\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
a, 3Fe + 2O2 \(\underrightarrow{t^o}\) Fe3O4
de: 0,2\(\rightarrow\) \(\dfrac{2}{15}\) \(\rightarrow\) \(\dfrac{1}{15}\) (mol)
b, \(V_{O_2}=22,4.\dfrac{2}{15}\approx2,9867l\)
c, \(m_{Fe_3O_4}=\dfrac{1}{15}.232\approx15,467g\)