONG vA CHAT GIAT RLTA TONG HOfP
1. Xa
phdng
la hdn hap cac
muoi
Na
hoac
muoi
cua
axit
beo va mot
so
chat
phu gia.
2. De san
xuat
xa

phdng
ngifdri
ta c6 the:
- Dun
nong
chat
beo vdi
dung
dich
kiem.
- Oxi hoa
ankan
(parafin)
cua ddu mo nhd oxi
khong
khi, d
nhiet
do
cao
CO
muoi
mangan
xiic tdc roi trung hoa
axit
bhng
NaOH
hoQc
KOH.
3.
Chat

giqt
rvta
tong
hap la
nhffng
chat
khong
phdi
Id
muoi
Na
hoac
K cua
axit
beo
nhiing
c6
tinh
nang
gigt
riia
nhiixd
phdng.
4.
Chat
gidt
rtia
tong
hap
difcfc

sdn
xuat
ti( cdc sdn
phdm
cua dau mo.
5. Xd
phdng
hi
gidm
tdc
dung
gigt
rita
trong
niidc
cvtng
con
chat
gidt
tdy
riia
tong
hap
khong
bi
gidm
tdc
dung
gigt
rifa.

6. Xd
phdng
cd liu
diem
Id
khong
1dm hgi da, it gay 6
nhiSm
moi
triidng.
Chat
gigt
riia
tong
hap gay hgi da do
trong
do cd
chat
tdy
trdng,
gay 6
nhiem
moi triidng.
BAI TAP AP
DUNG
lai
1. Ho^n th^nh cdc phiicfng
trinh
phan ufng
theo

scf do sau:
CH4
)A —^B
—i51_>C
—> E + B
lai
2. Viet phifcfng
trinh
phan ufng thifc hien day bien hda sau (viet cdc
chat
d\i6i
dang cong thufc cau tao):
C5H10O
^ CsHioBr^O ^
CsHsBra
^ C5H12O3
^WCnHisOe
CsHsOsNa
CH4
Biet
C5H10O la mot ancol bac ba.
ai 3. Viet cong thufc cau tao cac dong phan dcfn chufc, mach hor c6 the c6
cua
C4H6O2.
ai 4. Viet cong thufc cau tao cdc
chat
c6 ten sau day:
a) Isopropyl
axetat
b) Alylmetacrylat

c) Phenyl
axetat
d)
sec-Butyl
fomiat
BO'I
DI/ONG
HdA HQC 12
5- E^ot chdy ho^n toan 7,4 gam
este
X dcfn
chufc
thu
di/ac
6,72 lit khi
CO2
(dktc) va 5,4 gam
nU6c.
a) Xdc
dinh
cong thufc phan tuf cua X.
b) Dun 7,4 gam X trong dung dich NaOH vCfa du den khi phan ufng
ho^n toan thu diicrc 3,2 gam ancol va mot lugng muoi Z. Viet cong thufc
cau tao cua X vd
tinh
khoi lufcfng cua Z.
Bai 6. Dot
chay
hoan toan 1,76 gam mot
este

X thu diTOc 3,52 gam
CO2
va
1,44 gam
H2O.
Xdc
dinh
cong thufc phan
tijf
cua X?
Bai 7. E la
este
ciia mot axit don
chufc
va ancol don chufc. De thuy phan
hoan toan 6,6 gam
chat
E phai dung 34,1 ml dung dich NaOH 10%
(d
= 1,1 g/ml). Lirgng NaOH nay dung
di/
25% so
vdri
lufong NaOH phan
iJng.
Hay de xuat cong thufc cau tao dung cua E?
Bai 8. De xa phong hoa 17,4 gam mot
este
no don
chufc

c^n dung
300ml
dung
dich NaOH 0,5M. Tim cong thufc phan tuf cua
este
dem dung.
Bai 9. 0,05 mol
este
E phan ufng vi/a du 100 gam dung dich NaOH 6%, ta
thu
difcfc 10,2 gam muoi va 4,6 gam
rxiau.
Biet
rufou
hoac
axit tao
thanh E don chufc. Hay xac
dinh
cong thufc cau tao cua E.
Bai 10. Khi thiic hien phan ufng
este
h6a 1 mol
CH3COOH
va 1 mol
2
C2H5OH, Itfong
este
thu dufoc
lorn
nhat la - mol. De dat hieu

suat
CLTC
tj
dai
la 90% (tinh
theo
axit) khi tien hdnh
este
hoa 1 mol
CH3-COOH
thi
can so mol
C2H5OH
la bao
nhieu?
(biet cac phan ufng
este
hoa thiTc
hien d cung nhiet do).
Bai 11. Tron 1 mol axit
axetic
vdi 1 mol rUgu etylic. Khi s6' mol cac
chat
trong
hon hop khong thay ddi nufa, nhan thay lUOng
este
thu dUOc \k
- mol. Tinh hkng so can bkng (K).
3
Bai 12. Thuy phan hoan todn 444 gam m6t

lipit
thu difoc 46 gam
glixerol
(glixerin) va hai loai axit beo. Xdc
dinh
cong thufc cua hai
loai
axit beo do.
Bai 13. Hay viet phi^ong
trinh
phan ufng cua
chat
beo c6 cong thufc phan
tuf
nhtf sau:
CH2-OCO-(CH2X4CH3
CH - OCO - (CH2- CH = CH
(CH2),
CH3
CH2
- OCO -
(CH2
\H = CH - CH2 - CH - CH
(CH2
)^
CH3
a)
Vdri
dung dich KOH d nhiet do cao.
b)

Vdri
I2
CO
dU.
c)
Vdri
H2
du, c6 Ni xiic tdc, d nhiet do vd dp
suat
cao.
Bfii
DUONG
H6A HOC 12 . ^
tiii 14. Khi
thiiy
ph4n a gam m6t este X thu
dUcfc
0,92 gam
glixerol,
3,02 gam^
muoi
linoleat
CiTHsiCOONa vk m gam
natri
oleat
CnHaaCOONa.
Tinh
gia tri cua a, m.
Viet
cong

thijfc
cau tao c6 the c6 ciia X.
Bai
15. a) De trung h5a liicfng axit beo tif do c6 trong 14 gam mot mSu
chat
beo can 15ml dung dich KOH 0,1M. Hay cho
biet
chi so' axit cua
mau
chat
beo tren.
b) Tinh kho'i lacfng NaOH can thiet de trung h6a 10 gam
mpt
chfi't
b6o
c6
chi
so'
axit 1^ 5,6.
Bai
16. Hay tinh chi so xk phong h6a cua m6t
chS't
b6o,
biet
rkng
khi
x^ ph6ng hoa
hoan
toan 1,5 gam
ch§.'t

b^o d6 can 50 ml dung
dich
KOH
0,1M.
Bai
17. Hay tinh khoi lucfng NaOH can de trung hoa axit tii do c6 trong
5 gam
chat
beo v6i chi so axit
bang
7.
Bai
18. Hay tinh chi
so'
iot cua triolein.
Hlf6NG
DAN
GIAI
f
Bai
1.
Phan
ufng:
tach
nhanh.^im
knh
nhanh
^ ^aHg + SHg
C3H2
H- H,0 ) CH.

CHO
1
^ Mn^*
CH3CHO
+ ^02 — >
CH3COOH
CH3COOH
+
CH-CH
> CH3COOC=CH2
CH3COOC=CH2 + NaOH > CH3COONa + CH3CHO
Bai
2.
Phan
ufng:
CH3 CH3
(1)
H2C=CH—C-OH
+
Br2
>
H2C-CH—C-OH
CH3 Br Br CH3
^•^3 CH3
(2)
H^C-CH-C-Br
. HBr
H^C-CH-i-Br
.
H2O

Br Br CH3 Br Br CH
I3
CH3
CH3
(3)
H2C-CH—C-Br
+
SNaOH
>
H2C-CH—C-CHgi-
3NaBr
III III
Br Br CH3 OH OH OH
10 BO'I
DUflNG
H6A HOC 12
(4)H2C—CH—C—CH3
+
3CH3COOH
OH OH OH
CH3COO—CH2
->
CH3COO-CH
+
3H2O
CH3COO—C(CH3)2
(5)
CH3COO-CH +
3NaOH
^ H2C—CH—C-CH3 +

3CH3COONa
• OH OH OH
CaO, t°
->CH4
+
NaaCOg
CH3COO-C(CH3)2
(6) CH3COONa + NaOH
Bai
3.
C4H6O2
c6 A = 2 va hai nguyen tuf oxi.
=> dong
phan
este
dcfn
chuTc,
khong no c6 mot noi doi d gdc dong
phan
axit
cacboxylic
dcfn
chufc
khong no mot ndi doi d gdc.
-
DSng
phdn
este:
HC00CH=CH-CH3 ; HCOOCH2-CH=CH2
CH3COOCH=CH2

; CH2=CHCOOCH3
-
Bong
phdn
axit
cacboxylic:
CH2=CH-CH2-COOH
CH3-CH=CH-C00H
CH2
- C - COOH
Bai
4.
a)
CH3COOCH(CH3)2
C)
CH3COOC6H5
I
CH,
b) CH2 = C - COOCH2 - CH = CH2
CH3
d) HCOO-CH(CH3)CH2-CH3
Bai
5. a) Ta c6: n^o, = = 0,3 (mol) va n^^^ = M = o,3 (mol).
' 22,4 ' 18
Vi
khi ddt
chay
X thu dirge sd mol
H2O
bang

sd mol
CO2
nen X Ik
este
no, dcfn
chufc.
Goi
cong
thiJfc
cua
este
no, dcfn
chufc
\k :
CnH2n02
(n > 2)
(3n-2]
Phan
lifng:
(mol) I
CnH2n02 +
0,3
0,3
O2
nCOz + nHaO (D
<- 0,3
n
Theo
dl bai, ta c6: Mx = — (14n + 32) = 7,4 n = 3.
n

Vay
cong
thufc phan tijf cua X:
C3H6O2.
B6|
DirdNG
HOA
H0C12
11
b)
X&c
dinh
cong
thijfc
cau
tao
cua
X
kho'i
luang ciia
Z:
Ta c6:
nx = — =
0,1 (mol).
74
Phan ufng:
RCOOR'
+
NaOH
>

RCOONa
+ R'OH.
(mol)
0,1 -> 0,1 0,1 0,1
M^:
IHR-OH
=
0,1(R'
+ 17) = 3,2 R' =
15:
CHg-
Vay cong thufc cau
tao
dung cua
X: CH3COOCH3.
Va
khoi
lugng cua
Z:
0,1
x 82 = 8,2
(gam).
Bai
6. Ta
c6:
nco^ = =
0.08 (mol); nn^o
= =
0.08
(mol)

Do n^Og
=
i^HgO
X c6 do bat bao h6a cua
phan tuf
A = 1
=>
X
1^
este
no, dofn chufc
=> X
dang
CnH2n02
C„H2n02
)
nC02
nco2
0,08 1,76 .
^nx=
= -— => Mx =
14n
+ 32 = ^ n = 4
n
n 0,08
n
Vay cong thufc phan tuf cua
X:
C4H8O2.
m

.
34,1
X 1,1 X 10 0 ,
Bai
7. Ta c6:
m^^oHdemdtag
= ^QQ = 3.751 (gam)
3,751
X
100
^ ,
^mOU
phan .ng
= (lOQ + 25) ^ ^
^^^""^
3
Mat khdc:
IIE =
nNaOH
= — =
0,075
(mol)
40
=>
ME = 88 (gam) o R +
44
+ R' =
88
=o R + R' = 44
-IQii

R =
1=>R'
=
43(C3H,)
=>
CTCT
(E):
HCOOC3H7 (propyl fomiat)
- Khi
R = 15 ^ R' = 29 =^
CTCT
(E):
CH3COOC2H5 (etyl axetat)
Bai
8.
Gpi cong thufc cua
este
no, don chufc 1^ C^HanOg.
Khi
phbng h6a thi: n^^^^
=
^naou
ne.te
= 0,3
X
0,5 = 0,15 (mol) => M^^,^ =^ = ^16
14n
+ 32 =
116=>n
= 6

vay cong thufc phan tuf cua
este
1^ C^H^Pz
•
2 BO'I Di/dNG HdA Hnr
Bai
9. Ta
c6: nNaOH
= ^'^^^ =
0,15 (mol)
=
Sneste
=>
este
3
chufc.
100.40
Theo
de
bai,
se c6 1
trong
2
chat
la
dcfn chufc nen
c6 2
truTcfng hop:
Triidng
h^p 1:

axit don chufc
rLfcfu
ba
chufc
(RCOOsR'
+
3NaOH
>
3RC00Na
+
R'(OH)3
(mol) 0,05
-> 0,15 0,15 0,05
10
2
Ta c6:
MRcooNa
=
^r-^
=
68
(dvC) R +
67
=
68
=> R = 1 (la H-)
MR(OH)
= =^2 (dvC) ^ R' + 51 = 92 R' = 41 ^ R' 1^
C3H5-
y

