Đáp án:
450 g
Giải thích các bước giải:
\(\begin{array}{l}
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
A{l_2}{O_3} + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}O\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_{Al}} = \dfrac{{0,15 \times 2}}{3} = 0,1\,mol\\
{m_{Al}} = 0,1 \times 27 = 2,7g\\
{m_{A{l_2}{O_3}}} = 12,9 - 2,7 = 10,2g\\
{n_{A{l_2}{O_3}}} = \dfrac{{10,2}}{{102}} = 0,1\,mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{0,1 \times 3}}{2} + 0,1 \times 3 = 0,45\,mol\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{0,45 \times 98}}{{9,8\% }} = 450g
\end{array}\)
