Trong không khí chứa $20\%O_2$ về thể tích.
$n_{O_2}=\dfrac{22,4}{22,4}.20\%=0,2(mol)$
$n_{Fe}=\dfrac{22,4}{56}=0,4(mol)$
$3Fe+2O_2\buildrel{{t^o}}\over\to Fe_3O_4$
$\dfrac{0,4}{3}>\dfrac{0,2}{2}$
$\Rightarrow Fe$ dư
$n_{Fe\text{pứ}}=\dfrac{3}{2}n_{O_2}=0,3(mol)$
$\Rightarrow n_{Fe\text{dư}}=0,4-0,3=0,1(mol)$
$\to m_{Fe\text{dư}}=0,1.56=5,6g$