/x 0,05
Vay cong thufc cau
tao la
(HCOOgCgHs.
Trildng
hcfp
2:
axit
da
chufc
va
rifgfu dcfn chufc.
(RCOO
R')3
+
3NaOH
>
R(C00Na)3
+
3R'0H
(mol)
0,05^ 0,15 . 0,05 0,15
4
6
^
= n »c o =
30,67:
khong nguyen
(loai)
U,

Uo.o
Vay cong thufc cau
tao
ciia
este
(E)
la:
(HCOO)3C3H5
Bai
10. Can bang:
CH3COOH +
C,H,OH
^^"^
CH3C00C,Hs
+ H^O
Ban
ddu: IM IM 0.0
2
2 2 2
Phan
ling:
-M -M -M -M
3
3 3 o
Can
6^71^
(1 - 0,9)M (a - 0,9)M 0,9M 0,9M
/o^2
Cho
V = 1

lit
K =
v3,
=
4
I
RCoi
a la so
mol CgH^OH can dCing,
ta
c6:
Kc
=
j^^^^-4=>a
=
2,925 (mol).
ai
11.
Goi the
tich
he
phan ufng
V Git)
CH3COOH
+
C2H5OH
^
CH3COOC2H5
+
H2O

Ban
ddu: 1 1
Phan
ling:
x x
Sail
phan iCng:
1-x 1-x x x
I
DUdNG HOA HOC
12
13
2
Do Heste = — =>
X
=
1
3 3 3
K
=
"CH3COOC2H5"
H2O"
K
=
CH3COOH
C2H5OH
73
V
V
V

V
J
V
=
4
Bai
12. Ta c6:
n^,^^^„,
46
92
=
0,5 (mol)
3
Goi
cong
thijfc cua
lipit
c6 dang: C3H5(OCOR)3 v6i R =
Phan
ijfng:
^C3H5(OH)3+3RC00H
- 0,5
C3H5(OCOR)3+3H2O:
(mol) 0,5
Theo
de b^i, ta c6:
R =
238,333
=
(*)

"ihpit
=0'5(41+
132 +3R) = 444
o
Ma:
„ =239; „
=237
va M_ „ =213; „ =211
Ket
hop vdi (*) Cap nghiem thich hop:
C17H35-
va CnHag
Bai
13.
CH2-OCO-(CH2)^,CH3
CH
- OCO(CHa)^ - CH =
CH(CH2)7
CH3
CH2
- OCO -
(CH2),
CH = CH -
CH2
- CH = CH
(CH2
\3
Chat beo
tren
Ik trieste cua glixerol vdi ba axit panmitic, axit

oleic
vk
axit
linoleic.
a)
Khi cho tac dung vdi dung dich KOH d nhiet do cao se xay ra phan
LJfng
xa phong hoa
triglixerit
va ba muoi
kali
cua ba axit beo
tren.
b)
Khi cho tac dung vdi h c6 dtf thi c6 phan ufng
cong
I2 vao cac noi
doi trong goc axit beo khong no.
c)
Khi cho tac dung
vdri
H2 diT, xiic tac Ni, t°, p thi c6 phan iJng
cong
vao noi doi trong g6c axit beo khong no.
Bai
14. So d6: X + NaOH ^ CnHsiCOONa + CnHssCOONa +
C3H5(OH)3
3,02
Ta c6: n
•Ci,H3,COONa

302
=
0,01 (mol)
14
BO'I
OUONG
H6A HOC 12
=>
ne„„33cooNa
= 0,02 (mol) =>
m,_^„^^,„„,^
= 0,02 x 304 = 6,08 (gam)
0 92
M^:
nNaOH
= 3ngiixeroi = 3 X = 0,03 (mol).
=>
mNaOH
= 0,03 X 40 = 1,2 (gam).
Ap dung dinh luat bao to^n khoi luong, ta c6:
=
3,02 + 6,08 + 0,92 - 1,2 = 8,82 (gam).
Bai
15. a) Theo dinh nghia: chi so axit cua chat beo la so miligam KOH
can dung de
trung
hoa het cac axit b^o
tii
do c6 trong 1 gam chat beo.
Ta c6: mKon = 0.015 x 0,1 x

56000
= 84 (mg)
84
^ Chi so axit la: — = 6.
14
b)
Chi so axit la 5,6 nghia la de
trung
hoa 1 gam chat beo can
5,6mg
KOH.
Vay
trung
h5a 10 gam chat beo cdn
5,6mg
x 10 = 56mg KOH.
Hay = 0,001 mol KOH
•^56
Vi
NaOH la
bazo
don chiifc
nhtf
KOH nen can so mol bang nhau trong
phan ufng
trung
hoa.
Do vay so gam NaOH can c6 la: 40 x 0,001 = 0,04 (gam) NaOH
Bai
16. So mol KOH:

0,050
lit x 0,1
mol/lit
=
0,005
mol KOH
So gam KOH:
0,005
mol x 56 g/mol =
0,280g
hay 280mg KOH
280
Chi
so xa phong hoa: =
186,67.
1,5
Bai
17.
Theo
dinh nghia, chi so axit cua chat beo bkng 7 c6 nghia la muo'n
trung
hoa lucfng axit beo tii do trong 1 gam chat beo phai dung 7mg
KOH.
Vay muon
trung
hoa axit beo i\i do trong 5 gam chat beo c6 chi
so 7 thi phai dung 5x7 = 35mg KOH, hay ^^mol KOH
^
M^mol
OH" => ^^^moi NaOH ^ khol lifong NaOH c^n dl

56 56
trung
hoa axit i\i do trong 5 gam chat beo c6 chi so axit bkng 7 la:
niNaOH
=
>^
40g = 25 (mg) =
0,025g/5g
chat beo.
56
Bai
18. Phan ufng:
(C,,H33COO)3C3H5
+ 3I2
(C,,H33COOl2)3C3H,
(gam) 884 3 x 254
^ Chi so' iot la
i^^-^
x 100 = 86,2. .
884
Bfii
DUONG
HOA HQC 12 IS
C.
BAI TAP NANG CAO
Bai
1. De
thuy phan hoan toan
0,74 gam mot
h6n

hgp
este
dcfn
chufc
car,
7
gam
dung dich KOH
8%
trong nude. Khi dun nong h6n
hgp
este
noj
tren
vdi axit
H2SO4
80%
sinh
ra
khi
X.
Lam lanh
X,
diia
ve
dieu kie^
thudng
va dem can, sau do cho
khi
Igi tii tii qua

dung dich brom
dy
trong
nifdrc thi thay khoi lifgng khi giam
1/3,
trong
do
khoi lifgng rieng
cua khi gan nhu khong doi.
|
a)
Tinh
khoi liTgng mol
cua
hon hgp
este,
xac
dinh thanh phan h6n hgp
khi
sau khi
da
lam lanh
va
tinh
khoi lugng cua chung.
b)
Xac
dinh thanh phan hon hgp
este
ban dau.

c)
Neu phan ufng
de
phan biet
2
este
tren,
viet phtfong
trinh
phan ufng.
Bai
2.
Lam
bay hoi mot
chat hufu
co A
(chufa
cac
nguyen
to C, H, O)
diJgc
chat
hoi
CO
ti
kho'i
doi
vdi metan bang 13,5.
Lay 10,8 gam
chat

A va
19,2
gam O2 cho vao
binh kin, dung
tich
25,6
lit (khong doi).
Dot
chay
hoan toan
A, sau do
giuf nhiet
do
binh
163,8°C
thi
ap
suat trong binh
bang
1,26
atm. Lay toan
bg san
pham
chay
cho vao 160
gam dung dich
NaOH
15%
dirge dung dich
B c6

chijfa 41,1 gam
hon hgp
hai muoi. Khi
ra
khoi dung dich
c6
the
tich
Vi
lit (dktc).
a)
Xac
dinh
cong
thCfc phan tii, viet
cong
thufc
cau tao
cija
A
(biet
rkng
khi
cho A tac
dung vdi kiem
tao ra
mot
ancol
va 3
muoi).

b)
Tinh
Vi va C%
ciia
cac
chat trong dung dich
B.
c)
Cho 10,8
gam
A tac
dung vCra
du
vdi
V2
lit dung dich NaOH
3M
thu
dufgc
a
gam hSn hgp muoi.
Tinh
V2 va a.
Bai
3.
Thiiy
phan hoan toan
este
A cua mot
axit hufu

co don
chufc
va mot
ancol
don
chufc
bang lufgng dung dich NaOH vifa du. Lam
bay hoi
hoan
toan dung dich
sau
thiiy
phan. Phan
hoi do
dugc
dan qua
binh
1
dgng
CUSO4
khan di/,
hoi con lai
dugc
ngUng
tu het vao
binh
2
dgng
Na da
thay

CO khi
G bay
ra. DSn khi
G
qua binh dgng CuO dii, nung nong thi
6,4 gam
Cu
di/gc
giai phong. Lugng
este
ban dau
tac
dung vdi dung dich
brom dif thi
co 32 gam
brom tham
gia
phan ufng. Brom chiem
65,04%
khoi
lurgng phan tuf san pham sau khi
cong
hgp
vao A.
Hay:
a)
Xac
dinh
cong
thufc phan

tijf
va
cong
thufc cau
tao
ciia
A.
b)
Hoan thanh
so do
phan ufng:
^ trung htfp
^ g
+ NaOH
^
Q
_I_
Bai
4.
Dun 20,4
garri
mot hgp chat hufu
co A
don
chufc
vdri
300 ml
dung dich
NaOH IM thu
durgc

muoi
B va hgp
chat hOu
co C. C tac
dung
vdri
Na
dU
cho
2,24
lit H2 (dktg). Biet rang khi nung muo'i
B vdi
NaOH thu dirge
khi
K
CO ti khoi do'i vdi oxi bang 0,5.
C la
mot
hgp
chat
don
chufc
khi
hi
oxi
hoa
bang khong khi
tren
Cu
nong

tao ra san
pham
D
khSng phan
ufng
vdi dung dich AgNOa trong NH3.
16
- - • ' :
•
Bdl
DI/SNG
HOA
HOC12
a) xac dinh
cong
thufc cau
tao
cua
B, C, A
va
D.
b)
Sau phan ufng giffa
A va
dung dich NaOH thu dirge
F. C6 can F
dtfgc
1
cha't r^n.
Tinh

khoi lugng cha't r^n nay.
c)
Them
vao
10,2,gam
A
mot cha't
G don
chufc
cung
chufc
hoa hoc
vdi
A
vdi
so
mol no
=
0,5nA. Do't
chay
h5n
hgp A, G
thu dirge
33
gam
CO2 va
12,6
gam
H2O.
Xac

dinh
cong
thufc
phan ttf
va
cong
thufc
cau tao
bi§'t
rkng
khi dun
G
vdi dung dich NaOH
ta
thu
difgc
muoi
B
tren
va m6t
san
pham
co
phan
ufng
vdi AgNOa/NHs.
gai
5- Cho vao
binh kin
co

dung
tich
500 ml 2,64
gam
mot
este
A
roi
dem
nung binh
den
273°C,
toan
bg
este
hoa
hoi thi ap sua't b^ng
1,792
atfti.
a)
Xac
dinh
cong
thufc
phan tuf
cua A.
Tinh
nong
do mol
ciia dung dich

NaOH
can
thiet
de
thuy phan het
lirgng
este
noi
tren
biet rang the
tich
dung dich NaOH
la 50
ml.
b)
xac dinh
cong
thufc
cau tao cua A
va
tinh
khoi lifgng muoi thu dirge
sau phan
ufng
(vdi hieu suat 100%) trong
cac
triTomg
hgp sau:
- San pham thu dirge sau phan ufng la h8n hgp
2

muoi va
1
rirgu.
- San pham thu dirge
la 1
muoi va
2
rifgu
la
dong
d4ng lien tiep.
Bai
6. Mot hon hgp gom
hai
este
don
chufc,
co 3
nguyen
to C, H, O. Lay
0,25
mol
este
nay phan
ufng
vdi
250 ml
dung dich NaOH
2M
dun nong

thi
thu dirge mot andehit
no
mach
hd vk
28,6 gam hai muoi hufu cO.
Cho
biet kho'i lugng muoi
nay
hkng
4,4655
Ian
khoi lugng muoi kia.
De
phan ufng
het
vdri
NaOH
con
dU
can
dung
150 ml
dung dich HCl IM,
phan
tram
kho'i lugng
cua
oxi trong* andehit
la

27,58%.
Xac dinh
cong
thufc
ca'u
tao
eua hai
este.
' \
Bai
7.
Mot h6n
hgp 2
este
don
chufc
dugc
nung nong
vqri
-mot lugng NaOH
viTa
dij
tao ra
h6n hgp
dong
dang lien tiep
va
hon Hgp muo'ji.
a)
Dot

chay
2
rUgu thu
duge
CO2
va hoi H2O
theo
ti
le
the
tich
7 : 10.
Tim
cong
thUe phan tuf va thanh phan phan
tram
theo
s6'
mol
c^e
rUgu
trong
hon hgp.
b)
Cho
2
muoi
tac
duijg vdi lugng
H2SO4

viTa
dii
dugc
hon hop
2
axit
no.
La'y
2,08 gam h5n hgp 2
axit
(nguyen cha't)
cho vao
100ml dung dich
Na2C03
IM.
Sau
phan ufng lugng
Na2C03
du tac
djjng
viraMij
vdi''85
ml
dung dich HCl 2M.
Xac dinh
cong
thufc
phan tuf
eua 2
axit

