PHƯƠNG PHÁP QUY ĐỔI ESTE THÔNG QUA ĐỀ THI THPTQG 2019
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D c a a = ch n C<br />
ng d n gi i 2<br />
( C n H 2n-2 O 4 )<br />
( C m H 2m-10 O 6 )<br />
R 1 COO<br />
R 2 COO<br />
x mol<br />
R 3 COO<br />
R 3 COO C 3 H 5<br />
R 3 COO<br />
y mol<br />
C 3 H 6<br />
( Gi s ng ch t TN1 g p k l m 2)<br />
Ta d<br />
x + y = 0,16k<br />
2x + 3y =0,42 k<br />
x = 0,06k<br />
y = 0,1 k<br />
O 2<br />
0,5 mol<br />
NaOH<br />
0,42 k mol<br />
t = 4x + 6y + 0,1<br />
Theo s ch s mol CO2 2 s :<br />
x 5y<br />
CO 2 H 2O<br />
T (1)(2) gi i h c<br />
x<br />
y<br />
=<br />
3<br />
5<br />
(1)<br />
CO 2<br />
0,45 mol<br />
H 2 O<br />
t mol<br />
n n 0,45 - 4x - 6y - 0,1 = x + 5y 5x + 11y = 0,35 (2)<br />
x = 0,015<br />
y = 0,025<br />
Ch c c p nghi m :<br />
n = 10<br />
m = 12<br />
0,015.n + 0,025.m = 0,045<br />
D c a = 12,36<br />
202-2018): H n h p X g<br />
gam X thu c 1,56 mol CO 2 2O. M ng v v i 0,09 mol NaOH trong<br />
d ch ch ch a a gam h n h p mu i natri panmitat, natri stearat. c<br />
A. 25,86. B 26,40. C 27,70. D 27,30.<br />
ng d n gi i 1<br />
i X:<br />
C 15 H 31 COOH<br />
(C 15 H 31 COO) 3 C 3 H 5<br />
CH 2<br />
bi<br />
x<br />
y<br />
z<br />
C 15 H 31 COOH<br />
(C 15 H 31 COO) 3 C 3 H 5<br />
CH 2<br />
B<br />
D ch s mol CO2 2<br />
2y =1,56-<br />
B<br />
V y mu i :<br />
C 15 H 31 COONa 0,09<br />
CH 2 0,06<br />
C 15 H 31 COOH<br />
x mol<br />
C 17 H 35 COOH<br />
y mol<br />
(RCOO) 3 C 3 H 5<br />
z mol<br />
m=24,64 gam<br />
(C n H 2n - 4 O 6 )<br />
Ta de co : x + y + 3z = 0,09<br />
O 2<br />
CO 2<br />
NaOH<br />
0,09 mol<br />
H 2 O<br />
1,56 mol<br />
1,52 mol<br />
Muoi<br />
H 2 O<br />
c a =25,86 gam ch n A<br />
ng d n gi i 2<br />
O<br />
4,46 mol<br />
( 1)<br />
(2)<br />
c : nO = 4,46<br />
x<br />
C 3 H 5 (OH) 3<br />
CO 2<br />
1,56 mol<br />
H 2 O<br />
1,52 mol<br />
NaOH<br />
0,09 mol<br />
y<br />
C 3 H 8 O 3<br />
z<br />
H 2 O ( x + y)<br />
Muoi