PTHH: 3Fe + 2O2 → Fe3O4
\(n_{Fe}=\dfrac{10,8}{56}=\dfrac{27}{140}\left(mol\right)\)
a) Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=\dfrac{2}{3}\times\dfrac{27}{140}=\dfrac{9}{70}\left(mol\right)\)
\(\Rightarrow V_{O_2}=\dfrac{9}{70}\times22,4=2,88\left(l\right)\)
b) Xem lại đề.