Va
2
este
big't r^ng khi
dot
m6i
este
deu
thu
dugc
mot the
tich
khi
CO2 nho hon 6
Ian the tich hai
este
do
d
cung dieu kien
t° va P.
Bai
8. A va B la 2
chat hUu
co don
chufc
eo
ciing
cong
thUc phan tuf. Khi
dot

chay
hoan toan 10,2 gam h6n hgp
2
chat nay^ean 14,56 lit
O2
(dktc), khi
CO2 va hoi
nude
jbae-^featth-reo-the- tichaxhu nliavrtdo
«f
cung dieu kien).
BfiiDi/aNGH0AHgci2
THI/VIEN
TI41NH THUAN
,7
Mat
khdc
khi cho A, B tac
dung
vdi
dung
dich
NaOH
ngUffi ta tha'y:
- A tao dtrgc
muo'i
cua axit
hOfu
ca C va
riigu

D. Ti kho'i hoi cua C so vdi
H2 Ik 30. Cho hoi
rtrou
D di qua Qng
dUng
Cu dun nong dtfcfc chat E
khong tham gia phan
uTng
trang jgiiong.
- B tao ra dirge chat C va D'. Khi cho C tac dung vdi H2SO4
di^oc
E'
tham
gia phan
ijfng
trang gifong con khi D' tac dung vdi H2SO4 dac d t°
thich
hdp thi thu diidc 2 anken.
Xac dinh
cong
thufc
cau tao cua 2 chat A, B.
ai
9. Mot hdp chat
hiJu
cd X mach hd chi
chufa
1 loai nhom
chufc
difdc di4u

che tii axit no A va
rUdu
no B. Biet:
a)
Khi dot
chay
a gam X thu difdc n^^ = n^j ^ + 0,2 mol, a gam X khi
hda hdi chiem mot the
tich
bang the
tich
cua 5,8 gam khong khi d cung
dieu kien (lay
MKK
= 29).
b) Khi dot
chay
1 mol
riidu
no B can 2,5 mol O2.
c) a gam X tac dung vdi dung dich NaOH vCfa du tao ra 32,8 gam muoi
khan.
Hay cho biet
cong
thufc
cau tao cua X.
ai
10. Cho A la
este
cua glixerol vdi axit

cacboxylic
ddn
chufc
mach hd.
Dun
nong 7,9 gam A vdi NaOH cho tdi phan ufng hoan toan thu dtfdc
8,6 gam h6n hdp muo'i.
Cho h6n hdp muoi tac dung vdi H2SO4 dii diidc hon hdp 3 axit X, Y, Z
trong
do X, Y la dong phan ciia nhau, Z la dong dang ke tiep cua Y.
Lay mot phan hon hop axit do dem dot
chay
va cho CO2 thu
difoc
tac
dung vdi dung dich Ba(0H)2 dii cd 2,561 gam ke't tua.
a)
Tim
cong
thi'jfc
phan tuf va viet
cong
thufc cau tao cd the c6 cua A biet
Z
CO
mach
cacbon
khong phan nhanh.
b)
,Tinh

khoi
liidng
hon hdp axit da bi dot
chay.
ai
11. Cho 5,7 gam hdn hop 2
este
ddn
chufc,
mach hd dong phan cua
nhau
tac dung_vdi 50ml dung dich NaOH. Dun nhe, gia suf phan ufng
xay ra hoan toan. De
trung
hoa liJdng NaOH d\i can 50ml dung dich
H2SO4
0,5M ta thu
dddc
dung dich D.
a)
Tinh
tong so' mol 2
este
trong 5,7 gam hdn hdp biet rang de
trung
hoa 10 ml dung dich NaOH can 30 ml dung dich H2SO4 0,25M.
b) ChiJng cat D
diTOc
h6n hop 2
rUdu

c6 so' nguyen tuf
cacbon
trong phan
tuf
bang nhau. Hon hop 2 mau lam mat mau 6,4 gam Br2 trong dung
dich.
Ne'u cho Na tac dyng vdi h6n hdp 2
riidu
thu difdc x lit H2 (dktc).
Co can phan con lai sau khi chitog cat D roi cho tAc dung vdi
H2SO4
thu
dirge hSn hgp 2
axit.
H6n hdp nay lam mat mau dung dich chufa y gam
Br2. Tim
cong
thufc phan tuf cua cac
este.
c)
Tinh
X, y va kho'i lifgng mdi rqgu.
} Bdi
DirflNG
HOA HOC 12
ai
12. A Ik axit
hOfu
ca mach
th&ng,

B
\k
ancol
dcm
chufc
bac 1 c6 nh6nh. Khi
trung
hoa hoan toan A thi so mol NaOH can
trung
hoa gap doi so mol A.
Khi
dd't
chay
B tao ra CO2 vk H2O c6 ti le s6' mol
tqong
ufng Ik 4 : 5.
Khi
cho 0,1 mol A tac dung vdi B hieu suat 73,5 % thu
ddgc
14,847
gam
chat hufu
CO
E.
a)
Viet
cong
thufc cau tao
cija
A, B, E vagpi ten.

b)
Tinh
khoi
liTgng
ciia A, B da
phan
ufng de tao ra
lirgng
chat E nhq
tren.
ai
13. Lay 100ml dung dich chufa 2
este
A, B don
chufc
c6 nong do mol
chung la 0,8M. Cho hon hgp nay tac dung vdi 150ml dung dich NaOH
IM.
Sau
phan
ufng thu dirge 2 san
pham
hflu
cd la 2 muoi C, D c6 kho'i
M
41
lirong
la 10,46 gam (Ti le 2
phan
tuf khoi —- = —) dong

thdi
thu dirge
Mp 65
1
rirgu
E CO khoi
lirgng
la 2,9 gam. Rirgu nay khong ben bien thanh
andehit. Sau
phan
ufng
phai
diing 200ml dung dich HCl 0,2M de
trung
hoa NaOH dir. Xac dinh
cong
thufc
phan
tuf va
cong
thufc cau tao cua A
va B.
Bai
14. Dot
chay
17,6 gam mot hon hgp 3 chat A, B, C dgn
chufc
la dong
phan
va cho san

pham
chay
Ian
lirgt
qua binh 1 dirng P2O5, binh 2 dirng
KOH
dit
thi khoi
lirgng
binh 1 tang 14,4 gam va binh 2 tang 35,2 gam.
a)
Xac dinh
cong
thufc
phan
tuf va
cong
thufc cau tao c6 the cd cua A, B,
C biet r^ng ca 3 chat deu mach hd.
b) Lay 17,6 gam hon hgp A, B, C va chia ra lam 2
phan
bang nhau.
Phdn 1: Bi
trung
hoa bdi 0,5 lit dung dich NaOH 0,1M d nhiet dp
thirdng
{phan ling thuc hien trong thai gian ngdn).
Phan 2: Tac dung
vtra
du vdi 1 lit dung dich NaOH 0,1M {dun nong mot

thai
gian de phan ilng xdy ra hoan toan). Sau khi c6 can dirge chat ran D.
Hgi
E
dugc
lam ngirng tu va sau khi loai het nirdc chufa trong E con lai
mot chat long c6 khoi
lirgng
la 2,58 gam. Xac dinh
cong
thufc cau tao
dung cua A, B, C va thanh
phan phan
tram
hon hgp
theo
khoi
lirgng.
c) Them NaOH dir vao chat rSn D va nung hon hgp nay dirge m6t hon
hgp
khi F.
Tinh
ti khoi cua F doi vdi H2.
Bai
15. Mot hon hgp X gom 2
este
A, B dong
phan
{khong chica chicc hoa hoc
ndo khdc ngodi chdc este). Dot

chay
hoan toan X can 2,8 lit O2 (dktc).
Cho ha'p thu toan bg san
pham
chay
vao trong dung dich Ca(0H)2 diT
tao thanh 10 gam ke't tua va kho'i
lirgng
dung dich giam 3,8 gam.
a)
Xac dinh
cong
thufc
phan
tuf va cac
cong
thufc cau tao cd the cd cua A
va B.
B6|
DI/SNG
HOA HOC 12
b) Goi Y, Z Ik 2 d6ng phdn mach hd c6 cung s6'
nguy§n
tuf C vk O v6i A,
B
nhung it hcfn 2H. Cho h5n horp phan drng v6i NaOH difcfc 2 mudi va
rUcfu.
Cac axit tao ra 2 muo'i nay la 2 dong d^ng ke" tiep va c6 mach
cacbon
khong phan nhanh. Cho 2 axit nay phan uTng vdi 600ml dung

dich Na2C03 IM (axit them tif tU) thi khong c6 CO2 bay ra. Sau phan
ijfng
de tac dung het vdi lugng
cacbonat
phai dung 800ml dung dich HCl
IM.
Dot
chay
het 2 axit thu dirge
32,48
lit CO2 (dktc).
Xdc dinh
cong
thufc phan tuf va
cong
thufc cau tao cua 2
axit.
Tinh
thanh phan phan trSm theo khoi Itfgng cua h6n hcfp Y, Z.
1
16. Mot h6n hop X gom 2 chat hufu ccf A, B khong tac dung vdi dung
dich Br2 va deu tAc dung vdi dung dich NaOH. Ti khoi cua X do'i vdi H2
bang 35,6.
Cho X tac dung hoan toan
vdfi
dung dich NaOH thi thay phai diing
4 gam NaOH, phan ufng cho ta mot rugu don chiifc va 2 muol ctia axit
hufu
cof dcfn chufc. Neu cho toan bo li/cfng
ri/cfu

thu di/gfc tac dung vdi Na
da
CO
672ml khi (dktc) thoat ra.
Xac dinh
cong
thufc phan
tijf
va
cong
thufc cau tao cija A, B.
, 17. Mot h6n hop X gom 3
d6ng
phan A, B, C mach hd deu chuTa C, H,
0. Biet 4 gam h6n hcfp X d
136,5°C
va 2 atm c6 cung the
tich
vdi 3 gam
pentan d
273°C
va 2 atm.
a)
Xac dinh
cong
thufc phan tuf cua A, B, C.
b) Cho 36 gam hon hap tac dung viTa dij vdi dung dich NaOH c6 chufa
m
gam NaOH. Co can dung dich diicfc chat ran Y va h6n hop Z. Z tac
dung

vira
dii vdi dung dich AgNOs/NHa tao ra 108 gam Ag va dung dich
Z' chufa 2 chat hufu ccf. Dien phan dung dich Z' vdi dien chiic tro, c6
mang
ng&n
dugc
h5n hcfp khi F ben anot. Nung chat rdn Y vdi NaOH
diT
dagc
h6n h,ap khi G. Dun G vdi Ni xiic tac dUgfc h6n hop khi F' gom
2 khi
CO
so mol bkng nhau.
Trgn
Ikn F vdi F' roi cho qua ni/dc Br2 dif thi khoi li/cfng dung dich tang
len
1,75 gam. Khi
tron
Ian F, F' cac chat khong c6 phan ufng vdi nhau.
1.
X^c dinh
cong
thufc cau tao cua A, B, C, biet rSng moi chat chi chufa
mot loai nhom chufc.
2.
Tinh
thanh phan phan
tram
cua A, B, C trong hSrt hop X.
3.

