Alexei78
Дано m(CH3COOH)=120 g m(C2H5OH)=69 g --------------------------- m(CH3COOC2H5)-? 120 69 X CH3COOH+C2H5OH-->CH3COOC2H5+H2O 46 88 M(CH3COOH)=60 g/mol M(C2H5OH)=46g/mol n(CH3COOH)=m/M=120/60=2 mol n(C2H5OH)=m/M=69/46=1.5 mol n(CH3COOH)>n(C2H5OH) M(CH3COOC2H5)=88 g/mol 69 / 46 = X/88 X=132 g ответ 132 г
Answers & Comments
nCH3COOH=100/Mr=100/60=1,67 моль
nC2H5OH=69/Mr=69/46=1,5моль
Решаем по недостатку
1,5<1,67
nCH3COOC2H5=1,5
mCH3COOC2H5=1,5*Mr=1,5*88=132г масса эфира
m(CH3COOH)=120 g
m(C2H5OH)=69 g
---------------------------
m(CH3COOC2H5)-?
120 69 X
CH3COOH+C2H5OH-->CH3COOC2H5+H2O
46 88
M(CH3COOH)=60 g/mol
M(C2H5OH)=46g/mol
n(CH3COOH)=m/M=120/60=2 mol
n(C2H5OH)=m/M=69/46=1.5 mol
n(CH3COOH)>n(C2H5OH)
M(CH3COOC2H5)=88 g/mol
69 / 46 = X/88
X=132 g
ответ 132 г