Em tham khảo nha :
\(\begin{array}{l}
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{Al}} = \dfrac{{5,4}}{{27}} = 0,2mol\\
{n_{{H_2}S{O_4}}} = 0,2 \times 1,35 = 0,27mol\\
\dfrac{{0,2}}{2} > \dfrac{{0,27}}{3} \Rightarrow Al\text{ dư}\\
{n_{A{l_d}}} = 0,2 - \dfrac{{0,27 \times 2}}{3} = 0,02mol\\
{m_{A{l_d}}} = 0,02 \times 27 = 0,54g\\
b)\\
{n_{{H_2}}} = {n_{{H_2}S{O_4}}} = 0,27mol\\
{V_{{H_2}}} = 0,27 \times 22,4 = 6,048l
\end{array}\)