a) Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\end{matrix}\right.\) => 56a + 65b = 47,5 (1)
\(n_{NO_2}=\dfrac{47,04}{22,4}=2,1\left(mol\right)\)
Bảo toàn electron: \(n_{NO_2}=3n_{Fe}+2n_{Zn}\)
=> 3a + 2b = 2,1 (2)
Từ (1), (2) => a = 0,5; b = 0,3
=> \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,5.56}{47,5}.100\%=58,95\%\\\%m_{Zn}=100\%-58,95\%=41,05\%\end{matrix}\right.\)
b) mdd = 47,5 + 441 - 2,1.46 = 391,9 (g)
Bảo toàn nguyên tố Fe, Zn: \(\left\{{}\begin{matrix}n_{Fe\left(NO_3\right)_3}=n_{Fe}=0,5\left(mol\right)\\n_{Zn\left(NO_3\right)_2}=n_{Zn}=0,3\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}C\%_{Fe\left(NO_3\right)_3}=\dfrac{0,5.242}{391,9}.100\%=30,88\%\\C\%_{Zn\left(NO_3\right)_2}=\dfrac{0,3.189}{391,9}.100\%=14,7\%\end{matrix}\right.\)
