Final answer:
When aluminum reacts with oxygen to produce aluminum oxide, the limiting reactant must first be identified. In this case, oxygen is the limiting reactant with 3.00 moles present. The calculations result in 16.98 g of excess aluminum and 203.92 g of aluminum oxide produced, none of which match the provided answer choices.
Step-by-step explanation:
The reaction in question involves aluminum (Al) and oxygen (O2) to form aluminum oxide (Al2O3). To calculate both the amount of excess reactant and mass of aluminum oxide produced we first need to determine the limiting reactant. Using stoichiometry based on the balanced chemical equation 4Al + 3O2 → 2Al2O3, we compare the mole ratio of reactants to the provided masses. Converting the mass of each reactant to moles, we get:
- 124.9 g Al / (26.98 g/mol) = 4.63 moles Al
- 96.0 g O2 / (32.00 g/mol) = 3.00 moles O2
The mole ratio from the balanced equation suggests that 4 moles of Al would require 3 moles of O2 to react completely. Thus, by dividing the actual moles of each reactant by the theoretical ratio, the smallest result will indicate the limiting reactant:
- 4.63 moles Al / (4 moles Al) = 1.1575
- 3.00 moles O2 / (3 moles O2) = 1.00
Since 1.00 is the smaller number, O2 is the limiting reactant. We then calculate the mass of aluminum that would react with the available oxygen:
- 3.00 moles O2 × (4 moles Al / 3 moles O2) = 4.00 moles Al
- 4.00 moles Al × 26.98 g/mol = 107.92 g Al
The amount of excess aluminum is found by subtracting this mass from the original mass of aluminum:
- 124.9 g Al - 107.92 g Al = 16.98 g excess Al
To find the mass of aluminum oxide produced:
- 3.00 moles O2 × (2 moles Al2O3 / 3 moles O2) = 2.00 moles Al2O3
- 2.00 moles Al2O3 × 101.96 g/mol = 203.92 g Al2O3
Therefore, none of the provided answer choices (a, b, c, or d) are correct based on these calculations.