$n_{H^+}=n_{HCl}+2n_{H_2SO_4}=0,5.0,1+0,5.2a=0,05+a(mol)$
Bảo toàn e: $2n_{Mg}+3n_{Al}=n_{H^+}$
$\Rightarrow 0,05+a=0,18.3+0,12.2$
$\Leftrightarrow a=0,73$
$n_{Cl^-}=n_{HCl}=0,05(mol)$
$n_{SO_4^{2-}}=n_{H_2SO_4}=0,5.0,73=0,365(mol)$
$\to m_{\text{muối}}=m_{Al^{3+}}+m_{Mg^{2+}}+m_{Cl^-}+m_{SO_4^{2-}}$
$=m_{Al}+m_{Mg}+m_{Cl^-}+m_{SO_4^{2-}}$
$=0,18.27+0,12.24+0,05.35,5+0,365.96$
$=44,555g$