Tinh
m (khoi lirgng NaOH).
HlfCfNG
DAN
GIAI
1. Phan dng:
R-COO-R'
+ KOH > R-COOK + R'OH
=> So mol 2
este
= 0,01 (mol) v^ M = 74
Co 2 kha nSng xay ra :
- Ca 2
este
deu c6 phan tuf khoi Ik 74
(HCOO-C2H5
v^
CH3-COO-CH3)
- Mot trong hai
este
c6 phan tuf khoi < 74 do la
HCOO-CH3.
B6|
Dl/dNG
HOA HOC 12
NhiX
vay cA 2
Ijha
nSng diu c6 1
este

fomat, khi dun n6ng vdi
H2SO4
bi
phan huy tao ra CO (M = 28), ngoki ra con mot khi bi hap thu bdi nude •
brom, khi d6 phai Ik anken sinh ra khi phan ancol trong
este
bi tkch
ntfdc. Mat khac, khoi Itfgng rieng h6n hop khi khopg doi, tufc la khi d6
phai
c6 phan tuf khoi = 28, do 1^'
C2H4.
C2H4
+ Bra ——^ C2H4Br2
Neu trong h6n hop c6
H-COO-C2H5
> CO +
C2H4
+ H2O
thi
sau khi di qua nifdc brom kho'i liiong khi phai giam di 1/2
(trdi
gia
thiet).
Vay ckc gdc HCOO- va
C2H5-
phai thuoc ve 2
este
khdc nhau.
H8n
hop chufa

H-COO-CH3
(x mol) v^
R-COO-C2H5
(y mol).
Ta
CO
: x + y = 0,01 ; x = 2y (do CO = 2 x
C2H4
)
y =
0,01/3
va X =
0,02/3
Tac6:60x^
+ (R + 73) x Ml o,74
3 3
=> R = 29^
Cong
thufc phan tuf 1^ :
C2H5-COO-C2H5
Khoi
lugng hdn hop khi sau phan ufng vdi
H2SO4
Ik :
28 X 0,01 = 0,28 (gam)
=>
0,02/3
X 60 = 0,4 (gam)
HCOO-CH3
54,1%

0,34 (gam)
C2H5-COO-C2H5
45,9%
Phan biet 2
este
bang phan ufng vdi dung dich AgN03 trong NH3 :
HCOO-CH3
+
2Ag(Nli3);
+
2H2O
>(NH4)2C03
+ 2NH;
+ CH3OH + 2Ag>i.
Bki
2. a)
Theo
de bai, ta c6: MA = 13,5 x 16 = 216 (dvG)
Taco:
n^ = —^ = 0,05 (mol) ; = = 0,6 (mol)
n^o^ = 0,45 (mol)
=> So mol hon hop sau phan ufng chky \k 0,6 mol.
Khi
cho CO2 v^o dung dich NaOH
CO2 + NaOH > NaHCOg
CO2 + 2NaOH —^
NaaCOs
+ H2O
Do't ch^y A: C^HyO, + (x +
|-|)02

xCOa
+ |H20
(mol) 0,05 ^
0,05x
Ma:
nj,o=
o,05x
= 0,45 X = 9
=>
Cong
thufc cua A
di/gc
viet lai: CgHyO^
=i>
y + 16z = 216 - 108 = 108
^ Cap nghiem
phii
hop: z = 6 vk y = 12 => CTPT ciia A1^
C9H12O6
BAI mfflMR uA« unr 4«
Theo
de
ra
A
tdc
dung
vdi NaOH
cho
ancol
vk 3

muoi.
Yky
ancol
nky c6 ba
nhom
chufc
-OH
va la
glixerol,
ba
axit
khdc
nhau
c6
tong
so'
cacbon
la 6,
vay
CO
mot axit
c6 1
nguyen
tijf
cacbon,
mot axit
c6 2
nguyen
tCr
cacbon

va
mot axit
c6
3
nguyen
ttf
cacbon.
Cong
thdfc
cau tao cua A:
H2C—OCOH
HC-OCOCH3
H2C-OCOCH=CH2
b)TinhVi:
Phanilng:
CgHiaOe
+ 9O2 —>
9CO2
+
6H2O
(mol)
0,05 9 X
0,05
9 x
0,05
6 x
0,05
^
=
0,6

-
0,45
=
0,15 (mol)
^
Vi
=
0,15
x
22,4
=
3,36 (lit)
Nong
do
phan trSm cua
dung
dich
B:
^dung
dich
B ~
"^dung dich NaOH
"^00^ "^H^O
= 160
+
0,45
X
44 + 0,3
x
18

=
185,2 (gam)
Phan
Lfng:
CO2 +
NaOH
>
NaHCOg
(mol)
XXX
C02
+
2NaOH- )•
NasCOg
+
H2O
(mol)
y 2y y
x
+ y =
0,45
fx = 0,3
84x
+
106y
=
41,1
|y
=
0,15

Vay:
C%,,„,,^
= X
100%
=
13,6%;
Theo
de
bai, ta
c6 he:
C%^
CO = ^'^^ X
100%
=
8,58%
c)
Phan
Lfng:
C9H12O6
+
3NaOH
>
C3H5(OH)3
HCOONa
+
CHgCOONa
+
CH2=CHC00Na
(mol)
0,05 0,15 0,05 0,05 0,05

=^ Va
=
0,05 (lit)
=^ a =
n^HCOONa
+
"lcH3COONa
+
"^C,H3C00Na
= 12,2 (gam)
ai
3.
a) Lap
cong
thufc phan tuf
vk
cong
thufc
cefu
tao cua
este
A:
Goi
cong
thOfc
phan
tiuf
ciia
este
A

1^:
RCOOR'
Phan
Lfng:
RCOOR'
+
NaOH -——>
RCOONa
+
R'OH
(1)
B6|
DlJdNG
HOA H0C12
Ho'i bay
ra
gom
ancol
Ian hai
nLfcfc.
Dung CUSO4
de
ha'p thu hcfi nifdc,
c6n
lai
hoi
ancol
R'OH.
Ngung
tu hcfi

ancol
roi
cho
tac
dung
vdi Na
du
R'OH
+
Na
->
R'ONa+
^HaT
2
(2)
Khi
G
la
H2.
Do
tang khoi
iLfgng
binh
dLfng
Na: m^.^^
-
m^^
= 6,4 (3)
H2
+

CuO
—>
Cu
+
H2O
(4)
6,4
(mol)
0,1
64
=
0,1
TCf (3) =^
mR.0H
= 6,2 + 2 X 0,1 = 6,4 (gam)
TCf (2) =>
nR.oH
= Sn^ = 0,2 (mol)
6
4
0,2
R'
=
15
(CH,,-)
Cong
thiJc phan tuf cua
ancol
la
CH3OH

Ta c6: nA =
nancoi
= 0,2 (mol) = n
A
chufa
1
lien
ket 71 (C=C)
(mol)
=> %Br
=
Br^
phan ilng
RCOOCH3
+
Bra
>
RBr2COOCH3
0,2
-> 0,2 0,2
160
65,04
R
= 27
(C2H3-)
R
+
219
100
=> Cong thufc phan tLf cua

A
la:
C4H6O4
Cong thurc cau tao cua este
A:
CH2=CH-COO-CH3
b) Phan ifng:
nH2C=:CH
COOCH3
H2C—CH
H2C—CH
COOCH3
COOCH3
+
nNaOH
>
nCHaCOONa
-H2C—CH
-
I
OH
Bai
4. a) A
la este
vi
khi dun
A
vdi dung dich
NaOH
ta

thu dLfcfc muoi
B
va
1
rifau
C.
RCOOR'
+
NaOH
RCOONa
+
R'OH
R'OH
+
Na
R'ONa+
-H2
2
Bfil
DI/8NG H(5A HOC 12
23
2 24
Tac6:
HA = nR.QH = 2njj = 2 x = 0,2 (mol)
, ' 22,4
Vay MA = ^ = 102
U,
2
Khi
K la 1 hidrocacbon c6 MK =

32.0,5
= 16.
Vay, K la CH4 va muoi B 1^ CHsCOONa do
CHgCOONa + NaOH
CH4t
+ NaaCOg
A:
CH3COOR' ^
MA
= 15 + 44 + R' = 102
mu R' 1^
CJiy
=^ R' = 43 = 12x + y =^ Nghiem hop li: x = 3, y = 7
Vay R \k
C3H7
cong thufc cua
rifgu
C la
C3H7OH
Ydi
C3H7OH
ta
CO
2 dong phan rtfcru
CH3-CH2-CH2OH
CH3-CH2-CHO + H2O
CH3-CHOH-CH3
CH3-CO-CH3
+ H2O
Vi

sdn pham oxi hoa D khong cho phan
ijfng
vdri
AgN03/NH3, D 1^ xeton
CH3-CO-CH3
vk
rirgu
B la
CH3-CHOH-CH3.
C6ng thufc caiu tao cua
este
A:
CH3-COO-CH(CH3)2
(axetat
isopropyl).
b) Chat r^n thu di/cfc sau phan ufng
CH3COOC3H7 + NaOH > CH3C00Na + C3H7OH
(mol) 0.^ 0,2 0,2
=> nNaOH ban diu = 0,3' X 1 = 0,3 (mol)
Vay dif 0,3 - 0,2 = 0,1 (mol) NaOH.
Sau khi c6 can dung dich F ta thu dUgrc
chat
r^n gom 0,1 mol NaOH
0,2 mol CHgCOONa (Ri/gu C3H7OH 1^
chat
long bay di).
=> tCLrin = 0,1 X 40 + 0,2 X 82 = 20,4 (gam).
c)
Ta c6:
HA

= ^ =± 0,1 (mol); no = 0,05 (mol)
1U2
G cung la 1 dcto
este
nen ta c6
c6ng
thufc la CxHy02.
Ta
tinh
n^o^n^^^dl suy ra x v& y
Cong thufc cua A viet gon 1^
C5H10O2
C5H10O2
+^02 —>
5CO2
+
5H2O
(mol) 0,1 0,5 0,5
33
Ta c6: n^^^ = — = 0,75 (mol) ^ n^^^ = 0,75 - 0.5 = 0,25 (mol)
Va
n„^o(A,G,
= ^ = 0,7 (mol) ^ n^^^,^, = 0,7 - 0,5 = 0,2 (mol)
Bdl
DUONG
HdA
HOCI2
(mol)
CxHyOz +
0,05

X + ^ - 1
4
O2
->
xCOa
n
CO,
=
0,05x
= 0,25 =:> X = 5 va n^^ =
0,05x
0,05y
= 0,20
. |H20
0,05y
2
y
= 8
^ay
CTPT
cua G la
C5H8O2
thuoc CTTQ
C„H2„
+ 2 - 4O2 (co 2 lien ket n
\k
este
khong no).
Sir xa phong hoa G cho ra muoi CHsCOONa. Vay G la m6t
axetat

r:> G 1^
(CH3COO)3C3H5,
g6c
C3H5
CO ndi doi C=C
San pham G la 1 andehit (ttf 1
rifcfu
(e mol) khong ben chuyen th^nh)
Vay cong thufc cau tao cua (G) la
CH3-COO-CH=CH-CH3
CH3-COO-CH=CH-CH3 + NaOH CHgCOONa + CH3-CH=CH0H
Ri/gu nay khong ben chuyen thanh andehit:
CH3-CH2-CHO.
Bat 5. a) Cong thufc phan tuf cua
este
A
Phiicfng
phap:
A c6
cong
thijtc la CJiyO^ (Vdi z = 2, z ^ 4
(dieste),
z = 6 (tri
este)
v.v ) 4 an (x, y, z, nA) va 2
phiidng
trinh
(2,64 va so mol)
thieu
2

phiiOng
trinh
nen sau khi c6 MA tif cho z = 2,4
tinh
x, y, z,
v.v
PV 1,792 x 0,5
So' mol
este
nA =
RT
22,4
273
= 0,02 (mol)
X 2 X 273
9 fi4
Vay MA = =^ = 132
0,02
Neu
A
CO
cong thufc phan tuf la C^HyO, ta c6 12x + y + 16z = 132
+) Cho z = 2 => 12x + y = 100
X
5
6
7
8
y
40 28

16
4
Loai
c6ng
thufc
C7H16O2
vi
C7H16
thuoc CTTQ CnHgn ^ 2 hOp
cha't
no
trong
khi
este
phai c6 1 noi doi C=0.
+) Cho z = 4 12x + y = 68
X
4
5
6
y
20
8
am
C5H8O4
thuoc CTTQ CnHan + 2 -
4O4 CO
2 li§n ket n
+) Cho z = 6 => 12x + y = 36
X

2
3
y 12
0
Loai tri/c;ng hcfp nay
Vay chi c6n 1
nghiem
C5H8O4
Bfii
DI/SNG H6A HOC 12
25
N6ng
dp dung dich NaOH
Este
nay c6 2 chuTc
este
vay phan
ufng
vdi NaOH theo ti le mol 1 : 2
0,02 mol
este
A can 0,04 mol NaOH
CM(NaOH)
= ^ = 0,8M.
0, 05
b)
Cong thufc cau tao cua A
Trifcfng
iitfp
1: San pham thu dirge gom 1

rirgu,
2
muoi
thi
este
phai
phat xuat tCr 1 mgu 2 chiifc 2 axit khdc nhau
RiCOOH,
R2COOH, R3(OH)2
Este
A . ^^Ro
/
R2COO
Tdng
so
cacbon
trong Ri, R2, R3 la 5 - 2 = 3. Ri/Ou 2 chufc vay R3 tdi
thieu
phai c6 2 nguyen tuf cabon CH2OH-CH2OH:
Ri,
R2 con lai 1 nguyen tuf
cacbon
vay Ri = H (axit 1^ HCOOH) v^ R2 la
CH3- (axit la CH3COOH)
Vay cong thufc cau tao cua
este
la:
HCOO—CH2
H3C—COO—CH2
Chii y: Loai triTomg hop R3 c6 3

cacbon
vay Ri, R2 khong c6
cacbon
n^o
ca, Ri, R2 la H => chi c6 1
mudi
la HCOONa
(trai
vdfi
de)
Kho'i
liiong
2
muo'i
HCOO—CH2
u
n nr^r^ iu + 2NaOH -> HCOONa + CHgCOONa + (CH20H)2
(mol)
0,02 0,02 0,02
m
2
muoi
=
0,02(68
+ 82) = 3 (gam)
Trifcfng
hcfp
2: San pham thu difOc gom 1
muo'i
vk 2

rufOu.
Vay
este
ph^t
xuat tCr axit 2 chufc va 2
rirou
khac nhau.
Axit:
Ri(C00H)2, RLTOU: R2OH, R3OH
Ri,
R2 va R3
CO
tat ca 3 nguyen tuf
cacbon
R2, R3 phai c6 chufa
cacbon
vay chi c6 the R2 la CH3- va R3 la C2H5 Vay Ri se kh6ng c6
cacbon
vay axit la
COOH
COOH
V^
2
rirgu
\k CH3OH, C2H5OH do do A c6 c6ng
thiJc
cau tao:
OOC—CH3
OOC—C2H5
B6|

