Đáp án:
\(\begin{array}{l}
T{H_1}:{V_{NaOH}} = 1,2l\\
T{H_2}:{V_{NaOH}} = 2l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
{n_{AlC{l_3}}} = 0,2 \times 1,5 = 0,3\,mol\\
{n_{Al{{(OH)}_3}}} = \dfrac{{15,6}}{{78}} = 0,2\,mol\\
T{H_1}:\text{$AlCl_3$ dư}\\
AlC{l_3} + 3NaOH \to Al{(OH)_3} + 3NaCl\\
{n_{NaOH}} = 3{n_{Al{{(OH)}_3}}} = 0,2 \times 3 = 0,6\,mol\\
{V_{NaOH}} = \dfrac{{0,6}}{{0,5}} = 1,2l\\
T{H_2}:\text{NaOH dư}\\
AlC{l_3} + 3NaOH \to Al{(OH)_3} + 3NaCl(1)\\
Al{(OH)_3} + NaOH \to NaAl{O_2} + 2{H_2}O(2)\\
{n_{Al{{(OH)}_3}(1)}} = {n_{AlC{l_3}}} = 0,3\,mol\\
{n_{NaOH(1)}} = 3{n_{AlC{l_3}}} = 0,3 \times 3 = 0,9\,mol\\
{n_{Al{{(OH)}_3}(2)}} = 0,3 - 0,2 = 0,1\,mol\\
{n_{NaOH(2)}} = {n_{Al{{(OH)}_3}(2)}} = 0,1\,mol\\
{n_{NaOH}} = 0,9 + 0,1 = 1\,mol\\
{V_{NaOH}} = \dfrac{1}{{0,5}} = 2l
\end{array}\)
