a)
$n_{CH_3COOH} = \dfrac{50.5\%}{60} = \dfrac{1}{24}(mol)$
$CaO + 2CH_3COOH \to (CH_3COO)_2Ca + H_2O$
Theo PTHH :
$n_{(CH_3COO)_2Ca} = \dfrac{1}{2}n_{CH_3COOH} = \dfrac{1}{48}(mol)$
$\Rightarrow m_{muối} = \dfrac{1}{48}.158 = 3,292(gam)$
b) $n_{CaO} = \dfrac{1}{2}n_{CH_3COOH} = \dfrac{1}{48}(mol)$
$m_{dd\ sau\ pư} = m_{CaO} + m_{dd\ CH_3COOH} = \dfrac{1}{48}.56 + 50 = \dfrac{307}{6}(gam)$
$C\%_{(CH_3COO)_2Ca} = \dfrac{3,292}{\dfrac{307}{6}}.100\% = 6,43\%$