$\text{a)}\\Mg + 2HCl \to MgCl_2 + H_2 \\n_{Mg}=\dfrac{4,8}{24}=0,2\,(mol)\\\to n_{HCl}=2n_{Mg}=0,4\,(mol)\\\to m_{HCl}=0,4.36,5=14,6\,(g)\\\text{b)}\\n_{H_2}=n_{Mg}=0,2\,(mol)\\\to V_{H_2}=0,2.22,4=4,48\,(l)\\\text{c)}\\4H_2 + Fe_3O_4 \to 3Fe + 4H_2O\\n_{H_2}=0,2\,(mol)\\n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\,(mol)\\\text{Ta có: }\dfrac{n_{H_2}}{4}<\dfrac{n_{Fe_3O_4}}{1}\\\to \text{Chất rắn sau phản ứng gồm }Fe,\,{Fe_3O_4}_{\text{dư}}\\n_{{Fe_3O_4}_{\text{dư}}}=0,1-\dfrac{1}{4}n_{H_2}=0,05\,(mol)\\n_{Fe}=\dfrac{3}{4}n_{H_2}=0,15\,(mol)\\\to a=0,15.56+0,05.232=20\,(g)$