Em tham khảo nha:
\(\begin{array}{l}
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{Al}} = \dfrac{{5,4}}{{27}} = 0,2\,mol\\
{n_{{H_2}}} = 0,2 \times \frac{3}{2} = 0,3\,mol\\
{V_{{H_2}}} = 0,3 \times 22,4 = 6,72l\\
b)\\
{n_{{H_2}S{O_4}}} = 0,4 \times 1 = 0,4\,mol\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
\dfrac{{{n_{Al}}}}{2} < \dfrac{{{n_{{H_2}S{O_4}}}}}{3} \Rightarrow {H_2}S{O_4} \text{ dư tính theo Al }\\
{n_{{H_2}}} = 0,2 \times \dfrac{3}{2} = 0,3\,mol\\
{V_{{H_2}}} = 0,3 \times 22,4 = 6,72l
\end{array}\)
