\(Mg+2HCl\rightarrow MgCl_2+H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\\ n_{HCl}=\dfrac{29,2}{36,5}=0,8\left(mol\right)\\ n_{H_2}=\dfrac{0,8}{2}+0,2=0,5\left(mol\right)\\ V_{H_2\left(\text{đ}ktc\right)}=0,5.22,4=11,2\left(l\right)\)