Đặt \(\begin{cases} n_{Fe}=x(mol)\\ n_{Mg}=y(mol) \end{cases}\Rightarrow 56x+24y=8(1)\)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ Mg+2HCl\to MgCl_2+H_2\\ \Rightarrow x+y=0,2(2)\\ (1)(2)\Rightarrow \begin{cases} x=0,1(mol)\\ y=0,1(mol) \end{cases}\Rightarrow \begin{cases} m_{Fe}=0,1.56=5,6(g)\\ m_{Mg}=0,1.24=2,4(g) \end{cases} \)