\(n_{Fe}=a\left(mol\right),n_S=b\left(mol\right)\)
\(n_{O_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)
\(m_{hh}=56a+32b=20\left(g\right)\left(1\right)\)
\(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)
\(S+O_2\underrightarrow{t^0}SO_2\)
\(n_{O_2}=\dfrac{2}{3}a+b=0.3\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.3,b=0.1\)
\(\%Fe=\dfrac{0.3\cdot56}{20}\cdot100\%=84\%\)
\(\%S=16\%\)