Đáp án:
\(\begin{array}{l}
b)\\
{V_{{H_2}}} = 13,44l\\
c)\\
{C_M}{H_2}S{O_4} = 1,2M\\
{C_M}A{l_2}{(S{O_4})_3} = 0,4M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
b)\\
{n_{Al}} = \dfrac{m}{M} = \dfrac{{10,8}}{{27}} = 0,4\,mol\\
{n_{{H_2}}} = \dfrac{{0,4 \times 3}}{2} = 0,6\,mol\\
{V_{{H_2}}} = 0,6 \times 22,4 = 13,44l\\
c)\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,6\,mol\\
{C_M}{H_2}S{O_4} = \dfrac{{0,6}}{{0,5}} = 1,2M\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{0,4}}{2} = 0,2\,mol\\
{C_M}A{l_2}{(S{O_4})_3} = \dfrac{{0,2}}{{0,5}} = 0,4M
\end{array}\)