4Fe + 3$O_{2}$→ 2$Fe_{2}$$O_{3}$
a)$n_{Fe}$= $\frac{m}{M}$ =$\frac{1,86}{56}$≈0,03(mol)
Theo PTHH,$n_{Fe_{2}O_{3}}$= $\frac{1}{2}$.$n_{Fe}$=$\frac{1}{2}$.0,03=0,015(mol)
$m_{Fe_{2}O_{3}}$=n.M=0,015.160=2,4(g)
b)Theo PTHH,$n_{O_{2}}$= $\frac{3}{4}$.$n_{Fe}$=$\frac{3}{4}$.0,03=0,0225(mol)
$V_{O_{2}}$=n.22,4=0,0225.22,4=0,504(l)
c)$V_{kk}$=5$V_{O_{2}}$=5.0,504=2,52(l)