26.1 g of Fe were placed into a container and allowed to react with 78.4 g O2. What is the theoretical yield of Fe2O3?

Answer 1

This is a limiting reactant (LR) problem. You always get this scenario when you are given the amounts of both reactants and asked for a theoretical yield. The game plan here is to determine how many moles of Fe₂O₃ can be made from EACH reactant. The one that produces less will be your LR and will determine how much product is produced.

First, we need a balanced equation:

_4__Fe_(s) + ___3__O_2(g) ----> __2___Fe₂O_3(s)

26.1g 78.4g X mol or g

26.1 g Fe x ______1________mol Fe x ___2___________mol Fe₂O₃ = 0.221 mol Fe₂O₃

58.84 g Fe 4 mol Fe

78.4 g O₂ x ______1________mol O₂ x _____2_________mol Fe₂O₃ = 1.634 mol Fe₂O₃

32 g O₂ 3 mol O₂

The reaction will stop when the Fe runs out so Fe is the LR.

Now, the theoretical yield (in grams) can be determined:

0.221 mol Fe₂O₃ x ___164.85________g Fe₂O₃ = 36.43 g Fe₂O₃

1 mol Fe₂O₃_