a,
Gọi $x$, $y$ là số mol $C_2H_5OH$ và $CH_3COOH$ trong X.
- P1: $n_{C_2H_5OH}=0,25x (mol); n_{CH_3COOH}=0,25y (mol)$
$n_{H_2}=\dfrac{0,28}{22,4}=0,0125(mol)$
$2C_2H_5OH+2Na\to 2C_2H_5ONa+H_2$
$2CH_3COOH+2Na\to 2CH_3COONa+H_2$
$\Rightarrow 0,25x+0,25y=0,0125.2$ $(1)$
- P2: $n_{C_2H_5OH}=0,75x (mol); n_{CH_3COOH}=0,75y (mol)$
$n_{H_2O}=0,18(mol)$
$C_2H_5OH+3O_2\to 2CO_2+3H_2O$
$CH_3COOH+2O_2\to 2CO_2+2H_2O$
$\Rightarrow 3.0,75x+2.0,75y=0,18$ $(2)$
$(1)(2)\Rightarrow x=0,04; y=0,06$
$m_{C_2H_5OH}=46x=1,84g$
$m_{CH_3COOH}=60y=3,6g$
b,
$C_2H_5OH+CH_3COOH\buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5+H_2O$
Theo lí thuyết tạo $0,04$ mol este.
$\to m_{\text{este}}=0,04.88.80\%=2,816g$