\(n_{Fe}=\frac{11,2}{56}=0,2\left(mol\right)\)
\(n_{O2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
\(PTHH:3Fe+2O_2\rightarrow Fe_3O_4\)
Lập tỉ lệ: \(\frac{0,2}{4}>\frac{0,1}{2}\)
Nên Fe dư, O2 hết
\(n_{Fe_{dư}}=0,2.-\left(\frac{3.}{2}\right).0,1=0,05\left(mol\right)\)
\(m_{Fe_{dư}}=0,05.56=2,8\left(g\right)\)
\(m_{Fe3O4}=232.0,05=11,6\left(g\right)\)