Giải thích các bước giải:
\(\begin{array}{l}
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + {H_2}\\
A{l_2}{O_3} + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}O\\
2Al{(OH)_3} + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 6{H_2}O\\
{m_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{273,75 \times 21,863}}{{100}} = 59,85g\\
\to {n_{A{l_2}{{(S{O_4})}_3}}} = 0,175mol\\
\to {n_{Al}}(trong hỗn hợp ban đầu) = 2{n_{A{l_2}{{(S{O_4})}_3}}} = 0,35mol\\
{n_{{H_2}S{O_4}}} = \dfrac{3}{2}{n_{Al}} = 0,525mol\\
\to {m_{{H_2}S{O_4}}} = 51,45g\\
\to {m_{{H_2}S{O_4}}}{\rm{dd = }}\dfrac{{51,45 \times 100}}{{20}} = 257,25g\\
{n_{{H_2}}} = \dfrac{{5,04}}{{22,4}} = 0,225mol \to {m_{{H_2}}} = 0,45g\\
m + {m_{{H_2}S{O_4}}}{\rm{dd = }}{{\rm{m}}_{A{l_2}{{(S{O_4})}_3}}} + {m_{{H_2}}} \to m = 273,75 + 0,45 - 257,25 = 16,95g
\end{array}\)
