Ta có: \(n_{Fe}=\dfrac{6,72}{56}=0,12\left(mol\right)\)
PT: \(2Fe+O_2\underrightarrow{t^o}2FeO\)
Theo PT: \(\left\{{}\begin{matrix}n_{FeO}=n_{Fe}=0,12\left(mol\right)\\n_{O_2}=\dfrac{1}{2}n_{Fe}=0,06\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{FeO}=0,12.72=8,64\left(g\right)\)
\(V_{O_2}=0,06.24,79=1,4874\left(l\right)\)