Em tham khảo nha :
\(\begin{array}{l}
a)\\
{n_{{H_2}S{O_4}}} = 0,4 \times 3 = 1,2mol\\
{m_{{H_2}S{O_4}}} = 1,2 \times 98 = 117,6g\\
b)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{Al}} = \dfrac{2}{3}{n_{{H_2}S{O_4}}} = 0,8mol\\
{m_{Al}} = 0,8 \times 27 = 21,6g\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{{n_{{H_2}S{O_4}}}}}{3} = 0,4mol\\
{m_{A{l_2}{{(S{O_4})}_3}}} = 0,4 \times 342 = 136,8g\\
{n_{{H_2}}} = {n_{{H_2}S{O_4}}} = 1,2mol\\
{V_{{H_2}}} = 1,2 \times 22,4 = 26,88l
\end{array}\)