a) $n_{HCl} = \dfrac{109,5.10\%}{36,5} = 0,3(mol)$
$Mg + 2HCl \to MgCl_2 + H_2$
Theo PTHH :
$n_{Mg} = n_{H_2} = \dfrac{1}{2}n_{HCl} = 0,15(mol)$
$m_{Mg} = 0,15.24 = 3,6(gam)$
b) $V_{H_2} = 0,15.22,4 = 3,36(lít)$
c) $m_{dd\ sau\ pư} = m_{Mg} + m_{dd\ HCl} - m_{H_2} = 3,6 + 109,5 - 0,15.2 = 112,8(gam)$
d) $C\%_{MgCl_2} = \dfrac{0,15.95}{112,8}.100\% = 12,63\%$