Em tham khảo nha:

\(\begin{array}{l}
4)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,4\,mol\\
{m_{Fe}} = 0,4 \times 56 = 22,4g\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,8\,mol\\
{m_{HCl}} = 0,8 \times 36,5 = 29,2g\\
5)\\
\text{ Gọi a là khối lượng của Mg }\\
{m_{Al}}:{m_{Mg}} = 9:4 \Rightarrow {m_{Al}} = \dfrac{9}{4} \times a = 2,25a\,g\\
{m_{hh}} = 46,8g \Leftrightarrow a + 2,25a = 46,8 \Rightarrow a = 14,4\\
2Mg + {O_2} \xrightarrow{t^0} 2MgO\\
4Al + 3{O_2} \xrightarrow{t^0} 2A{l_2}{O_3}\\
{n_{Mg}} = \dfrac{{14,4}}{{24}} = 0,6\,mol\\
{n_{Al}} = \dfrac{{14,4 \times 2,25}}{{27}} = 1,2\,mol\\
{n_{{O_2}}} = \dfrac{{0,6}}{2} + 1,2 \times \dfrac{3}{4} = 1,2\,mol\\
{V_{{O_2}}} = 1,2 \times 22,4 = 26,88l
\end{array}\)