Đáp án:
\(\begin{array}{l}
{C_\% }{H_2}S{O_4} \text{ dư }= 4,79\% \\
{C_\% }A{l_2}{(S{O_4})_3} = 16,7\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{Al}} = \dfrac{{5,4}}{{27}} = 0,2\,mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{200 \times 19,6\% }}{{98}} = 0,4\,mol\\
\dfrac{{{n_{Al}}}}{2} < \dfrac{{{n_{{H_2}S{O_4}}}}}{3} \Rightarrow {H_2}S{O_4} \text{ dư }\\
{n_{{H_2}S{O_4}}} \text{ dư }= 0,4 - 0,2 \times \dfrac{3}{2} = 0,1\,mol\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{0,2}}{2} = 0,1\,mol\\
{n_{{H_2}}} = 0,2 \times \dfrac{3}{2} = 0,3\,mol\\
{m_{{\rm{dd}}spu}} = 5,4 + 200 - 0,3 \times 2 = 204,8g\\
{C_\% }{H_2}S{O_4} \text{ dư } = \dfrac{{0,1 \times 98}}{{204,8}} \times 100\% = 4,79\% \\
{C_\% }A{l_2}{(S{O_4})_3} = \dfrac{{0,1 \times 342}}{{204,8}} \times 100\% = 16,7\%
\end{array}\)
