Cho `Mg, Fe` lần lượt là `x, y ` mol.
Ta có: `24x+56y=5,2g(1)`
`n_{H_2}=\frac{3,36}{22,4}=0,15(mol)`
`1)` `Mg+2HCl\to MgCl_2+H_2`
`Fe+2HCl\to FeCl_2+H_2`
`=> n_{H_2}=x+y=0,15(mol)(2)`
`(1),(2)=> x=0,1(mol), y=0,05(mol)`
`%m_{Mg}=\frac{24.0,1.100%}{5,2}\approx 46,15%`
`%m_{Fe}=\frac{56.0,05.100%}{5,2}\approx53,84%`
`2)` Ta nhận thấy `n_{HCl}=2n_{H_2}=0,3(mol)`
`=> V_{HCl}=\frac{n}{CM}=\frac{0,3}{1}=0,3(l)`