Đáp án:
5,1 g
Giải thích các bước giải:
${n_{NaCl}} = 0,05mol;{n_{AlC{l_3}}} = 0,5.0,3 = 0,15mol$
$\begin{gathered}
2NaCl + 2{H_2}O\xrightarrow[{cmn}]{{dp{\text{dd}}}}2NaOH + {H_2} + C{l_2} \hfill \\
0,05{\text{ }} \to {\text{ 0,05 0,025 0,025}} \hfill \\
2AlC{l_3} + 6{H_2}O\xrightarrow[{cmn}]{{dp{\text{dd}}}}2Al{(OH)_3} + 3{H_2} + 3C{l_2} \hfill \\
0,15{\text{ }} \to {\text{ 0,15 0,225 0,225}} \hfill \\
{\text{NaOH + Al(OH}}{{\text{)}}_3} \to NaAl{O_2} + 2{H_2}O \hfill \\
{\text{ 0,05 }} \to {\text{0,05}} \hfill \\
\end{gathered} $
Chất rắn không tan là: $Al{(OH)_3}(0,1{\text{ mol)}}$
$2Al{(OH)_3}\xrightarrow{{{t^o}}}A{l_2}{O_3} + 3{H_2}O$
${n_{A{l_2}{O_3}}} = \dfrac{1}{2}{n_{Al{{(OH)}_3}}} = 0,1.\dfrac{1}{2} = 0,05mol$
$ \Rightarrow {m_{A{l_2}{O_3}}} = 0,05.102 = 5,1\left( g \right)$
