6.5 Moles of Al reacts with 7.2 Moles of H2O What is the limiting reactant and calculate the Theoretical Yield?

2AL + 3H2O - -> AL203 + 3H2

H₂O is the limiting reactant

Theoretical yield of 240 g Al₂O₃ and 14 g H₂

Explanation:

Find how many moles of one reactant is needed to completely react with the other.

6.5 mol Al * (3 mol H₂O / 2 mol Al) = 9.75 mol H₂O

We need 9.75 mol of H₂O to completely react with 6.5 mol of Al. But we only have 7.2 mol of H₂O. Therefore, H₂O is the limiting reactant.

Now find the theoretical yield:

7.2 mol H₂O * (1 mol Al₂O₃ / 3 mol H₂O) * (102 g Al₂O₃ / mol Al₂O₃) ≈ 240 g Al₂O₃

7.2 mol H₂O * (3 mol H₂ / 3 mol H₂O) * (2 g H₂ / mol H₂) ≈ 14 g H₂

Since the data was given to two significant figures, we must round our answer to two significant figures as well.