Đáp án:
\(C{\% _{Cu{{(N{O_3})}_2}}} = 13,65\% \)
Giải thích các bước giải:
\(\begin{array}{l}
3Cu + 8HN{O_3} \to 3Cu{(N{O_3})_2} + 2NO + 4{H_2}O\\
{n_{Cu}} = 0,15mol\\
\to {n_{HN{O_3}}} = \dfrac{8}{3}{n_{Cu}} = 0,4mol\\
\to {m_{HN{O_3}}} = 25,2g\\
\to {m_{{\rm{dd}}}}_{HN{O_3}} = \dfrac{{25,2}}{{12,6\% }} \times 100\% = 200g\\
{n_{Cu{{(N{O_3})}_2}}} = {n_{Cu}} = 0,15mol\\
\to {m_{Cu{{(N{O_3})}_2}}} = 28,2g\\
{n_{NO}} = \dfrac{2}{3}{n_{Cu}} = 0,1mol \to {m_{NO}} = 3g\\
{m_{{\rm{dd}}}} = {m_{Cu}} + {m_{{\rm{dd}}}}_{HN{O_3}} - {m_{NO}} = 206,6g\\
\to C{\% _{Cu{{(N{O_3})}_2}}} = \dfrac{{28,2}}{{206,6}} \times 100\% = 13,65\%
\end{array}\)
