Đặt : nCH3OH=a(mol); nCH3COOH=b(mol)
PTHH: CH3OH + Na -> CH3ONa + 1/2 H2
a__________a________a____0,5a(mol)
CH3COOH + Na -> CH3COONa + 1/2 H2
b__________b______b_________0,5b(mol)
Ta có hpt:
\(\left\{{}\begin{matrix}32a+60b=20\\0,5a+0,5b=0,225\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,25\\b=0,2\end{matrix}\right.\)
=>nCH3OH=0,25(mol); nCH3COOH=0,2(mol)
15 gam X sẽ có: 0,15 mol CH3COOH và 0,1875 mol CH3OH.
PTHH: CH3COOH + CH3OH ---H+,to--> CH3COOCH3 + H2O
Vì: 0,15/1 < 0,1875/1
=> Tính theo nCH3COOH.
nCH3COOCH3(LT)= nCH3COOH=0,15(mol)
Mặt khác: nCH3COOCH3(TT)= 7,215/74=0,0975(mol)
=>H=(0,0975/0,15).100=65%