For the following reaction, 0.220 moles of iron are mixed with 0.187 moles of oxygen gas. iron (s) + oxygen (g) iron(II) oxide (s)?

2Fe(s) + O₂(g) → 2FeO(s)
Mole ratio Fe : O₂ = 2 : 1

Initial number of moles of Fe = 0.220 mol
Initial number of moles of O₂ = 0.187 mol

If Fe completely reacts :
Number of moles of O₂ needed = (0.220 mol) × (1/2) = 0.110 mol < 0.187 mol
Hence, O₂ is in excess, and Fe is the limiting reagent.

The formula of the limiting reagent : Fe


According to the above equation, mole ratio Fe : FeO = 1 : 1
Number of moles of Fe reacted = 0.220 mol
Number of moles of FeO = 0.220 mol