PTHH: \(CuO+CO\underrightarrow{t^o}Cu+CO_2\)
\(Fe_xO_y+yCO\underrightarrow{t^o}xFe+yCO_2\)
\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)
\(n_{CaCO_3}=\dfrac{40}{100}=0,4\left(mol\right)\)
Theo PTHH: \(n_{CO_2}=n_{CaCO_3}=n_{CuO}+y.n_{Fe_xO_y}=0,4\left(mol\right)\)
TH1: FexOy là FeO
Ta có: \(\left\{{}\begin{matrix}80.n_{CuO}+72.n_{FeO}=24\\n_{CuO}+n_{FeO}=0,4\end{matrix}\right.\) => Nghiệm âm
TH2: FexOy là Fe3O4
Ta có: \(\left\{{}\begin{matrix}80.n_{CuO}+232.n_{Fe_3O_4}=24\\n_{CuO}+4.n_{Fe_3O_4}=0,4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}n_{CuO}=\dfrac{2}{55}\left(mol\right)\\n_{Fe_3O_4}=\dfrac{1}{11}\left(mol\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{\dfrac{2}{55}.80}{24}.100\%=12,12\%\\\%m_{Fe_3O_4}=\dfrac{\dfrac{1}{11}.232}{24}.100\%=87,88\%\end{matrix}\right.\)
TH3: FexOy là Fe2O3
Ta có: \(\left\{{}\begin{matrix}80.n_{CuO}+160.n_{Fe_2O_3}=24\\n_{CuO}+3.n_{Fe_2O_3}=0,4\end{matrix}\right.\)
=> \(n_{CuO}=n_{Fe_2O_3}=0,1\left(mol\right)\)
\(\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,1.80}{24}.100\%=33,33\%\\\%m_{Fe_2O_3}=\dfrac{0,1.160}{24}.100\%=66,67\%\end{matrix}\right.\)
