\(n_{H2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{HCl}=2n_{H2}=2.0,2=0,4\left(mol\right)\)
BTKL: m kim loại+mHCl=m muối+mH2
\(\Rightarrow8+0,4.36,5=m_{muoi}+0,2,2\)
\(\Rightarrow m_{muoi}=22,2\left(g\right)\)