$n_{NO}=\dfrac{0,672}{22,4}=0,03(mol)\\ Đặt\ \begin{cases}n_{Fe}=x(mol)\\n_{Cu}=y(mol)\end{cases}\\ \to 56x+64y=2,48\ (1)\\ \xrightarrow{\text{BT e}}\ 3.n_{Fe}+2.n_{Cu}=3.n_{NO}\\ \to 3x+2y=0,09\ (2)\\ \xrightarrow{\text{Từ (1),(2)}}\begin{cases}x=0,01\\y=0,03\end{cases}\\ \to m_{Fe}=0,01.56=0,56(g)\\ \to m_{Cu}=0,03.64=1,92(g)\\ n_{NO_3^-}=n_e=3.n_{NO}=0,09(mol)\\ m_{Muối}=m_{\text{kim loại}}+m_{NO_3^-}=2,48+0,09.62=8,06(g)\\ n_{HNO_3}=4.n_{NO}=0,12(mol)\\ \to C_{M\ dd\ HNO_3}=\dfrac{0,12}{0,2}=0,6(M)$

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