a) $n_{HCl} = \dfrac{219.5\%}{36,5} = 0,3(mol)$
$ZnO + 2HCl \to ZnCl_2 + H_2O$
Theo PTHH : $n_{ZnO} = \dfrac{1}{2}n_{HCl} = 0,15(mol)$
$m = 0,15.81 = 12,15(gam)$
b) $m_{dd\ sau\ pư} = 12,15 + 219 = 213,15(gam)$
$n_{ZnCl_2} = n_{ZnO} = 0,15(mol)$
$C\%_{ZnCl_2} = \dfrac{0,15.136}{213,15}.100\% = 8,82\%$
c) $ZnO + H_2SO_4 \to ZnSO_4 + H_2$
$n_{H_2SO_4} = n_{ZnO} = 0,15(mol)$
$V_{dd\ H_2SO_4} = \dfrac{0,15}{1,5} = 0,1(lít)$