1. Write the balanced chemical equation:

4 Al + 3 O₂ → 2 Al₂O₃

2. Convert the masses of reactants to moles:

* Aluminum (Al):

* Molar mass of Al = 26.98 g/mol

* Moles of Al = 2.20 g / 26.98 g/mol = 0.0815 mol

* Oxygen (O₂):

* Molar mass of O₂ = 32.00 g/mol

* Moles of O₂ = 1.95 g / 32.00 g/mol = 0.0609 mol

3. Determine the limiting reactant:

* Using the mole ratio from the balanced equation:

* 4 moles of Al react with 3 moles of O₂

* Calculate the moles of O₂ needed to react with the given moles of Al:

* (0.0815 mol Al) * (3 mol O₂ / 4 mol Al) = 0.0611 mol O₂

* Since we only have 0.0609 mol O₂, oxygen is the limiting reactant.

4. Calculate the moles of aluminum oxide (Al₂O₃) produced:

* Using the mole ratio from the balanced equation:

* 3 moles of O₂ produce 2 moles of Al₂O₃

* Calculate the moles of Al₂O₃ produced:

* (0.0609 mol O₂) * (2 mol Al₂O₃ / 3 mol O₂) = 0.0406 mol Al₂O₃

5. Convert moles of Al₂O₃ to grams (theoretical yield):

* Molar mass of Al₂O₃ = 101.96 g/mol

* Theoretical yield:

* (0.0406 mol Al₂O₃) * (101.96 g/mol) = 4.15 g Al₂O₃

Therefore, the theoretical yield of aluminum oxide is 4.15 grams.