Answer:
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.
