Em tham khảo nha:

\(\begin{array}{l}
4)\\
a)\\
5Zn + 12HN{O_3} \to 5Zn{(N{O_3})_2} + {N_2} + 6{H_2}O\\
CuO + 2HN{O_3} \to Cu{(N{O_3})_2} + {H_2}O\\
{n_{{N_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_{Zn}} = 5{n_{{N_2}}} = 0,5\,mol\\
{m_{Zn}} = 0,5 \times 65 = 32,5g\\
{m_{CuO}} = 34 - 32,5 = 1,5g\\
d)\\
{n_{CuO}} = \dfrac{{1,5}}{{80}} = 0,01875\,mol\\
{n_{HN{O_3}}} = 0,1 \times 12 + 0,01875 \times 2 = 1,2375\,mol\\
{V_{{\rm{dd}}HN{O_3}}} = \dfrac{{1,2375}}{2} = 0,61875l\\
c)\\
{n_{Cu{{(N{O_3})}_2}}} = {n_{CuO}} = 0,01875\,mol\\
{n_{Zn{{(N{O_3})}_2}}} = {n_{Zn}} = 0,5\,mol\\
{C_M}Cu{(N{O_3})_2} = \dfrac{{0,01875}}{{0,61875}} = 0,03M\\
{C_M}Zn{(N{O_3})_2} = \dfrac{{0,5}}{{0,61875}} = 0,81M\\
5)\\
a)\\
Cu + 4HN{O_3} \to Cu{(N{O_3})_2} + 2N{O_2} + 2{H_2}O\\
Al + 6HN{O_3} \to Al{(N{O_3})_3} + 3N{O_2} + 3{H_2}O\\
{n_{N{O_2}}} = \dfrac{{2,912}}{{22,4}} = 0,13\,mol\\
hh:Cu(a\,mol),Al(b\,mol)\\
\left\{ \begin{array}{l}
64a + 27b = 2,09\\
2a + 3b = 0,13
\end{array} \right.\\
 \Rightarrow a = 0,02;b = 0,03\\
\% {m_{Cu}} = \dfrac{{0,02 \times 64}}{{2,09}} \times 100\%  = 61,24\% \\
\% {m_{Al}} = 100 - 61,24 = 38,76\% \\
b)\\
{n_{HN{O_3}}} = 2{n_{N{O_2}}} = 0,26\,mol\\
{m_{HN{O_3}}} = 0,26 \times 63 = 16,38g
\end{array}\)