In a reaction of iron and oxygen to produce iron (III) oxide, 12.5 grams of iron was consumed. The limiting reagent is (compound) if 5.2 grams of oxygen was present. The theoretical yield (amount that is produced) is (mass to the second decimal) Blank 1:
Blank2:

The limiting reagent is O₂ if 5.2 grams of oxygen was present. The theoretical yield is 17.25g of Fe₂O₃

Explanation:

The reaction is:

4Fe + 3O₂  →  2Fe₂O₃

We convert the mass of each reactant to moles:

12.5 g . 1mol / 55.85g = 0.224 moles of Fe

5.2g . 1 mol/ 32g = 0.1625 moles of oxygen.

We analyse the stoichiometry.

4 moles of Fe react to 3 moles of O₂

Then 0.224 moles of Fe will react to (0.224 . 3)/4 = 0.1678 moles of O₂

I need 0.1678 moles of O₂ and I only have 0.1625. Oxygen is the limiting reactant.

Now we go to the products:

3 moles of O₂ can produce 2 moles of Iron (III) oxyde

Then, 0.1625 moles of O₂ may produce (0.1625 . 2)/3 = 0.108 moles of Fe₂O₃

We convert the moles to mass:

0.108 mol . 159.7g /1mol = 17.25 g

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