\(n_{Mg}=\dfrac{0.48}{24}=0.02\left(mol\right)\)
\(n_{O_2}=\dfrac{0.672}{22.4}=0.03\left(mol\right)\)
\(2Mg+O_2\underrightarrow{t^0}2MgO\)
\(0.02...0.01\)
\(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)
\(0.03...\left(0.03-0.01\right)\)
\(m_{Fe}=0.03\cdot56=1.68\left(g\right)\)
\(m_{hh}=1.68+0.48=2.16\left(g\right)\)