a,
Gọi $3x$, $3y$ là số mol $C_2H_5OH$ và $CH_3COOH$
Mỗi phần có $x$ mol $C_2H_5OH$, $y$ mol $CH_3COOH$
- P2:
$n_{NaOH}=0,1.2=0,2(mol)$
$CH_3COOH+NaOH\to CH_3COONa+H_2O$
$\to y=n_{CH_3COOH}=0,2(mol)$
- P1:
$n_{Na}=\dfrac{9,2}{23}=0,4(mol)$
$2CH_3COOH+2Na\to 2CH_3COONa+H_2$
$2C_2H_5OH+2Na\to 2C_2H_5ONa+H_2$
$\to x+y=0,4$
$\to x=0,4-0,2=0,2(mol)$
Vậy $m=46.3x+60.3y=63,6g$