a) $Zn + 2HCl \to ZnCl_2 + H_2$
Theo PTHH : $n_{H_2} = n_{Zn} = \dfrac{16,25}{65} = 0,25(mol)$
$V_{H_2} = 0,25.22,4 = 5,6(lít)$
b) $n_{HCl} = 2n_{Zn} = 0,5(mol)$
$C\%_{HCl} = \dfrac{0,5.36,5}{150}.100\% = 12,17\%$
c) $m_{dd\ sau\ pư} = m_{Zn} + m_{dd\ HCl} - m_{H_2} = 16,25 + 150 - 0,25.2 = 165,75(gam)$
$C\%_{ZnCl_2} = \dfrac{0,25.136}{165,75}.100\% = 20,51\%$