DI/SNG
»6k HQC 12
Khdi
liiOng
muo'i
nnr—r\A COONa
uuy ^2NaOH-> I + C2H5OH + CH3OH
oo(:-C2H5
^oof^^
(mol)
0,02 0,02
=>
^N.,c,o,
= 0.02 X 134 = 2,68
(gam),
pal 6. Ta c6:
nNaOH
= 0,5 (mol) va nHci = 0,15 (mol)
Phan iJng: NaOH + HCl > NaCl + H2O
(mol)
0,15 0,15
HNaOH
thay phan = 0,5 - 0,15 = 0,35 (mol) > 0,25 (mol) => Mot trong hai
mudi
1^ do hop chat phenol tao thanh tiep tuc bi
trung
hda bdi
NaOH.
Andehit
no, mach hd xuat phat bdi

este
don chufc nen cung don chufc
Cong thufc
tdng
qudt la: CnH2nO
Ta c6: ^ = <=> n = 3 (C3H6O) ^ CTCT: CH3-CH2-CHO
14n
72,42
Dat cong thufc
tdng
quAt
cua 2
este
la:
RC00-CH=CH-CH3 (a mol) va
RCOO-R'
(b mol)
=> a + b = 0,25 (mol) (D
RC00CH=CH-CH3 + NaOH ^ RCOONa + CH3-CH2-CHO
(mol)
a a a
RCOOR'
+ NaOH > RCOONa + R'OH
(mol)
b b b b ,
R'OH
+ NaOH > R'ONa + H2O (chufa nhan
benzen)
(mol)
b b b

HNaOH
= a + 2b = 0,35 (2)
TCr (1) va (2) suy ra: a = 0,15; b = 0,1
=>
m . =
0,25(R
+ 67) + (R' +
39).0,1
= 28,6 => 5R + 2R' = 159 (3)
muoi
'
Xet hai triTcfng hop sau:
+)
Trifcfng
hcfp
1: Neu mRcooNa =
4,4655
x
mR-pNa
=>
0,25(R
+ 67) =
4,4655
x 0,1 x (R' + 30) =>
2,391R'
= R +
220,691
(4)
Giai
(3) va (4), ta difgc: R = 1 (-H); R' = 7 (CeHg-)

. +;
Trifcfng
hap 2: Neu m^oNa =
4,4655
x mRcooNa => V6 nghiem
vay
CTCT cua 2
este:
HC00-CH=CH-CH3
_Bfl|
DUSNG
HOA HOC 12
Bai 7. a) X^c dinh
cong
thuTc phan ttf cua 2 rUcfu
Goi
cong
thufc cua 2
este
A, B la A: RiCOOR'i (a mol)
B:
RaCOOR'a
(b mol)
Este A, B dan
chufc
ndn axit RiCOOH,
R2COOH
2 ruau Ri'OH, Rg'Oli
cung dan
chufc

^co, •• VH^O = 7 : 10 vay n.^^ : n^^^ = 7 : 10
"coj < "H^O "^^y giong nha
trife/ng
horp ankan 2
ri/au
R'lOH R'20Ii
deu la rugu no.
Dl xAc dinh
cong
thufc phan tuf cua 2
rifgu
nay ta dung 1 rugu duy nhat
C-H - - OH de
tinh
n
n
2ii + 1
(mol) (a + b) (a + b)n (n+l)(a + b)
'^H.o
n + 1 10 -
7„„„
=> —— = - = — => n = - = 2,33 =>n<n<m = n+ l
^cp,
n 7 3
vay n = 2 Ri'OH m
C2H5OH
m
= 3 =^ R'aOH 1&
C3H7OH
Thanh phan %

theo
s6 mol cua h6n hgp 2
riigu
C2H5OH
+ 3O2 > 2CO2 +
3H2O
(mol) a 2a 3a
C3H7OH
+ ^02 >
3CO2
+
4H2O
(mol) b 3b 4b
'^H,D 3a + 4b 10
n^^Q^ 2a + 3b
a = 2b
vay: %C2H.0H = = =
66,67%
a + b 3b
%C3H70H = 100% -
66,67%
=
33,33%
b)
Xac dinh
cong
thufc phan tuf cua 2 axit
RiCOOR'i + NaOH > RiCOONa + R'lQH
(mol) a a
RsCOOR's'+NaOH

^^RgCOONa
+ R'20H
(mol) b b
B6\G HI^A Hnr 17
+)
Phan
ufng giufa 2 mu6'i
vdri
H2SO4
2RiCOONa
+ H2SO4
(mol) a
2R2COONa
+ H2SO4
(mol) b
+)
Phan
ufng giufa 2 axit vori Na2C03
2R1COOH
+
NaaCOa
a
2
2R2COOH
+ NaaCOg
b
^
2
-> 2R1COOH +
Na2S04

a
-> 2R2COOH +
Na2S04
b
(mol)
(mol)
->
2RiC00Na
+ CO2 + H2O
2R2COONa
+ CO2 + H2O
n
a + b
NajCOj
phan
ling
(mol)
n
Na,CO,
ban dau
'^Na,C03
da ~ 0,1
,„ = 0,1 X 1 (mol)
a + b
(mol)
+)
Phan
ufng giufa Na2C03 du vdi HCl
Na2C03 + 2HC1 > 2NaCl +
C02t

+ H2O
nHci
= 2 X
0,085
= 0,17 (mol) => n
0,17
Na^COj
du
- =
0,085
(mol)
o
0,085
= 0,10 -
a
+ b
a
+ b
=
0,015 ^ a + b = 0,03
Va a = 2b a = 0,02 (mol); b = 0,01 (mol)
Vay
cong
thufc phan tuf cua 2 axit
ma axit = 0,02(Ri + 45) +
0,01(R2
+ 45) = 2,08
R2 + 2Ri = 73
Vi
2 axit nay deu no nen Ri =

CnH2n
+1 va R2 = CmHam + 1
Vay 2(14n + 1) + 14m + 1 = 73 o 28n + 14m = 70 2n + m = 5
n
0
1
2
m
5
3
1
Co 3 cap nghiem
• n = 0 ^ HCOOH va m = 5 => CgHnCOOH
. n = 1=^ CH3COOH va m = 3 C3H7COOH
. n = 2 C2H5COOH va m = 1 => CH3COOH
Bfil
DUONG H6A HOC 12
Vely
cdc
cong
thufc
c6
the
c6
cua
2
este
ijfng
vdri
moi cfip

nghi§m
tren
HCOOC2H5
va
C5H11COOC3H7
CH3COOC2H5
va
C3H7COOC3H7
C2H5COOC2H5
va
CH3COOC3H7
Khi
dot
chay
mSi
este
CxHy02 thi
n^,^^
=
xneste-
Vay
neu n^^^
<
Sn^ste
thi
X
< 6,
este
chufa dvtdi
6

nguyen
tijf
cacbon.
Vay
chi c6 cap
nghiem cuoi
C2H5COOC2H5
va
CH3COOC3H"
(so C = 5)
la
thoa man dieu kien
tren.
Bai
8. A
chac
chan
la
este
con
B
tac
dung
vdfi
NaOH
tao
duoc
2
chat hOfu
co

C, D'. Vay
B
cung
la
este.
Khi
dot
chay
A,
B
ta
difgc
n^^^
=
n^^^^
Vay
cong
thijfc
cau tao cua
A, B
deu
chufa
1
lien ket
n (cua
nhom
C = O).
Vay mach
cacbon
cua

A, B
khong
c6
lien
ket
n, do do A, B la
este
no,
Axit
va
nfgfu
tao ra A,
B
cting
la hop
chat
no.
+) Xac
dinh
cong thu'c cua
A,
B
Phiidng
phdp:
Ta xdc
dinh cong thu'c
cdu tao cua
axit
C va rMu D, ti(
do viet cong thifc

cdu tgo cua
este.
Ti
khdi
d^n^ = 30
=^
Mc = 2 x 30 = 60
Do
C la
axit
no,
C co
cong
thufc
CnHgn
+ i-COOH
Hay 14n
+ 46 = 60
=>
n = 1 va C la
CH3COOH
A,
B
CO
cung
cong
thufc phan
tiJf
la
CnH2n02

nen
khi
do't
chay
h5n
hop
A,
B
CO
the xem
nhtf
dot
chay
1
chat duy nhat
co
cung
cong
thufc phan
tuf
vdri
A,
B
C„H2„02
+ " O2 >
nC02
+
nHaO
10,2
. (3n - 2) 10,2 14,56 ^

^_
nA
B = =>
n,,
= X - =
0,65
' 14n
+ 32 2
14n
+ 32 22,4
=>
n = 5
Vay
A,
B co
cung
cong
thufc phan tuf
la
C5H10O2.
+) Xac
dinh
cong thu'c cau tao cua
A
A
la
este
cua
CH3COOH
vay

A co
cong
thufc
la
CH3COOR,
C5H10O2
vay
R-C3H7
Vay
A
CO
the co 2
cong
thufc
cau tao
tuy
theo
C3H7
mach th^ng
hay
phan nhanh
CH3-COO-CH2-CH2-CH3
H3C—COO—CH—CH3
CH3
JO
B6|
Dl/flNG
HflA
HOC
12

Dg'
chon
cong
thufc ca'u
tao
dung cua
A
giufa
2
cong
thufc n^y
ta
dUa
tren
cong
thufc
cau tao cua
2
san
pham oxi hda cua rUOu
D.
Khi
xa
phong hoa
este
A, ta
duoc
vdri
C3H7
mach th^ng

CH3COOCH2CH2CH3
+
NaOH
->
CH3COONa
+
CH3CH2CH2OH
Rifcfu
E
thu
dufoc
la
rtfcfu
bac 1,
chat
nay
khi
bi oxi hoa
tren
Cu
nung
nong
cho
andehit
CH3CH2CH2OH
+ IO2
^—>
CH3CH2CHO
+
H2O

Andehit
co
phan
ting
trang gifong,
trai
vofi
de
vay loai
trLfomg
hop nay
1
+) Neu
C3H7-
phan nhanh
1 H3C—COO—CH—CH3 +
NaOH
^
CH3C00Na
+
CH3CHOHCH3
CH3
E
la
rLfgru
bac
2,
khi
bi
oxi hoa

cho
xeton
CH3-CHOH-CH3
+ IO2
^—>
CH3-CO-CH3
+
H2O
Xeton khong
co
phan ufng trang giJofng, dung vdi
de.
Vay
A co
cong
thufc
ca'u tao
la
:
H3C—COO—CH—CH3
CH3
+) Xac
dinh
cong thifc cau tao
ciia
B
B
khi
bi
xa

phong
hoa tao ra
C va
D'.
D'
la
riJgu
vi
khi
bi
khuf nifdfc (bcfi
H2SO4)
cho
ra
anken vay
C la
mudi RCOONa
V(Ji
H2SO4,
muo'i nay
co
phan ufng
2RC00Na
+
H2SO4
>
2RC00H
+
Na2S04
Axit

RCOOH
CO
phan ufng trang gifomg vay axit nay
la
HCOOH
B
la
este
cua
HCOOH
va co
cau tao
la
HCOOR, vcfi
cong
thuTc phan
tii
la
C5H10O2
ta
thay ngay
R
la
C4H9
va
ri/cfu
D'
la
C4H9OH.
Vdi

C4H9OH
ta
CO
3
dong phan (rifcfu
bac 1, 2 va 3) ta
chon
cong
thufc
ca'u
tao nao ma
khi
bi
khuf
nu'cfc
ta
thu diicfc
2
anken.
H3C—CH2—CH—CH2
>
CH3-CH2-CH=CH2
+
H2O
H OH
Chi
CO
1
dong phan anken duy nhat (loai)
1234

H2C—CH—CH—CH3
H OH H
Dl/flNO
HOA
HOCI2
31
Ttiy
theo
H
tdch
ra
v6i
OH
(dl tab
HgO)
1^
d
Ci hay
C3,
ta se
diTgrc
2
anken:
CH2=CH-CH2-CH3
(khii
H d
Ci)
CH3-CH=CH-CH3
(khi^ H d
C3)

Triidng
hcfp n^y phu hgrp
vdri
de.
OH
H3C
C CH3
CH3
H2C—C
CH3
CH3
Vay D'
1^
CH3-CHOH-CH2-CH3
CTCT
cua
este
B Ik:
HOOC—CH—CH2—CH3
<^H3
Bai
9. X la 1
este
ca
chufa
1 hay
nhieu chiifc
este.
Gia
sijT

cong
thufc
cua X la
CxHyOz
Phitcfng
phap:
De
biet
chiia
1 hay
nhieu
chvtc
este
ta tim 1 he
thitc
giQa
y va
X.
+) Neu
y = 2x => 1
lien
ket n (1
chufc
este)
y
= 2x - 2 => 2
lien
ket n (2
chufc
este)

Chti
y:
Khong
c6
lien
ket C=C do
axit
A va
ri/gu
B tao ra X
deu
la
hgfp
chat
no
vay
X
1^
1-este
no.
,
Goi
t = nx
ufng
vofi
a
gam
X
X
+

Oo
(mol)
>
XCO2
+ ^HaO
tx
^
=>
n - n = t
x-
^
2
=
0,2
(1)
a
gam X
chig'm ciing
the
tich vdi
5,8 gam
khong khi
vay so
mol
X = so
mol
khong khi
5,8
=>
t =

29
=
0,2
(mol) (2)
Giai
(1) (2) ^
X
- ^ = 1 => y = 2x - 2
- 2
Vay
X
CO
2
lien ke't
n (2
noi doi
0=0)
Vay
X
CO
2
chufc
este.
Do do
riicfu
B
hoSc
axit
A da
chufc.

