Đáp án:
\(\begin{array}{l}
a)\\
\% Cu = 70,59\% \\
\% CuO = 29,41\% \\
b)\\
{C_{{M_{HN{O_3}d}}}} = 0,2M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
3Cu + 8HN{O_3} \to 3Cu{(N{O_3})_2} + 2NO + 4{H_2}O\\
CuO + 2HN{O_3} \to Cu{(N{O_3})_2} + {H_2}O\\
{n_{NO}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{n_{Cu}} = \dfrac{3}{2}{n_{NO}} = 0,15mol\\
{m_{Cu}} = 0,15 \times 64 = 9,6g\\
{m_{CuO}} = 13,6 - 9,6 = 4g\\
\% Cu = \dfrac{{9,6}}{{13,6}} \times 100\% = 70,59\% \\
\% CuO = 100 - 70,59 = 29,41\% \\
b)\\
{n_{CuO}} = \dfrac{4}{{80}} = 0,05mol\\
{n_{HN{O_3}}} = 1,25 \times 0,6 = 0,75mol\\
{n_{HN{O_3}cd}} = \dfrac{8}{3}{n_{Cu}} + 2{n_{CuO}} = 0,5mol\\
{n_{HN{O_3}d}} = 0,75 - 0,5 = 0,25mol\\
{C_{{M_{HN{O_3}d}}}} = \dfrac{{0,25}}{{1,25}} = 0,2M
\end{array}\)