Final answer:
To find the percent yield of the reaction, we first determine the theoretical yield of aluminum oxide from 75.0 g of aluminum (141.79 g) and then calculate the percent yield using the actual yield (75.0 g), resulting in a percent yield of approximately 52.9%, with the closest answer choice being 50.0% (A).
So the correct answer is option (A).
Step-by-step explanation:
The question asks us to calculate the percent yield of aluminum oxide in the reaction between aluminum and oxygen. First, we must find the theoretical yield by using stoichiometry. Using the molar masses of Al and Al2O3 (26.98 g/mol and 101.96 g/mol, respectively), we can calculate the moles of aluminum starting at 75.0 g.
Moles of Al = 75.0 g/26.98 g/mol ≈ 2.78 mol Al.
Following the stoichiometry of the reaction, 4 mol Al would produce 2 mol Al2O3, meaning our 2.78 mol Al would fully react to produce 1.39 mol Al2O3 (since Al is the limiting reactant), which is equivalent to 141.79 g Al2O3 (1.39 mol * 101.96 g/mol). However, only 75.0 g Al2O3 is produced.
The percent yield can be calculated using the actual yield and the theoretical yield. Percent yield = (actual yield / theoretical yield) * 100% = (75.0 g / 141.79 g) * 100% ≈ 52.9%.
Since 52.9% is not one of the answer choices, we can assume a typo in the question or the answer choices. However, the closest answer choice to the correct percent yield is 50.0%, so the correct answer to the question is A) 50.0%.