Đáp án:

 \(\begin{array}{l}
a)\\
\% {m_{Cu}} = 63,16\% \\
\% {m_{Fe}} = 36,84\% \\
b)\\
{C_M}HN{O_3} = 4,7M
\end{array}\)

Giải thích các bước giải:

 \(\begin{array}{l}
a)\\
3Cu + 8HN{O_3} \to 3Cu{(N{O_3})_2} + 2NO + 4{H_2}O\\
Fe + 4HN{O_3} \to Fe{(N{O_3})_3} + NO + 2{H_2}O\\
{n_{NO}} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
hh:Cu(a\,mol),Fe(b\,mol)\\
64a + 56b = 30,4\\
\frac{{2a}}{3} + b = 0,4\\
 \Rightarrow a = 0,3;b = 0,2\\
\% {m_{Cu}} = \dfrac{{0,3 \times 64}}{{30,4}} \times 100\%  = 63,16\% \\
\% {m_{Fe}} = 100 - 63,16 = 36,84\% \\
b)\\
HN{O_3} + NaOH \to NaN{O_3} + {H_2}O\\
{n_{NaOH}} = \dfrac{{150 \times 20\% }}{{40}} = 0,75\,mol\\
{n_{HN{O_3}}} = 0,75 + \frac{{0,3 \times 8}}{3} + 0,2 \times 4 = 2,35\,mol\\
{C_M}HN{O_3} = \dfrac{{2,35}}{{0,5}} = 4,7M
\end{array}\)