Em tham khảo nha :
\(\begin{array}{l}
a)\\
3Cu + 8HN{O_3} \to 3Cu{(N{O_3})_2} + 2NO + 4{H_2}O\\
CuO + 2HN{O_3} \to Cu{(N{O_3})_2} + {H_2}O\\
{n_{NO}} = \dfrac{{0,448}}{{22,4}} = 0,02mol\\
{n_{Cu}} = \dfrac{3}{2}{n_{NO}} = 0,03mol\\
{m_{Cu}} = 0,03 \times 64 = 1,92g\\
{m_{CuO}} = 3,52 - 1,92 = 1,6g\\
b)\\
{n_{CuO}} = \dfrac{{1,6}}{{80}} = 0,02mol\\
{n_{HN{O_3}}} = \dfrac{8}{3}{n_{Cu}} + 2{n_{CuO}} = 0,12mol\\
{m_{HN{O_3}}} = 0,12 \times 63 = 7,56g\\
c)\\
Cu{(N{O_3})_2} + 2NaOH \to Cu{(OH)_2} + 2NaN{O_3}\\
{n_{Cu{{(N{O_3})}_2}}} = {n_{Cu}} + {n_{CuO}} = 0,03 + 0,02 = 0,05mol\\
{n_{Cu{{(OH)}_2}}} = {n_{Cu{{(N{O_3})}_2}}} = 0,05mol\\
{m_{Cu{{(OH)}_2}}} = 0,05 \times 98 = 4,9g
\end{array}\)
