Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Cu}} = 44,44\% \\
\% {m_{CuO}} = 55,56\% \\
b)\\
C{\% _{HN{O_3}(dư)}} = 28,23\% \\
C{\% _{Cu{{(N{O_3})}_2}}} = 18,05\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
Cu + 4HN{O_3} \to Cu{(N{O_3})_2} + 2N{O_2} + 2{H_2}O\\
CuO + 2HN{O_3} \to Cu{(N{O_3})_2} + {H_2}O\\
{n_{N{O_2}}} = 0,4mol\\
\to {n_{Cu}} = \dfrac{1}{2}{n_{N{O_2}}} = 0,2mol\\
\to {m_{Cu}} = 12,8g\\
\to {m_{CuO}} = 28,8 - 12,8 = 8g \to {n_{CuO}} = 0,1mol\\
a)\\
\% {m_{Cu}} = \dfrac{{12,8}}{{28,8}} \times 100\% = 44,44\% \\
\% {m_{CuO}} = 100\% - 44,44\% = 55,56\% \\
b)\\
{m_{{\rm{dd}}HN{O_3}}} = 302g\\
{n_{HN{O_3}}} = 2,4mol\\
{n_{HN{O_3}(pt)}} = 4{n_{Cu}} + 2{n_{CuO}} = 1mol\\
\to {n_{HN{O_3}(dư)}} = 1,4mol\\
\to {m_{HN{O_3}(dư)}} = 88,2g\\
{n_{Cu{{(N{O_3})}_2}}} = {n_{Cu}} + {n_{CuO}} = 0,3mol\\
\to {m_{Cu{{(N{O_3})}_2}}} = 56,4g\\
{m_{{\rm{dd}}}} = {m_{hh}} + {m_{{\rm{dd}}HN{O_3}}} - {m_{N{O_2}}} = 312,4g\\
\to C{\% _{HN{O_3}(dư)}} = \dfrac{{88,2}}{{312,4}} \times 100\% = 28,23\% \\
\to C{\% _{Cu{{(N{O_3})}_2}}} = \dfrac{{56,4}}{{312,4}} \times 100\% = 18,05\%
\end{array}\)
