Question

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)

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Answer 1

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

Step-by-step 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