Giải thích các bước giải:
\(\begin{array}{l}
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
F{e_2}{O_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}O\\
{n_{{H_2}}} = 0,15mol\\
\to {n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,1mol\\
\to {m_{Al}} = 0,1 \times 27 = 2,7g\\
\to {m_{F{e_2}{O_3}}} = 18,7 - 2,7 = 16g \to {n_{F{e_2}{O_3}}} = 0,1mol\\
a)\\
\% {m_{Al}} = \dfrac{{2,7}}{{18,7}} \times 100\% = 14,44\% \\
\% {m_{F{e_2}{O_3}}} = \dfrac{{16}}{{18,7}} \times 100\% = 85,56\% \\
b)\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \frac{1}{2}{n_{Al}} = 0,05mol\\
{n_{F{e_2}{{(S{O_4})}_3}}} = {n_{F{e_2}{O_3}}} = 0,1mol\\
\to C{M_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{0,05}}{{0,5}} = 0,1M\\
\to C{M_{F{e_2}{{(S{O_4})}_3}}} = \dfrac{{0,1}}{{0,5}} = 0,2M\\
\end{array}\)
