If 16.0 grams of aluminum oxide were actually produced,what is the percent yield of the reaction below given that you start with 10.0 g of Al and 19.0 grams of O₂?
Reaction: 4Al + 3O₂ → 2Al₂O₃

Question

Quizplus · Accepted Answer

First, we need to calculate the theoretical yield of aluminum oxide using stoichiometry. We start with 10.0 g of Al and 19.0 g of O₂, and determine which reactant limits the reaction (i.e. which reactant is completely consumed and therefore determines the maximum amount of product that can be formed). Using the mole ratio of Al to O₂ (4:3), we find that 10.0 g Al is equivalent to 0.372 mol Al. Using the mole ratio of Al to Al₂O₃ (4:2), we find that 0.372 mol Al is equivalent to 0.744 mol Al₂O₃. Using the mole ratio of O₂ to Al₂O₃ (3:2), we find that 19.0 g O₂ is equivalent to 0.594 mol O₂. However, we only need 0.496 mol of O₂ to react completely with 0.372 mol of Al (according to the 3:4 mole ratio), so O₂ is the limiting reactant. Using the mole ratio of Al₂O₃ to aluminum oxide (2:1), we can calculate the theoretical yield of aluminum oxide to be 0.372 mol Al₂O₃ x (1 mol Al2O3 / 2 mol Al₂O₃) x (101.96 g Al2O3 / 1 mol Al2O3) = 19.1 g Al2O3. Finally, we can calculate the percent yield by dividing the actual yield (16.0 g) by the theoretical yield (19.1 g) and multiplying by 100%. This gives us a percent yield of (16.0 g / 19.1 g) x 100% = 84.7%. Therefore, the answer is C, 84.7%.

If 160 Grams of Aluminum Oxide Were Actually Produced,what Is the the Percent