\(n_{Fe}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)
\(0,2\) → \(\dfrac{2}{15}\) → \(\dfrac{1}{15}\) ( mol )
a) \(m_{Fe_3O_4}=n.M=\dfrac{1}{15}.\left(56.3+16.4\right)=\dfrac{232}{15}\left(g\right)\)
b) \(V_{O_2}=n.22,4=\dfrac{2}{15}.22,4=\dfrac{224}{75}\left(l\right)\)