Em tham khảo nha :
\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{Al}} = \dfrac{{5,4}}{{27}} = 0,2mol\\
{n_{{H_2}}} = \dfrac{3}{2}{n_{Al}} = 0,3mol\\
{V_{{H_2}}} = 0,3 \times 22,4 = 6,72l\\
b)\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
{n_{{H_2}S{O_4}}} = 0,15 \times 1 = 0,15mol\\
{n_{Fe}} = \dfrac{{{n_{{H_2}S{O_4}}}}}{3} = 0,05mol\\
{m_{Fe}} = 0,05 \times 56 = 2,8g
\end{array}\)
