Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Cu}=b\left(mol\right)\end{matrix}\right.\left(ĐK:a,b>0\right)\)
Ta có: \(n_{SO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
=> 56a + 64b = 17,6 (1)
PTHH:
\(2Fe+6H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)
\(Cu+2H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}CuSO_4+SO_2+2H_2O\)
Theo PT: \(n_{SO_2}=\dfrac{3}{2}n_{Fe}+n_{Cu}\)
=> 1,5a + b = 0,4 (2)
Từ (1), (2) => \(\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,2.56}{17,6}.100\%=63,64\%\\\%m_{Cu}=100\%-63,64\%=36,36\%\end{matrix}\right.\)
