Em tham khảo nha :

\(\begin{array}{l}
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
hh:Fe(a\,mol),Mg(b\,mol)\\
a + b = 0,15(1)\\
56a + 24b = 6,8(2)\\
 \text{Từ (1) và (2)}\Rightarrow a = 0,1;b = 0,05\\
{m_{Fe}} = 0,1 \times 56 = 5,6g\\
\% Fe = \frac{{5,6}}{{6,8}} \times 100\%  = 82,35\% \\
\% Mg = 100 - 82,35 = 17,65\% \\
b)\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
Mg + 2{H_2}S{O_4} \to MgS{O_4} + S{O_2} + 2{H_2}O\\
{n_{S{O_2}}} = \dfrac{3}{2}{n_{Fe}} + {n_{Mg}} = 0,2mol\\
{V_{S{O_2}}} = 0,2 \times 22,4 = 4,48l\\
{n_{{H_2}S{O_4}pu}} = 3{n_{Fe}} + 2{n_{Mg}} = 0,4mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{0,4 \times 110}}{{100}} = 0,44mol\\
{m_{{H_2}S{O_4}}} = 0,44 \times 98 = 43,12g\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{43,12 \times 100}}{{75}} = 57,493g
\end{array}\)