To determine the grams of oxygen needed to produce 95.6 g of aluminum oxide, we need to use the chemical formula for aluminum oxide, which is Al2O3.

From the chemical formula, we can see that for every 2 moles of aluminum (Al), there are 3 moles of oxygen (O).

First, we need to calculate the number of moles of aluminum oxide produced:

Moles of Aluminum Oxide = Mass / Molar Mass

Moles of Aluminum Oxide = 95.6 g / 101.96 g/mol

Moles of Aluminum Oxide = 0.937 mol

Since there are 3 moles of oxygen for every 2 moles of aluminum oxide, we can calculate the moles of oxygen needed:

Moles of Oxygen = (3 moles O / 2 moles Al2O3) × Moles of Aluminum Oxide

Moles of Oxygen = (3 / 2) × 0.937 mol

Moles of Oxygen = 1.406 mol

Finally, we can convert the moles of oxygen to grams using the molar mass of oxygen (16 g/mol):

Mass of Oxygen = Moles of Oxygen × Molar Mass

Mass of Oxygen = 1.406 mol × 16 g/mol

Mass of Oxygen = 22.5 g

Therefore, approximately 22.5 grams of oxygen are needed to produce 95.6 grams of aluminum oxide.