Em tham khảo nha :

\(\begin{array}{l}
1)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{Al}} = \dfrac{m}{M} = \dfrac{{5,4}}{{27}} = 0,2mol\\
{n_{{H_2}S{O_4}}} = {C_M} \times V = 0,2 \times 3 = 0,6mol\\
\frac{{0,2}}{2} < \frac{{0,3}}{3} \Rightarrow {H_2}S{O_4}\text{ dư}\\
{n_{{H_2}}} = \dfrac{3}{2}{n_{Al}} = 0,3mol\\
{V_{{H_2}}} = n \times 22,4 = 0,3 \times 22,4 = 6,72l\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{{n_{Al}}}}{2} = 0,1mol\\
{C_{{M_{A{l_2}{{(S{O_4})}_3}}}}} = \dfrac{n}{V} = \dfrac{{0,1}}{{0,2}} = 0,5M\\
{n_{{H_2}S{O_4}d}} = {n_{{H_2}S{O_4}}} - \dfrac{3}{2}{n_{Al}} = 0,3mol\\
{C_{{M_{{H_2}S{O_4}}}}} = \dfrac{n}{V} = \dfrac{{0,3}}{{0,2}} = 1,5M\\
2)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
{n_{{H_2}}} = \dfrac{{5,04}}{{22,4}} = 0,225mol\\
hh:Al(a\,mol),Mg(b\,mol)\\
\frac{3}{2}a + b = 0,225(1)\\
27a + 24b = 4,5(2)\\
 \text{Từ (1) và (2)}\Rightarrow a = 0,1;b = 0,075\\
{m_{Al}} = 0,1 \times 27 = 2,7g\\
{m_{Mg}} = 4,5 - 2,7 = 1,8g\\
3)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{Fe}} = \dfrac{{0,56}}{{56}} = 0,01mol\\
{n_{{H_2}S{O_4}}} = {n_{Fe}} = 0,01mol\\
{m_{{H_2}S{O_4}}} = 0,01 \times 98 = 0,98g\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{0,98 \times 100}}{{16,9}} = 5,8g\\
{n_{FeS{O_4}}} = {n_{{H_2}}} = {n_{Fe}} = 0,01mol\\
{m_{FeS{O_4}}} = 0,01 \times 152 = 1,52g\\
{m_{{\rm{dd}}spu}} = 0,56 + 5,8 - 0,01 \times 2 = 6,34g\\
C{\% _{FeS{O_4}}} = \dfrac{{1,52}}{{6,34}} \times 100\%  = 23,97\% \\
{V_{{H_2}}} = n \times 22,4 = 0,01 \times 22,4 = 0,224l
\end{array}\)