Câu hỏi 4

\(\begin{gathered}
{n_{O{H^ - }}} = 2{n_{{H_2}}} = 2.0,6 = 1,2mol \hfill \\
{m_O} = 86,3.\frac{{19,47}}{{100}} = 16,8gam \to {n_O} = 1,05mol \hfill \\
\to {n_{A{l_2}{O_3}}} = \frac{{{n_O}}}{3} = 0,35mol \hfill \\
A{l_2}{O_3} + 2O{H^ - } \to 2Al{O_2}^ - + {H_2}O \hfill \\
0,35 \to 0,7 \to \,\,\,\,\,\,\,\,0,7 \hfill \\
DD\,Y\left\{ \begin{gathered}
O{H^ - }du:1,2 - 0,7 = 0,5 \hfill \\
Al{O_2}^ - :0,7 \hfill \\
\end{gathered} \right.\xrightarrow{{ + HCl:2,4}} \hfill \\
O{H^ - } + H + \to {H_2}O \hfill \\
0,5 \to 0,5 \hfill \\
Al{O_2}^ - + {H^ + } + {H_2}O \to Al{(OH)_3} \hfill \\
0,7 \to 0,7 \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,7 \hfill \\
3{H^ + } + Al{(OH)_3} \to A{l^{3 + }} + 3{H_2}O \hfill \\
1,2 \to 0,4 \hfill \\
\to {n_{Al{{(OH)}_3}sau\,pu}} = 0,7 - 0,4 = 0,3 \hfill \\
\to {m_ \downarrow } = 0,3.78 = 23,4gam \hfill \\
\end{gathered} \)

Đáp án B