\(n_{Fe}=0,1\left(mol\right)\)
\(3Fe+2O_2\xrightarrow[]{t^\circ}Fe_3O_4\)
0,1 → \(\dfrac{1}{30}\)
\(\Rightarrow n_{Fe_3O_4}\left(\text{thực tế}\right)=\dfrac{1}{30}\cdot80\%=\dfrac{2}{75}\left(mol\right)\)
\(\Rightarrow m_{Fe_3O_4}\left(\text{thực tế}\right)=\dfrac{2}{75}\cdot232=6,1867\left(g\right)\)