a)
$n_{BaO} = \dfrac{30,6}{153} = 0,2(mol)$
$BaO + 2HCl \to BaCl_2 +H_2O$
Theo PTHH, $n_{HCl} = 2n_{BaO} = 0,2.2 = 0,4(mol)$
$\Rightarrow m_{dd\ HCl} = \dfrac{0,4.36,5}{15\%} = 97,33(gam)$
b) $n_{BaCl_2} =n_{BaO} = 0,2(mol)$
$m_{dd\ sau\ pư} = m_{BaO} + m_{dd\ HCl} = 30,6 + 97,33=127,93(gam)$
$C\%_{BaCl_2} = \dfrac{0,2.208}{127,93}.100\% =32,52\%$