Đáp án:
6,5l hay 4,5 l
Giải thích các bước giải:
TH1: Kết tủa tan 1 phần
\(\begin{array}{l}
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{Al}} = \dfrac{{2,7}}{{27}} = 0,1\,mol\\
{n_{HCl}} = 0,3 \times 2 = 0,6\,mol\\
\dfrac{{0,1}}{2} < \dfrac{{0,6}}{6} \Rightarrow\text{ HCl dư}\\
{n_{HCl}}\text{ HCl dư} = 0,6 - \dfrac{{0,1 \times 6}}{2} = 0,3\,mol\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,1\,mol\\
HCl + NaOH \to NaCl + {H_2}O(1)\\
AlC{l_3} + 3NaOH \to 3NaCl + Al{(OH)_3}(2)\\
Al{(OH)_3} + NaOH \to NaAl{O_2} + 2{H_2}O(3)\\
2Al{(OH)_3} \to A{l_2}{O_3} + {H_2}O(4)\\
{n_{A{l_2}{O_3}}} = \dfrac{{2,55}}{{102}} = 0,025\,mol \Rightarrow {n_{Al{{(OH)}_3}}}(4) = 0,05\,mol\\
{n_{Al{{(OH)}_3}}}(3) = 0,1 - 0,05 = 0,05\,mol\\
{n_{NaOH}}(1) = {n_{HCl}}\text{ HCl dư} = 0,3\,mol\\
{n_{NaOH}}(2) = 0,1 \times 3 = 0,3\,mol\\
{n_{NaOH}}(3) = {n_{Al{{(OH)}_3}}}(3) = 0,05\,mol\\
{n_{NaOH}} = 0,3 + 0,3 + 0,05 = 0,65\,mol\\
{V_{NaOH}} = \dfrac{{0,65}}{{0,1}} = 6,5l
\end{array}\)
TH2: Kết tủa chưa tan
\(\begin{array}{l}
HCl + NaOH \to NaCl + {H_2}O(5)\\
AlC{l_3} + 3NaOH \to 3NaCl + Al{(OH)_3}(6)\\
2Al{(OH)_3} \to A{l_2}{O_3} + {H_2}O(7)\\
{n_{A{l_2}{O_3}}} = \dfrac{{2,55}}{{102}} = 0,025\,mol \Rightarrow {n_{Al{{(OH)}_3}}}(7) = 0,05\,mol\\
{n_{NaOH}}(5) = {n_{HCl}} = 0,3\,mol\\
{n_{NaOH}}(7) = 0,05 \times 3 = 0,15\,mol\\
{n_{NaOH}} = 0,3 + 0,15 = 0,45\,mol\\
{V_{NaOH}} = \dfrac{{0,45}}{{0,1}} = 4,5M
\end{array}\)
