Đáp án:

\(\begin{array}{l}
a)\\
\% Al = 21,95\% \\
\% Cu = 78,05\% \\
b)\\
{V_{Ba{{(OH)}_2}}} = 69ml
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
4)\\
a)\\
Al + 6HN{O_3} \to Al{(N{O_3})_3} + 3N{O_2} + 3{H_2}O\\
Cu + 4HN{O_3} \to Cu{(N{O_3})_2} + 2N{O_2} + 2{H_2}O\\
{n_{N{O_2}}} = \dfrac{{2,688}}{{22,4}} = 0,12mol\\
hh:Al(a\,mol),Cu(b\,mol)\\
3a + 2b = 0,12(1)\\
27a + 64b = 2,46(2)\\
 \text{Từ (1) và (2)}\Rightarrow a = 0,02;b = 0,03\\
{m_{Al}} = 0,02 \times 27 = 0,54g\\
{m_{Cu}} = 2,46 - 0,54 = 1,92g\\
\% Al = \dfrac{{0,54}}{{2,46}} \times 100\%  = 21,95\% \\
\% Cu = 100 - 21,95 = 78,05\% \\
b)\\
Cu{(N{O_3})_2} + Ba{(OH)_2} \to Cu{(OH)_2} + Ba{(N{O_3})_2}\\
2Al{(N{O_3})_3} + 3Ba{(OH)_2} \to 2Al{(OH)_3} + 3Ba{(N{O_3})_2}\\
{n_{Ba{{(OH)}_2}(b)}} = {n_{Cu{{(N{O_3})}_2}}} + \dfrac{3}{2}{n_{Al{{(N{O_3})}_3}}} = {n_{Cu}} + \dfrac{3}{2}{n_{Al}} = 0,06mol\\
{n_{Ba{{(OH)}_2}}} = \dfrac{{{n_{Ba{{(OH)}_2}(b)}} \times 115\% }}{{100\% }} = 0,069mol\\
{V_{Ba{{(OH)}_2}}} = \dfrac{{0,069}}{1} = 0,069l = 69ml
\end{array}\)