\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ a.Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{Mg}=n_{H_2}=0,2\left(mol\right)\\ \Rightarrow m_{Mg}=0,2.24=4,8\left(g\right)\\ b.n_{HCl}=0,2.2=0,4\left(mol\right)\\ m_{HCl}=0,4.36,5=14,6\left(g\right)\\ C\%_{ddHCl}=\dfrac{14,6}{50}.100=29,2\%\)