$nFe=1,86/56=0,03mol$
$a/$
$PTHH :$
$4Fe + 3O2 → 2Fe2O3$
$Theo$ $pt:$
$nFe2O3=1/2nFe=1/2.0,03=0,015mol$
$⇒mFe2O3=0,015.160=2,4g$
$b/$
$Theo$ $pt:$
$nO2=3/4.nFe=3/4.0,03=0,0225mol$
$⇒V_{O_{2}}=0,0225.22,4=0,504l $
$c/$
$V_{kk}=5.V_{O_{2}}=5.0,504=2,52l$