Đáp án:
a) 13,44l
b) 294g
c) 22,53%
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{Al}} = \dfrac{{10,8}}{{27}} = 0,4\,mol\\
{n_{{H_2}}} = 0,4 \times \frac{3}{2} = 0,6\,mol\\
{V_{{H_2}}} = 0,6 \times 22,4 = 13,44l\\
b)\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,6\,mol\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{0,6 \times 98}}{{20\% }} = 294g\\
c)\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{0,4}}{2} = 0,2\,mol\\
{m_{{\rm{dd}}spu}} = 10,8 + 294 - 0,6 \times 2 = 303,6g\\
{C_\% }A{l_2}{(S{O_4})_3} = \dfrac{{0,2 \times 342}}{{303,6}} \times 100\% = 22,53\%
\end{array}\)