Đáp án:
\(\begin{array}{l}
\% Cu = 63,2\% \\
\% Fe = 36,8\% \\
m = 52,4g\\
{C_{{M_{HN{O_3}}}}} = 1,6M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
3)\\
3Cu + 8HN{O_3} \to 3Cu{(N{O_3})_2} + 2NO + 4{H_2}O\\
Fe + 4HN{O_3} \to Fe{(N{O_3})_3} + NO + 2{H_2}O\\
{n_{NO}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
hh:Cu(a\,mol),Fe(b\,mol)\\
64a + 56b = 15,2(1)\\
\dfrac{2}{3}a + b = 0,2(2)\\
\text{Từ (1) và (2)}\Rightarrow a = 0,15mol;b = 0,1mol\\
{m_{Cu}} = 0,15 \times 64 = 9,6g\\
{m_{Fe}} = 0,1 \times 56 = 5,6g\\
\% Cu = \dfrac{{9,6}}{{15,2}} \times 100\% = 63,2\% \\
\% Fe = 100 - 63,2 = 36,8\% \\
{n_{Cu{{(N{O_3})}_2}}} = {n_{Cu}} = 0,15mol\\
{m_{Cu{{(N{O_3})}_2}}} = 0,15 \times 188 = 28,2g\\
{n_{Fe{{(N{O_3})}_3}}} = {n_{Fe}} = 0,1mol\\
{m_{Fe{{(N{O_3})}_3}}} = 0,1 \times 242 = 24,2g\\
m = 28,2 + 24,2 = 52,4g\\
{n_{HCl}} = \dfrac{8}{3}{n_{Cu}} + 4{n_{Fe}} = \dfrac{8}{3} \times 0,15 + 4 \times 0,1 = 0,8mol\\
{C_{{M_{HN{O_3}}}}} = \dfrac{{0,8}}{{0,5}} = 1,6M
\end{array}\)
