For the following reaction, 89.5 grams of iron are allowed to react with 36.9 grams of oxygen gas. iron (s) oxygen (g) iron(III) oxide (s) What is the maximum amount of iron(III) oxide that can be formed

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

the answer is b. dilution