Em tham khảo nha:

Em xem lại đề câu 3 nha

\(\begin{array}{l}
3)\\
3Cu + 8HN{O_3} \to 3Cu{(N{O_3})_2} + 2NO + 4{H_2}O\\
Fe + 4HN{O_3} \to Fe{(N{O_3})_3} + NO + 2{H_2}O\\
hh:Cu(a\,mol),Fe(b\,mol)\\
{n_{NO}} = \dfrac{{11,2}}{{22,4}} = 0,5\,mol\\
\left\{ \begin{array}{l}
64a + 56b = 12\\
\dfrac{{2a}}{3} + b = 0,5
\end{array} \right.\\
 \Rightarrow a = 0,9;b =  - 0,6\\
4)\\
3Cu + 8HN{O_3} \to 3Cu{(N{O_3})_2} + 2NO + 4{H_2}O\\
Fe + 4HN{O_3} \to Fe{(N{O_3})_3} + NO + 2{H_2}O\\
hh:Cu(a\,mol),Fe(b\,mol)\\
{n_{NO}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
\left\{ \begin{array}{l}
64a + 56b = 24,8\\
\dfrac{{2a}}{3} + b = 0,3
\end{array} \right.\\
 \Rightarrow a = 0,3;b = 0,1\\
\% {m_{Cu}} = \dfrac{{0,3 \times 64}}{{24,8}} \times 100\%  = 77,42\% \\
\% {m_{Fe}} = 100 - 77,41 = 22,59\% 
\end{array}\)

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