Giải thích các bước giải:
\(\begin{array}{l}
3.\\
AlC{l_3} + 3NaOH \to Al{(OH)_3} + 3NaCl\\
2Al{(OH)_3} \to A{l_2}{O_3} + 3{H_2}O\\
{n_{NaOH}} = 0,8mol\\
{n_{A{l_2}{O_3}}} = 0,05mol\\
\to {n_{Al{{(OH)}_3}}} = 2{n_{A{l_2}{O_3}}} = 0,1mol\\
\to {n_{Al{{(OH)}_3}}} < {n_{NaOH}} \to {n_{NaOH}}dư\\
\to {n_{AlC{l_3}}} = {n_{Al{{(OH)}_3}}} = 0,1mol\\
\to C{M_{AlC{l_3}}} = \dfrac{{0,1}}{{0,2}} = 0,5M
\end{array}\)
