Câu hỏi 4

$\begin{gathered}
{m_O} = 84.0,2 = 16,8\,gam\, \Rightarrow {n_O} = 1,05mol \Rightarrow {n_{A{l_2}{O_3}}} = 0,35mol. \hfill \\
{n_{OH}} = 2{n_{{H_2}}} = 1,2mol \hfill \\
A{l_2}{O_3} + 2O{H^ - }\xrightarrow{{}}2Al{O_2}^ - + {H_2}O \hfill \\
\end{gathered} $

Vậy OH^- dư.

${n_{Al{O_2}^ - }} = 0,35.2 = 0,7\,mol$ , n_OH^- dư = 1,2- 0,35.2=0,5 mol.

 ${n_{{H^ + }}} = 3,2.0,75 = 2,4mol;\,{H^ + } + O{H^ - }\xrightarrow{{}}{H_2}O$

=> n_H⁺dư = 2,4-0,5=1,9mol.

$\begin{gathered}
{H^ + } + Al{O_2}^ - + {H_2}O\xrightarrow{{}}Al{(OH)_3} \downarrow ;Al{(OH)_3} + 3{H^ + }\xrightarrow{{}}A{l^{3 + }} + 3{H_2}O. \hfill \\
\Rightarrow {n_{Al{{(OH)}_3}}} = (4.0,7 - 1,9)/3 = 0,3mol \Rightarrow m = 0,3.78 = 23,4gam \hfill \\
\end{gathered} $

Đáp án A