Đáp án:
\({V_{{H_2}}} = 2,24l\)
\({m_{Al{{(OH)}_3}}} = 5,2g\)
Giải thích các bước giải:
\(\begin{array}{l}
K + {H_2}O \to KOH + \dfrac{1}{2}{H_2}\\
6KOH + A{l_2}{(S{O_4})_3} \to 3{K_2}S{O_4} + 2Al{(OH)_3}\\
{n_K} = 0,2mol\\
\to {n_{KOH}} = {n_K} = 0,2mol\\
\to {n_{{H_2}}} = \dfrac{1}{2}{n_K} = 0,1mol\\
\to {V_{{H_2}}} = 2,24l\\
{m_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{200 \times 25}}{{100}} = 50g\\
\to {n_{A{l_2}{{(S{O_4})}_3}}} = 0,146mol\\
\to \dfrac{{{n_{KOH}}}}{6} < {n_{A{l_2}{{(S{O_4})}_3}}}
\end{array}\)
Suy ra \(A{l_2}{(S{O_4})_3}\) dư
\(\begin{array}{l}
\to {n_{Al{{(OH)}_3}}} = \dfrac{1}{3}{n_{KOH}} = \dfrac{{0,2}}{3}mol\\
\to {m_{Al{{(OH)}_3}}} = 5,2g
\end{array}\)
