Final answer:
To produce 84.5 grams of Al₂O₃, you would need 39.792 grams of O₂. This is determined through stoichiometry and mole-to-mass conversions using the balanced chemical reaction and molar masses of the reacting substances.
Step-by-step explanation:
To calculate the mass in grams of oxygen gas (O₂) required to produce 84.5 grams of aluminum oxide (Al₂O₃), we need to follow several steps involving mole-to-mass conversions and stoichiometry from the balanced chemical reaction: 4Al + 3O₂ → 2Al₂O₃
First, determine the molar mass of Al₂O₃. Using periodic table values, Al₂O₃ has a molar mass of approximately 101.96 g/mol. Next, convert the mass of Al₂O₃ to moles by dividing by its molar mass: 84.5 g Al₂O₃ ÷ 101.96 g/mol = 0.829 moles of Al₂O₃. The mole ratio between O₂ and Al₂O₃ in the balanced equation is 3:2. Therefore, for every 2 moles of Al₂O₃ produced, 3 moles of O₂ are required. We use this ratio to calculate the moles of O₂ needed: 0.829 moles Al₂O₃ × (3 moles O₂ / 2 moles Al₂O₃) = 1.2435 moles O₂. Finally, convert moles of O₂ to grams using the molar mass of O₂ which is 32.00 g/mol: 1.2435 moles O₂ × 32.00 g/mol = 39.792 grams of O₂. Therefore, 39.792 grams of O₂ are needed to produce 84.5 grams of Al₂O₃.