a,
`n_{HCl}=\frac{150.2,34%}{100.36,5%}=0,096\ (mol)`
`Mg+2HCl\to MgCl_2+H_2`
`Fe+2HCl\to FeCl_2+H_2`
Theo PT: `n_{H_2}=1/2 n_{HCl}=0,048\ (mol)`
`=>V_{H_2}=0,048.22,4=1,0752\ (l)`
b,
Đặt: $\left\{\begin{matrix}n_{Mg}=x\ (mol)\\n_{Fe}=y\ (mol)\end{matrix}\right.$
`=>x+y=n_{H_2}=0,048\ \ (1)`
Mặt khác: $\left\{\begin{matrix}n_{MgCl_2}=n_{Mg}=x\ (mol)\\n_{FeCl_2}=n_{Fe}=y\ (mol)\end{matrix}\right.$
`=>95a+127b=2,16\ \ (2)`
Giải hệ `(1)`, `(2)` ra số âm, bạn xem lại đề.