Đáp án:
\(\begin{array}{l}
\% Cu = 44,44\% \\
\% CuO = 55,56\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
Cu + 4HN{O_3} \to Cu{(N{O_3})_2} + 2N{O_2} + 2{H_2}O\\
CuO + 2HN{O_3} \to Cu{(N{O_3})_2} + {H_2}O\\
{n_{N{O_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
\Rightarrow {n_{Cu}} = \dfrac{{{n_{N{O_2}}}}}{2} = 0,1mol\\
{m_{Cu}} = 0,1 \times 64 = 6,4g\\
{m_{CuO}} = 14,4 - 6,4 = 8g\\
\% Cu = \dfrac{{6,4}}{{14,4}} \times 100\% = 44,44\% \\
\% CuO = \dfrac{8}{{14,4}} \times 100\% = 55,56\%
\end{array}\)