\(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ 3Fe+2O_2\underrightarrow{^{to}}Fe_3O_4\\ Vì:\dfrac{0,6}{2}>\dfrac{0,3}{3}\Rightarrow O_2dư\\ n_{Fe_3O_4\left(LT\right)}=\dfrac{0,3}{3}=0,1\left(mol\right)\\ n_{Fe_3O_4\left(TT\right)}=\dfrac{18,56}{232}=0,08\left(mol\right)\\ H=\dfrac{0,08}{0,1}.100=80\%\)