2
.
B6\G
HdA HQC
12
+)
C6ng
thtfc ciia
rtfofu
B
la rugu
no c6 the
dcfn
hay da
chufc
B CO
CT Ik
C„H2n + 2O,.
CnH2n
+
20z'
+
3n
+ 1 - z'
02
->
nC02
+ (n +
DHaO
Vofi

1
mol
B can
3n
+ 1 - z'
mol
O2.
Vay
3n
+ 1 - z'
=
2,5 ^ 3n + 1 - z' = 5 => 3n - z' = 4
z'
1
2
3
n
5
7
—
2
—
3 3
V(5i
z' = 2, n = 2
B
la
ri/gru
da
chufc

OH2OH-CH2OH
+) Cong thu-c
cua B
Dl CO
the
CO
X la 1
dieste,
axit
A c6 the
don
chufc
hay da
chufc
•
A
CO
2
chufc
axit
OnH2n(OOOH)2
X
se la 1
dieste
mach vong
.COOH
HO-CH2
^C00-CH2
+
H2O

HO—CH2
CnH2n\
COO—CH2
Loai
trirdng
hop
nay
vi
theo
de X la 1
este
mach hd.
A
la
don axit ROOOH. Cong thufc
cau tao cua
este
X la:
RCOO—CH2
RCOO—CH2
Khi
x^
phbng hoa:
RCOO—CH2
+ 2NaOH
^
2R0OONa
+
RCOO—CH2
Cuf

0,2
mol
X tao ra 0,4
mol muoi
m„,u6-i
= 0,4(R + 67) = 32,8
R
+ 67 = 82 R = 15
Vaiy
R la
CH3
va
axit
A
1^
CH3COOH.
Do
do
cong
thufc
cau tao cua X
1^:
H3C—COO—CH2
HO—CHo
I
HO—CH2
H3C—COO—CH2
Bai 10. a) Gid stjT X, Y, Z c6 cong thufc Ik RjCOOH,
R2COOH
vk

R'COOH
vdri
Ri = R2 = R (vi X, Y dong phan) va R' = R + 14 (do Z la dong dang
ke tiep cua Y (hcfn Y mot -CH2-)
Vay cong thijfc cau tao cua
este
A xuat phat tU 3 axit tren va glixerol la:
Rl—COO—CH2
R^_COO—CH2
R2—COO—CH
'
R'—COO—CH2
R'—COO—CH
I
R2—COO—CH2
Khi
xk phong hoa A, ta thu dtfoc glixerol va 3 muo'i
Rl—COO—CH2
HO—CH2
I I
Ro—COO—CH
+3NaOH ^
HO—CH
+
R'—COO—CH2
HO—CH2
RjCOONa
R^COONa
R'COONa
Goi a = nA, phan uTng phong hoa cho ta 3 muoi

vdri
so mol moi
mudi
deu
bhng
a
mA
= a(Ri + R2 + R' + 173) = 7,9 hay a(2R + R' + 173) = 7,9 (1)
Kh6'i
ItfOng 3
mudi
ma
muo-i
= a(2R + R' + 201) = 8,6 (2)
Lay (2) trCr (1), ta dugc: 28a = 0,7 =^ a = — mol
Thay vao (1):
0,1
(2R + R' + 173) = 7,9
2R + R' + 173 = = 316
Vi
R' = R + 14
2R + R + 14 = 3R + 14 = 143
Neu
R \k CxHy => 12x + y = 43
0,1
3R = 129 => R = 43
2R + R' = 143
X
2
3

4
y
19
7
am
Vay R 1^ -C3H7 va R' la -C4H9
Axit
Z
CO
mach th^ng vay c6 cong thufc cau tao la :
CH3-CH2-CH2-CH2-COOH
Y dong dang vdi Z nen cung c6 mach thing CH3-CH2-CH2-COOH
X dong phan
vdri
Y c6 mach phan nhanh:
H3C—CH—COOH
CH3
Ghep
cong thufc cua 3 axit vdi cong thufc cua glixerol ta c6 cong thufc cau
tao cua
este
A:
I
'
H3C—CH—COO—CH2
H3C—CH2—CH2-COO—CH
H3C—CH2—CH2—CH2—COO-CH2
CH3
hoac
H3C—CH—COO—CH2

H3C—CH2—CH2—CH2-COO—CH
H3C-CH2—CH2—COO—CH2
b) H5n hgfp 3
mudi
vdri
H2SO4
2C3H7-COONa +
H2SO4
> 2C3H7-COOH + Na2S04
2C3H7-COONa + H2SO4 > 2C3H7COOH + Na2S04
2C4H9COONa
+ H2SO4 > 2C4H9COOH +
Na2S04
Do so' mol 3
mudi
bang
nhau, so mol 3 axit X, Y, Z cung
bang
nhau.
Goi X sd mol m6i axit trong hon hop
daoc
dem dd't
chay
C3H7-COOH
+ 5O2 ~—>
4CO2
+ 4H2O
(mol) x • 4x
C3H7-COOH
+ 5O2 —>

4CO2
+ 4H2O
(mol) x 4x
C4H9-COOH
+
6,502
^—-^
5CO2
+ 5H2O
(mol) x 5x
=> n^Q^ = (4 + 4 + 5)x = 13x
^ n„„ = n„ I - ^'^^^ = 0,013 (mol) .
CO2
BaCOg-L
197 J ^ '
=^ 13x = 0,013 hay x = 0,001 (mol)
Khdi
lifong hon hop 3 axit dem dot
chay
0,001(88
+ 88 + 102) =
0,278
(gam)
Bai 11. a)
Este
A: RCOORi' (a mol) B:
R2COOR2'
(b mol)
Tinh
nNaOH trong 50 ml dung dich NaOH (diia tren 10 ml dung dich

NaOH
trung hoa 30 ml dung dich H2SO4
0,25M)
va nNaOH du
Suy ra nNaOH phan ufng
vdfi
este
HNaGH
= n2
este
=> a + b = 0,05 (mol)
B6|
DliflNn
Hfii
Hfir
19 as
b)
ma
este
=
M(a
+ b) = 5,7 =>
MA
=
MB
=
114
Ri
+
R'l

= R2 +
R'2
= 70.
2
rugru
vk
axit
d4u l^m mat mau
rnidc
Br2 vSy
trong
2
rUcru cung
nhij
trong
2
axit phai
c6 it
nhat
1
chat khong
no.
2
rtfou
axit
c6
cCing
so
nguyen
tijf

cacbon
vay 2
axit cung
c6
cung
s6'
nguyen
tuT
cacbon
(do 2
este
nay
dong phan)
Cong
thuTc
chung
cua 2
este:
CnHsn
+
2
-
2kCOOCniH2m
+
1
-
2k'
(k,
k' so
lien

ket n c6
tren
gdc
R,
R')
ta c6
R
+
R'
= 70 7(n + m) - (k + k') = 34
k
+
k'=l=^n
+ m = 5
n,
m
phai
> 2 (vi vdi 1
cacbon,
khong
the c6 2
riidu,
2
axit khac nhau)
k
= 0, k' = 1
C2H5COOC3H5
(goc
CO
noi doi

trong
R'OH
phai
toi
thieu
CO
3
nguyen ttf
cacbon
thi
riiOu
mdi
ben)
k
= 1, k' = 0 ^
C2H3COOC3H7
C2H5COO-CH2-CH=CH2
; propionat propennil
CH2=CH-COOC3H7
;
acrilat propil.
c)
2
rtrgu
la
C3H5OH
hay
CH2=CH-CH20H
va
C3H7OH

chi c6
rifou
dau
'*
lam mat mau
nUdc
Br2 vay
a =
n^^H^oj,
=
ng^^
= 0,04 (mol)
^=^o,n,ou
=0,01(mol)
=
2,32
(gam);
m^^„ = 0,60 (gam)
"H,
= r.,. =
0>025
(mol)
^ = 0,56
(lit).
Chi
CO
axit
CH2=CH-C00H
lam mat
mau rnidc

Brz
^
n3_,
= b =
0,01, (mol)
=> y = 1,60
(gam).
Bai
12.
a)
Theo
de
bai,
ta c6: nA :
nNaOH
= 1 : 2
=>
A la
axit hai chufc.
Goi
cong
thufc tong quat
cua A la
R(C00H)2
Ma
n^^o^ :
n^^^=
4:5=^ n^^<
n^^^^B
la

ancol
no
Theo
de ra B la
ancol dcrn
chufc
=> Goi
cong
thufc tong quat cija
B la
CnH2n
+
lOH
CnHzn.
lOH
+ ^02 ^—>
nC02
+ (n +
DHgO
"co n 4
^
_co^ ^ __u_ ^ ^ ^ 4
(C4H9OH)
^H,o
n + 1 5
)6 Bdl DlidNG HOA HOC
12
Vi
ancol 1^ ancol
bac

hai c6 nhdnh
n§n B
c6
cong
thufc
cau tao la:
H3C—CH—CH2—OH
•
ai^'^ol
isobutylic
CH3
h^n
iJfng
este
\k:
H,SO,
R(C00H)2
+
nC4H90H
^
0,0735
(mol)
0,0735n
(mol)
> nH20 + R(COOH), _
^
(COOC^H^)
0,0735
(mol)
9

'n
14
847
ME
= R + 45(2 - n) +
lOln
= ' = 202 R + 56n = 112
0,0735
)
Khi n = 1 =^ R = 56
(~C4H8-)
Cong
thufc
cau tao cua A :
HOOC-(CH2)4-COOH: axit adipic
=>
Cong
thufc
cSi'u
tao cua E:
HOOC—CH2—CH2-COO—CH2-CH-CH3
CH3
Isobutylhidro
adipat
+)Khin
=
2=>R
= 0=>
CTCT
cua A:

HOOC-COOH: axit oxalic
Cong
thufc ca'u
tao
cua
E:
H3C—OH—CHo-OOC—COO—CH2—CH—CH3
3
I 2 2 I 3
CH3
CH3
b)
Tinh
khoi liTcfng
cua A, B:
+) Khi
n = 1:
mA
=
0,0735
x (56 + 90) =
5,439
(gam)
mB
=
0,0735
X 74 =
10,31
(gam)
+) Khi

n = 2;
mA
=
0,0735
x 74 =
10,878
(gam);
me
= 2 X
0,0735
x 74 = 6,615 (gam)
+)
X6t
trifcfng
hcfp
3:
Loai
vi
nNaOH
khong du thuy phan
so mol
este.
Bai
13. a)
ChiJtng
minh rdng
1
trong
2
este

A, B cd
chvta
goc
benzen
bang
each
so
sdnh
so mol
NaOH phan ling
vdi
tong
so mol
este.
Neu
HNaOH
> n2
este
thi
86 CO 1
este
phan
ijfng
v6i
NaOH theo
ti le mol
1
: 2,
este
n^y c6

dang
R-COO-CeHs
A:
RiCOOR'i
(a mol)
B:
RgCOOR'a (b mol) vdi
R'2=C6H5
Phan dfng giufa
A, B
vdi NaOH
(mol)
RiCOOR'i
+
NaOH
a
a
R2COOR'2
+ 2NaOH
^mol)
b 2b
DU8NG HdA HOC
12
-> RiCOONa
+
R'lOH
a
a
->
RgCOONa

+
R'20Na
+ H2O
b
b
37
=>
n2este=
a + b = 0,08 (1)
(do n = C X V = 0,8 X 0,1 = 0,08 mol)
So mol NaOH ban dau =
0,15.1
= 0,15 (mol)
HNaOH
du =
HHCI
= 0,2 X 0,2 = 0,04 (mol)
=> nNaOH pt» vdi A,c = 0,15 - 0,04 = 0,11 (mol)
a + 2b = 0,11 (2)
Tii (1) (2) ^ a = 0,05 (mol) A; b = 0,03 (mol) B
Chi
c6
este
A phan ufng cho ra
rifcfu
E
nE = HA = 0,05 (mol) ME =
MR.QH
= = 58
0,05

Do mqu R'OH khong ben bien thanh andehit R"CHO ta c6
ME
= R" + 29 = 58 =i> R" = 29
Vay R" 1^
C2H5
va andehit la CH3-CH2-CHO bien doi tii rUgfu khong
bin
CH3-CH=CH0H.
Phan ufng xa ph6ng h6a A, B chi cho ra 2
mudi
vSy b^t bu6t trong 3
mudi
RiCOONa, RgCOONa va
R'aONa
c6 2
mudi
gidng nhau.
Yky Ri trung vdi R2 -
Khdi
iLfgtng
2
mudi:
m2
muol
=
niRcooNa
+ ^n\om
(a + b) (b)
0,08(Ri
+ 67) +

