`m_{\text{tăng}}=m_{\text{hh kim loại}}-m_{H_2}`
`=> m_{\text{tăng}}=7,8-m_{H_2}`
`=> 7=7,8-m_{H_2}`
`=> m_{H_2}=0,8g`
`=> n_{H_2}=\frac{0,8}{2}=0,4(mol)`
Cho `Mg, Al` lần lượt là `x, y` mol
Ta có: `24x+27y=7,8g(2)`
Bảo toàn `e`
$\mathop{Mg}\limits^{0}\to \mathop{Mg}\limits^{+2}+2e \ \ || \ \ \mathop{2H}\limits^{+1}+2e \to \mathop{H_2}\limits^{0}$
$\mathop{Al}\limits^{0} \to \mathop{Al}\limits^{+3}+3e$
Ta có: `n_{H_2}=x+1,5y=0,4(mol)(2)`
`(1),(2)=> x=0,1(mol), y=0,2(mol)`
`=> %m_{Mg}=\frac{0,1.24.100%}{7,8}\approx 30,77%`
`=> %m_{Al}=\frac{27.0,2.100%}{7,8}\approx69,23%`