Em tham khảo nha:
\(\begin{array}{l}
a)\\
3Cu + 8HN{O_3} \to 3Cu{(N{O_3})_2} + 2NO + 4{H_2}O\\
CuO + 2HN{O_3} \to Cu{(N{O_3})_2} + {H_2}O\\
{n_{NO}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{n_{Cu}} = 0,3 \times \frac{3}{2} = 0,45\,mol\\
\% {m_{Cu}} = \dfrac{{0,45 \times 64}}{{30}} \times 100\% = 96\% \\
\% {m_{CuO}} = 100 - 96 = 4\% \\
b)\\
{n_{CuO}} = \dfrac{{30 - 0,45 \times 64}}{{80}} = 0,015\,mol\\
{n_{HN{O_3}}} = \dfrac{{0,45 \times 8}}{3} + 0,015 \times 2 = 1,23\,mol\\
{m_{HN{O_3}}} = 1,23 \times 63 = 77,49g\\
{m_{{\rm{dd}}HN{O_3}}} = \dfrac{{77,49}}{{10\% }} = 774,9g
\end{array}\)