0,03(R'2
+ 39) = 10,46
8R1 + 3R'2 = 1046 - 653 = 393 (3)
^
M^^coo^^
^ +67 ^41.
• M, -
M,,^,,,^
- R;
+39-65
<=>
41R'2 + 1599 =
65Ri
+
4355
41R'2 -
65Ri
=
2756
(4)
Ttr
(3) v^ (4) Ri = 15 vay Ri Ik CH3
R'2 = 91 R'2 la C6H4-CH3
+) Cong thtfc cau tao cua A va B
A
este
cua axit
CH3COOH
va
rirgu

CH3-CH=CH0H c6 cong thufc cau tao
la:
CH3-COO-CH=CH-CH3
B
este
cua axit CH3COOH va phenol nSn B c6 cong thufc cau tao \k:
H3C—COO
B6I
DUSNG
HOA HOC 12
pai
14. a) Cong
thufc
phan tuf cua A, B, C.
A,
B, C la
dong
phan nen c6
cung
CTPT.
P2O5
hut
nude
(P2O5
+
3H2O
>
2H3PO4)
14,4
N§n mjj^o = 14,4 (gam) => mn =

KOH
hut CO2 n§n:
m^Q = 35,2 (gam) => mc =
= 1,6 (gam)
3.35,2
11
= 9,6 (gam)
Kh6ng c6 nguyen td n^o
khac
ngoai oxi.
Suy ra: mo = 17,6 - (1,6 + 9,6) = 6,4 (gam)
12x y 16z
9,6 1,6 6,4
hay
x _ y _ z
r96^
16
r64^
112;
[lej
X
^ 8 =
16
= — hay — = — = - = n
4-^241
CTN
la (C,H^O)„
Vi
A, B, C dorn chufc, sd nguyen tuf oxi chi c6 the b^ng 1
(rLfgu,

ete,
andehit,
xeton)
hay bkng 2 (axit,
este).
+) Khi z = 1 => Cong thufc phan tuf la C2H4O. Chi c6 the c6 1
c6ng
thiJc
cafu tao 1^ CH3-CHO (loai vi phai c6 3 dong phan).
+) Khi z =.2 =>
C4H8O2.
Vay c6 lien ket n. A, B, C chi c6 the la axit hay
este.
Axit
CH3CH2CH2COOH
CH3-CH(CH3)-COOH
Este
HCOOC3H7 CH3COOC2H5 C2H5COOCH3
Co 3
triidng
hop: +) Trufcfng hcfp 1: 3
este
+) Triicfng hgp 2: 1 axit, 2
este
+) Trtrdng hop 3: 2 axit, 1
este.
PhiTOng
phdp:
de
chon

tritdng
hcfp
dung,
ta
difa
tren
so mol NaOH
phdn
ling d
nhiet
do
thitdng
va khi dun
nong
va
tren
khoi
liidng
2,58 gam cua
chat
long.
HNaOH
d t°
thifdng:
0,5.0,1
= 0,05 (mol)
nNaOH
khi dun nong: 1 x 0,1 = 0,10 (mol)
+)
Trififng

hcfp
1: A, B, C deu 1^
este.
Do
este
gan nhtr khSng phan ufng vdi NaOH trong
thdi
gian ng^n, ta
loai
trifdng
hcfp n^y.
Bfii
Di/flNG HOA HOC 12 39
+)
Trirtfng
hap 2; 1
axit,
2
este.
So' mol axit
= so'
mol NaOH phan ufng
cf t°
thLfomg
= 0,05 mol
So' mol axit
+ so mol 2
este
=
nNaOH

phan ufng khi dun nong
= 0,10
mol.
n2 este
=
0,10 - 0,05 = 0,05
(mol).
2
este
khi x^
phbng
hoa cho ra 2
muo'i
va 2
rtfcfu.
Vay khi E bay len
gom
2
rtfcfu
va
nifdc.
Sau
khi loai
het
ntfdc
con lai h6n
hcfp
2
riXcfu
vai

tong
so'
mol bkng
so'
mol
2
este
(0,05
mol).
M2ruau
= = 51,6
0,05
.
Mrugu
1 < 51,6 <
Mri,cru2
Rtfcfu
nang nhat trong
ttf 3
este
HCOOC3H7, CH3COOC2H5 va
C2H5COOCH3
\h
C3H7OH
CO
M = 60 >
51,6.
Cac
rtfgu
kia

C2H5OH
(M = 46)
va
CH3OH
(M = 32) deu c6 M < 51,6.
vay
rtfcfu
II
la C3H7OH va
este
tuong
ijfng
la HCOOC3H7.
Rtfcfu
I c6 the la
CH3OH
hay
C2H5OH
• Cap C3H7OH va CH3OH
Goi
a =
n^H^ojj;
b = n^
a
+ b = 0,05 va mg = 32a + 60b = 2,58
=>
a = 0,015
(mol)
CH3OH va b =
0,035

(mol)
C3H7OH
CH3OH
phat xua't ttf
este
C2H5COOCH3
Vay
hon
hcfp khi
dau gom: 0,1
(mol)
C3H7COOH
=i>
50%
0,03
(mol)
C2H5COOCH3
=>
15%
0,07 (mol)
HCOOC3H7
35%
•
Cap
C2H5OH
va
C3H7OH
Goi
c =
n^^^H^ojj;

d =
n^^^^
c + d = 0,05 va 46c + 60d = 2,58
^
c = 0,03 (mol)
C2H5OH
va d = 0,02
(mol)
C3H7OH
Hon hcfp
dau gom: 0,1
(mol)
C3H7COOH
=> 50%
0,06
(mol)
CH3COOC2H5
=> 30%
0,04 (mol)
HCOOC3H7
^ 20%
Chu
y: Do A, B, C c6
cung
cong
thufc phan tuf (cung
M) nen
thanh phan
phan
tram

theo
khoi Itfcfng cung
la
thanh phan phan
tram
theo
so mol.
+) TrttUng
hap 3: 2
axit,
1
este.
n2
axit
=
HNaOH
d
nhiet
do
thifdng
= 0,05 (mol)
neste
+
Haxit
=
nNaOH
a
nhift dp cao
= 0,10 (mol)
:=> neste

= 0,10 - 0,05 = 0,05 (mol)
Vi
chi CO 1
este,
phan ufng
vdri
NaOH
chi cho ra 1
rtfcfu
C3H7OH,
C2H5OH,
CH3OH.
0 B6I Dt/dNG HdA HOC 12
0
KQ
Do
Mnrau
= = 51,6
khac
vdfi
M cua 3
rtfcfu
tren,
vdy ta
loai
trtfdng
0,05
hcfp
nay.
Tom

lai chi c6
triidng
hap 2 la
dung.
c)
Sau
phan ufng
xa
phong
hoa, ta
dufcfc
3
muoi C3H7COONa
(0,05 mol),
CHsCOONa
va
HCOONa (phat xua't ttf
2
este).
Khi
nung
3
muo'i
nay vdi
NaOH,
ta c6:
CsHTCOONa
+
NaOH
>

CgHgt
+
Na2C03
(mol)
0,05 0,05
CH3COONa
+
NaOH
>
CH4t
+
Na2C03
(mol)
0,03 0,03
HCOONa
+
NaOH
> Hgt +
Na2C03
(mol)
0,02 0,02
Khoi
Itfcrng
chung
cua 3
khi:
0,05
X 44 + 0,03 X 16 + 0,02 x 2 = 2,72 (gam)
M3khi
=

^=27,2
97
9
Ti
khoi
doi vdi
Hg:
d = = 13,6. '
2
Bai
15. a)
Cong
thufc phan tuf
va
cong
thufc
ca'u tao cua A, B.
A,
B la
dong
phan
nen c6
ctmg
cong
thufc phan tuf.
CO2
+
Ca(0H)2
>
CaC034

+.
H2O
Ta
c6: n„„ = n„ „„ , = = 0,1 (mol)
Dung dich nhan
CO2
va HgO
nhtfng ma't
10 gam ket
tua.
.
Do giam khoi Itfcfng dung dich
=
VCI^^^^Q^^
-
(m^.^^
+ m^^^)
10-(0,1 x44+ mjj^o)
= 3,8
=>
m^j^o = 1,8
(gam)
va
n^^
„ =0,1 (mol)
Ta
de y
rkng
n^^^^
=

nj.^^.
Vay hop
chat
A, B nay chi
chufa
1
lien
ket n.
Do
A, B chi
chufa
este
nen A, B chi c6 the
chufa
1
chufc
este
n§n c6 2
nguyen tuf oxi.
Cong
thtfc phan tuf
cua A, B la
CxH2x02
CxH2x02
+ O2 —>
XCO2
+
XH2O
B61 Dl/flNG H6A
HOC

12 41
Ta c6: n„ = = 0,125 (mol)
22,4
^o, 3x -2 0,125
' neo^ 2x 0,1
Vay
cong
thufc phan
tijf
cua A, B la
C4H8O2
Cong
thufc cau tao c6 the c6 cua A, B:
HCOOCgHv,
CH3COOC2H5,
C2H5COOCCH3.
'
b)
Y va Z CO cung so nguyen
tijf
C va O nhi/ng it hon 2 nguyen tuf H.
Vay
cong
thufc phan tuf cua Y va Z la
C4H6O2.
- Y, Z phan ufng vdi
NaOH
cho muoi va rufcfu vay Y, Z chi c6 the la axit
hoac
este.

- Y va Z khong the cung 1^
este
vi sii xa phong h6a 2
este
dong
ph^n
phai cho 2 rtfgfu trong khi do de cho biet phan ufng chi cho ra 1
rifgru.
Vay Y phai la
este,
Z la axit.
Vdri
cong
thufc phan tuf
C4H6O2
hay
CnHsn
+ 2 - 4O2, Y v^ Z c6 2 lien ket
TI.
vay Y va Z 1^
este
hoac
axit khong no (c6 1 noi doi C=C)
Z chi CO the la:
CH2=CH-CH2-COOH
hay
CH3-CH=CH-C00H
hay
CH3-C(CH3)=C-COOH
(loai vi

theo
de mach
cacbon
khong phan nhdnh)
Y la
este
c6
cong
thufc cau tao la
CH2=CH-COO-CH3
Goi y = ny va z = nz
Phan
ufng cua Y, Z vdi
NaOH
CH2=CH-COO-CH3
+
NaOH
^
CH2=CH-C00Na
+
CH3OH
C3H5COOH
+
NaOH
->
CsHgCOONa
+ H2O
2 muoi nay phat xuat tif 2 axit.
CH2=CH-C00H
va

CH3-CH=CH-C00H
(d6ng
ding ke'
ti§p),
v<Ji so'
mol cung la y va z.
Hai axit nay phan ufng vdri
Na2C03
ma kh6ng c6 khi CO2 bay ra nghla
\h. phan ufng mdi den giai doan tao ra
NaHC03
CH2=CH-C00H
+
Na2C03
>
CH2-CH-C00Na
+
NaHCOg
y mol y mol y mol
C3H5COOH
+
Na2C03
C3H5C00Na
+
NaHCOa
z mol z mol z mol
2 trtfdng hgp c6 th§' c6:
Na2C03
phan ufng vifa du
hoac

dif
Na2C03.
B6I
DUflNG
H6A HOC 12
+) Trifdng
hofp
NazCO^
phan iJng vijfa du.
Ta c6:
n^^^^^^
= y + z =
n^^^.^^
= 0,6 (mol)
So mol HCl can thiet de phan ufng vdri
NaHC03
HCl +
NaHC03
>
NaCl
+ CO2 + H2O
(mol) (y + z) (y + z) nnci = y + z = 0,6 (mol)
Trai vdri de vi nnci = 0,8 x 1 = 0,8 mol
Vay diT
NazCOa
Taco:
n^^^.^^=
0,6 - (y + z) (mol)
Vdi HCl,
CO

2 phan ufng:
Na2C03
+ 2HC1 >
2NaCl
+ C02t + H2O
(mol) 0,6 - (y + z) 1,2 - 2(y + z)
NaHC03
+ HCl >
NaCl
+ C02t + H2O
(mol) (y + z) (y + z)
^ nnci Chung = 1,2 - 2(y + z) + y + z = 0,8 (mol)
<^ 1,2 - (y + z) = 0,8 y + z = 0,4 (mol) (D
Bot
chdy:
C2H3COOH
+ 3O2 —>
SCOs
+
2H2O
(mol) y 3y
C3H5COOH
+ §02 ^
4CO2
+
3H2O
(mol) z 4z
Tif (1) va (2) =^ y = 0,15 (mol) Y; z = 0,25 (mol) Z
Thknh phan phan trSm
theo

khdi
lufgrng
mc,H3CooH
= 0,15 X 72 = 10,8 (g);
m,^„^,,,,
= 0,25 x 86 = 21,5 (g)
Vay:
%C2H5COOH
= IMiLM =
33,43%
va
%C3H5COOH
= 66, 57%
10,8 + 21,5
Bai 16. A, B khong tac dung vdfi dung dich Brg, vay A, B 1^ hgp chat no.
A, B tac dung vdi
NaOH
cho ra ri/gu va muoi, vay A, B 1^ axit hay
este.
+) Trtfofng
hgfp
A, B deu la este.
A, B phai
CO
cong
thiJc
RiCOOR
vk
R2COOR
(cung R M phan ufng v<Ji

NaOH
chi cho ra mot rufou duy nhat).
RiCOOR
+
NaOH
>
RiCOONa
+ ROH
(mol) a a a a
R2COOR
+
NaOH
>
R2C00Na
+ ROH
(mol) b b b b
661DI/8NG
H6A HOC 12, • 43
4
=>
HNaOH
= a + b = — =0,1 (mol)
40
Vay
HROH
= a + b = 0,1
(mol).
Rifcfu
ROH
tac

dung
vdri
Na
ROH
+ Na
RONa
+ -Hat
2
Theo
phan ufng: cuf
0,1
mol ROH
cho 0,05
mol Hg
Khong phu hcfp vi
n^ =
0,672
22,4
=
0,03
(mol).
+)
Tru-ofng
hgfp
A la
axit,
B la
este.
A:
RiCOOH

Phan ufng:
RiCOOH
+
NaOH
a
a
R2COOR3
+
NaOH
b
b
a
+ b = 0,1
(mol)
b
B:
R2COOR3
(mol)
(mol)
RiCOONa
+ H2O
a
a
->
R2C00Na
+
R3OH
b
b
n.

