Ta có: \(n_{Fe}=\dfrac{35}{56}=0,625\left(mol\right)\)
PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Fe}=0,9375\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,9375.22,4=21\left(l\right)\)
PT: \(Fe_2O_3+3CO\underrightarrow{t^o}2Fe+3CO_2\)
Theo PT: \(n_{CO}=\dfrac{3}{2}n_{Fe}=0,9375\left(mol\right)\)
\(\Rightarrow V_{CO}=0,9375.22,4=21\left(l\right)\)