Ta có: \(n_{Fe_2O_3}=\dfrac{24}{160}=0,15\left(mol\right)\)
PT: \(Fe_2O_3+3CO\underrightarrow{t^o}2Fe+3CO_2\)
Theo PT: \(\left\{{}\begin{matrix}n_{Fe}=2n_{Fe_2O_3}=0,3\left(mol\right)\\n_{CO}=3n_{Fe_2O_3}=0,45\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow V_{CO}=0,45.22,4=10,08\left(l\right)\)
\(m_{Fe}=0,3.56=16,8\left(g\right)\)