\(3Fe+2O_2\rightarrow Fe_3O_4\)
\(n_{Fe}=\frac{8,4}{56}=0,15\left(mol\right)\)
\(n_{O2}=\frac{8,92}{22,4}=0,39\left(mol\right)\)
Xét tỉ lệ \(\frac{Fe}{3}=0,05>\frac{O_2}{2}=0,195\)
Vậy O2 dư
\(n_{O2_{du}}=0,39-0,15=0,24\left(mol\right)\)
\(m_{Fe3O4}=232.0,05=11,6\left(g\right)\)