`n_{HCl}=0,5.1=0,5(mol)`
`n_{H_2SO_4}=0,5.0,28=0,14(mol)`
`n_{H_2}=\frac{8,736}{22,4}=0,39(mol)`
Cho $\begin{cases} Mg: x(mol)\\ Al: y(mol)\\\end{cases}$
BTe:
$\mathop{Mg}\limits^{0}\to \mathop{Mg}\limits^{+2}+2e$
$\mathop{Al}\limits^{0}\to \mathop{Al}\limits^{+3}+3e$
$\mathop{2H}\limits^{+}+2e\to \mathop{H_2}\limits^{0}$
Do $n_{\mathop{H}^{+}}=n_{H_2}( =0,78)$
`=>` Phản ứng xảy ra đủ
`=> 2x+3y=0,78(mol)(2)`
`(1),(2)=> x=0,12(mol), y=0,18(mol)`
`m_{\text{muối}}=m_{\text{hh kim loại}}+ m_{Cl^{-}}+m_{SO_4^{2-}}`
`=> m_{\text{muối}}= 7,74+0,14.96+0,5.35,5=38,93g`