Answer:
122.3 g of Fe₂O₃ is the maximum amount formed
Explanation:
Our reactants for the reaction:
89.5 g of Fe
36.9 g of O₂
We convert the mass of each to moles:
89.5 g of Fe . 1 mol / 55.85 g = 1.60 moles
36.9 g of O₂ . 1mol / 32g = 1.15 moles
The reaction is 4Fe(s) + 3O₂(g) → 2Fe₂O₃ (s)
We determine the limiting reactant:
4 moles of Iron can react with 3 moles of O₂
Therefore 1.60 moles of Iron will react with (1.60 .3) / 4 = 1.2 moles
Oxygen is the limiting reactant; we need 1.2 moles, and we only have 1.15.
Then we work with the stoichiometry again.
3 moles of oxygen can produce 2 moles of Fe₂O₃
1.15 moles of oxygen may produce (1.15 . 2) / 3 = 0.766 moles of Fe₂O₃
We convert the moles to mass: 0.766 mol . 159.7 g /1mol = 122.3 g of Fe₂O₃