1
= —
n
b
= 0,06
(mol)
B a = 0,04
(mol)
A
Do
A, B la
axit,
este
no (chi c6 1
lien
ket n) nen A c6
cong
thvtc
la
CnH2n92,
B
CO
cong thufc
la
CmH2m02,
M = 2
X
35,6 = 71,2
M
=

Q'Q4(14n
+ 32) +
0,06(14m
+ 32)
<=>
56n +
84m
= 392
0,1
2n
+ 3m = 14
=
71,2
n
1
2
3
4 5
m
4 10/3
8/3
2 4/3
+) Vdi
n = 1, m = 4, ta c6:
A:
CH2O2
CO
cong thufc
cau tao Ik:
HCOOH

B: C4H8O2
CO
4
cong thufc
cau tao la:.
HCOOCH2CH2CH3
HCOOCH(CH3)2
CH3COOC2H5
C2H5COOCH3
+)
Vdri
n = 4, m = 2, ta c6:
A:
C4H8O2 vdi
2
c6ng
thufc
cau tao cua
axit
CH3-CH2-CH2-COOH
va
CH3-CH(CH3)-COOH
B: C2H4O2
c6
cong thufc
cau tao la:
HCOOCH3.
Bdl
DUdNG
HOA

HOCI2
ai
17. a)
Phifcfng
phdp:
Ta
tinh
MA
=
MB
= Mc de cd mot he
thvtc
giffa
x,
y,
z
(CMyOz).
Cho z= 1, 2,
tinh
x, y.
X
vk
pentan khong
d
cung
t° nen ta
dufa
ve
cung dieu kien
t° =

273°C,
p
= 2
atm.
DUa
ve
dieu kien
t", P
nay,
the
tich mdi
cua
hon horp
X la:
_
-
273
+ 273
273
+ 136,5 1,5
V.
i
vay d
cung
t°, P, 4 gam
hon hcfp
X c6 the
tich b^ng
Ian the
tich

1,5
cua 3 gam
C5H12.
Vay so" mol nx =
n„„
(mo
1,5
^="'2
nx
=
1,5
Mx
= 4 :
72
72
(mol)
72
=
72 =
MA
=
MB
= Mc
(A,
B, C
dong phan
c6
cung
c5ng
thufc phan tuf, cung

M nen Mx =
MA
=
MB
=
Mc)
Vori
cong thufc 1^ C^HyO^,
ta c6: 12x + y + 16z = 72
1
phifong
trinh,
3 an nen ta
Ian lucft
cho z = 1, 2. Suy ra x, y.
+)
Khi
z = 1: 12x + y + 16 = 72 => 12x + y = 56
X 3
4
5
y
20
8
am
Iky cong thufc phan tuf
la
C4H8O.
I)
Khi

z = 2: 12x + y + 32 = 72 => 12x + y = 40
X 2
3
4
y
16
4
am
Iky cong thufc phan tuf
la
C3H4O2.
b)
De
chon giufa
2
cong thufc C4H8O
va
C3H4O2
ta
diia tren
tinh
chat
la
X
tac
dung vdi NaOH
cho ra
chat
ran
(muo'i)

va
hoi
c6
chufa andehit (tCr
rifou
khong ben bien thanh).
Do do A,
B, C
phai
c6 2
nguyen tCf oxi
vk
cong thufc phan tuf dung
la
C3H4O2.
Cong thufc
cau tao
C3H4O2 thuoc cong thufc tong quat
la
C„H2n
+ 2
-
4O2
nen A,
B, C c6 2
lign
ket
TT.
Vay
CO

3
cong thufc
cau tao:
A:
CH2=CH-C00H (axit khong
no)
B: H-C00-CH=CH2
(este
khong
no)
C: CHO-CH2-CHO
(2
chufc andehit)
Loai CH3-CO-CHO vi
c6 2
loai nhom chufc
.^.dlfil
DUSNn
HhA Hnn
19
45
+)
Phan
tfng
vdri
NaOH.
Dat a =
HA,
b = n^, c = nc
CH2=CH-C00H

+ NaOH
(mol) a a
HCOOCH=CH2 + NaOH -
(mol) b b
-> CH2=CH-COONa + HgO
a
HCOONa +
CH3CHO
b
b
Vay chat ran Y gom 2 mudi CH2=CH-C00Na v^ HCOONa, con h6n
hop hori Z chufa 2 andehit CH3CHO va CHO-CH2-CHO
Vdri
dung dich AgNOa/NHs, 2 andehit cho phan
ijfng:
(thu gon)
CH3CHO > CH3COONH4 + 2Ag;
b
• b 2b
CHO-CH2-CHO
> CH2(COONH4)2 + 4Agi
c c 4c
(mol)
(mol)
nAg
= 2b + 4c = = 1
^ 108
(1)
Dung dich Z' chijfa 2 chat hOfu ccf la
CH3COONH4

va
CH2(COONH4)2
Khi
dien phan d anot ta c6 phan
ijfng
oxi hoa 2 anion.
2CH.,C00
- 2e
(mol)
(mol)
-> 2C02t + CzHet
0,5b
2 OOC - CH„ - COO
2e
->
4CO2
t + C2H4T
0,5c
HSn hgp khi F gom 2
chat
hufu
ccf
CgHe
(- mol); C2H4 (- mol) va CO2.
2 2
Khi
nung
chat
rSn Y
vdri

NaOH, ta c6:
• CH2=CH-C00Na + NaOH
(mol) a a
HCOONa +
NaOH
> Hgt + Na2C03
(mol) b b
>
C2H4t
+ Na2C03
Hon hap khi G gom C2H4 (a mol) v^ Hz (b mol). Khi dun G vdi Ni, ta c6:
C2H4 + H2 > C2H6
+) Neu b > a, ta
dtfoc
(I): C2H6 (a mol) va H2 dif (b - a) mol
+) Neu b < a, ta difcfc (H): C2H6 (b mol) va C2H4 dii (a - b) mol
Meu la triidng hap 7; a = b - a b = 2a (2)
iVeu la triidng /jgfp //; b = a - b => a = 2b (2')
Khi
tron chung F va F' roi cho qua dung dich Br2 chi c6 C2H4 bi giCf lai.
1
75
mc,H.
= 1-75 (gam) ^ n,^„^ = =
0,0625
(mol)
+) Neu la trtfofng
hofp
I (dif H2), C2H4 chi do F cung cap.
n„ „ = - =

0,0625
(mol) ^ c = 0,125 (mol)
TO
(1)
^
b
=
=
1^140025
^
Q^25
(mol)
B
(3)
TCr (2) ^ a = - = ^ = 0,125 (mol) A
36
Kiem
chtJng: a + b + c = 0,125 + 0,25 + 0,125 = 0,5 = —
(phti
hgp)
/ z
yay hon hgp X chufa: 0,125 mol CH2=CH-C00H
0,25 mol H-C00CH=CH2
0,125 mol CHO-CH2-CHO
0,125 X 100
%A = %C =
1,5
=
25% va %B = 50%
)

Trtfofng hdp II
C2H4
do F va G cung cap nen n^, „^ = a - b + | =
0,0625
(3')
ir
(2') va (3') ^ b + - =
0,0625
(mol)
2
;: hg phiWng
trinh:
2b + c = 0,125 va 2b + 4c = 1
c =
0,292
va b am loai triJcfng hgfp nay.
c) nNaOH = a + b -
0,375
(mol) => mNaOH =
0,375
x 40 = 15 (gam).
A7
CHl/dNG
II.
CACBOHIORAT
A.
KIEN
THLfC
CAN NH6
L

GLUCOZa
1.
Tinh
chat cua ancol da chvtc:
a)
Tdc dung vdi Cu(OH)2:
O nhiet do thifdng, glucozd da phdn vtng vdi Cu(OH)2 cho phifc dong
glucoza
CU(CBHUOB)2
titang tii nhiiglixerol.
2C6Hi20e
+ Cu(OH)2 >
(CBHi20e)2Cu
+
2H2O
b)
Phdn ling tao
este:
Glucoza
CO
the tao
este
chiia 5 goc axit axetic trong phdn tit khi tham
gia
phdn itng vdi anhidrit axetic [CHsCOJzO khi c6 mat
piridin.
2.
Tinh
chat cua andehit
dOn

ch^c:
a)
Oxi hoa glucozd bang dung dich AgNOj trong amoniac (phdn ling
trdng
bgc):
Dung dich AgNOj trong NH3 dd oxi hoa glucoza tao thanh muoi amoni
gluconat vd bgc kim logi bdm vdo thanh dng nghiem.
HOCH2[CHOH]4CHO
+ 2AgN03 +
3NH3
+ H2O —>
HOCH2[CHOH]4COONH4
+ 2Agi + 2NH4NO3
Amoni gluconat
b)
Oxi hoa bdng Cu(OH)2:
HOCH2lCHOH]4CHO
+
2Cu(OH)2
+ NaOH —>
HOCH2[CHOH]4COONa
+
Cu^ol
(do ggch) +
3H2O
natri
gluconat
c)
Khiiglucoza bdng hidro:
CH20H[CHOH]4CHO

+ H2 —>
CH20H[CHOH]4CH20H
sobitol
3. Phan ling len men: Khi cd enzim xiic tdc, glucoza trong dung dich
len men cho ancol etylic vd khi cacbonic:
CoHM
> 2C2H,OH +
2C02"[
•
ll SACCAROZa - C,2H220„
1. Phan
iJtng
cua ancol da chvtc v&i mot so hidroxit kim logi:
Trong
dung dich, saccaroza phdn itng vdi Cu(OH)2 cho dung dich ddng
saccarat mau xanh lam. Saccaroza tdc dung vdi vdi sifa cho cai^xi
sacarat tan trong nifdc.
Tinh
chat nay du'cfc dp dung trong qua
trinh
sdn
xudt vd tinh che difdng.
2. Phdn ihig thuy phan:
C,2H220u
+ H2O >
CaHj20o
+
CoH,20r,
glucoza fructoza
Phdn

ii'ng thuy phdn saccaroza cQng xdy ra khi c6 xuc tdc enzim.
III.
TINH
BOT,
(CeHu,Os)n
1. Phan vtng thuy phan:
Dun
nong tinh bot trong dung dich axit vd ca lodng se thu diiac glucoza
(CrMwO,}n
+
nH20
> nCoH,20,
Trong
ca the ngitai vd dong vat, tinh hot bi thuy phan tgo thanh
glucoza nhd cdc enzim.
2.
Phdn
vtng
mau v&i iot:
Do cdu tgo mgch d dgng xodn cd 16 rong, tinh bot hap thu iot cho
mail
xanh luc. Khi dun nong thi mau xanh bi mat, de ngugi thi mau
xanh
xudt hien.
IV.
XENLULOZa
1. Phan vfng thiiy phan
(CoH^oOs),, +
nH20
^"'"^ >

nCBHj20B
glucoza
2. Phdn
ii'ng
este hoa vdi axit
nitric:
[CsHy02(OH)3]n +
SuHNOj
, ^"•"•^^"4dac , [c,Hy02(ON02)3]n + 3nH20
xenluloza trinitrat
Xenluloza trinitrat rat de chdy vd no mgnh khong sinh ra khdi nen no
du'gc dimg 1dm thudc sung khong khdi.
B.BAI
TAP AP
DUNG
^^i
1. Viet cac phan
tfng
theo sof do chuyen ddi sau day:
Saccaroza
canxi
saccarat
saccarozcf
glucozo
n/gfu etylic
axit
axetic —>
natri
axetat metan andehit fomic.
^ai

2. Viet phi/ang
trinh
phan ufng theo scf do tao thanh va
chuy§'n
hoa
tinh
hot sau day:
,0
CO2
—(CeHioOsX,
—^
C12H22OU
—^
C6H12O6
—C2H5OH
^>
DUONG
HrtA
Hfir.
19