Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Al}} = 31,91\% \\
\% {m_{Cu}} = 68,09\% \\
b)\\
T{H_1}:\\
{V_{{\rm{dd}}\,NaOH}} = 0,36l\\
T{H_2}:\\
{V_{{\rm{dd}}\,NaOH}} = 0,76l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_{Al}} = \dfrac{{0,15 \times 2}}{3} = 0,1\,mol\\
\% {m_{Al}} = \dfrac{{0,1 \times 27}}{{8,46}} \times 100\% = 31,91\% \\
\% {m_{Cu}} = 100 - 31,91 = 68,09\% \\
b)\\
{n_{HCl}} = 0,15 \times 2 \times 10\% = 0,03\,mol\\
HCl + NaOH \to NaCl + {H_2}O(1)\\
{n_{NaOH(1)}} = {n_{HCl}} = 0,03\,mol\\
{n_{Al{{(OH)}_3}}} = \dfrac{{3,9}}{{78}} = 0,05\,mol\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,1\,mol\\
{n_{AlC{l_3}}} > {n_{Al{{(OH)}_3}}} \Rightarrow \text{ $AlCl_3$ dư hoăc kết tủa tan một phần}\\
T{H_1}:\text{ $AlCl_3$ còn dư}\\
AlC{l_3} + 3NaOH \to Al{(OH)_3} + 3NaCl(2)\\
{n_{NaOH(2)}} = 3{n_{Al{{(OH)}_3}}} = 0,05 \times 3 = 0,15\,mol\\
{n_{NaOH}} = {n_{NaOH(1)}} + {n_{NaOH(2)}} = 0,03 + 0,15 = 0,18\,mol\\
{V_{{\rm{dd}}\,NaOH}} = \dfrac{{0,18}}{{0,5}} = 0,36l\\
T{H_2}:\text{ Kết tủa tan 1 phần}\\
AlC{l_3} + 3NaOH \to Al{(OH)_3} + 3NaCl(3)\\
Al{(OH)_3} + NaOH \to NaAl{O_2} + 2{H_2}O(4)\\
{n_{Al{{(OH)}_3}(3)}} = {n_{AlC{l_3}}} = 0,1\,mol\\
{n_{Al{{(OH)}_3}(4)}} = 0,1 - 0,05 = 0,05\,mol\\
{n_{NaOH(3)}} = 3{n_{AlC{l_3}}} = 0,3\,mol\\
{n_{NaOH(4)}} = {n_{Al{{(OH)}_3}(4)}} = 0,05\,mol\\
{n_{NaOH}} = {n_{NaOH(1)}} + {n_{NaOH(3)}} + {n_{NaOH(4)}}\\
\Rightarrow {n_{NaOH}} = 0,03 + 0,3 + 0,05 = 0,38\,\,mol\\
{V_{{\rm{dd}}\,NaOH}} = \dfrac{{0,38}}{{0,5}} = 0,76l
\end{array}\)
