\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
Theo PT: \(n_{Fe_3O_4\left(LT\right)}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{30}\left(mol\right)\)
\(\Rightarrow m_{Fe_3O_4\left(LT\right)}=\dfrac{1}{30}.232=\dfrac{116}{15}\left(g\right)\)
\(\Rightarrow H=\dfrac{\dfrac{116}{15}}{11,6}.100\%\approx66,67\%\)