Em tham khảo nha :
\(\begin{array}{l}
4)\\
2Al + 6{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
{n_{Al}} = \dfrac{{5,4}}{{27}} = 0,2mol\\
{n_{S{O_2}}} = \dfrac{3}{2}{n_{Al}} = 0,3mol\\
{V_{S{O_2}}} = 0,3 \times 22,4 = 6,72l\\
5)\\
Fe + 4HN{O_3} \to Fe{(N{O_3})_3} + NO + 2{H_2}O\\
3Mg + 8HN{O_3} \to 3Mg{(N{O_3})_2} + 2NO + 2{H_2}O\\
{n_{NO}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
hh:Fe(a\,mol),Mg(b\,mol)\\
a + \dfrac{2}{3}b = 0,3(1)\\
56a + 24b = 15(2)\\
\text{Từ (1) và (2)}\Rightarrow a = 0,21;b = 0,135\\
{m_{Fe}} = 0,21 \times 56 = 11,76g\\
{m_{Mg}} = 0,135 \times 24 = 3,24g
\end{array}\)
