1. Write the balanced chemical equation:

4 Al + 3 O₂ → 2 Al₂O₃

2. Determine the molar mass of each reactant:

* Aluminum (Al): 26.98 g/mol

* Oxygen (O₂): 32.00 g/mol

3. Calculate the moles of oxygen gas:

* Moles of O₂ = mass of O₂ / molar mass of O₂

* Moles of O₂ = 192 g / 32.00 g/mol = 6 moles

4. Use the mole ratio from the balanced equation to find the moles of aluminum:

* The balanced equation shows that 4 moles of Al react with 3 moles of O₂.

* So, the mole ratio of Al to O₂ is 4:3

* Moles of Al = (4 moles Al / 3 moles O₂) * 6 moles O₂ = 8 moles Al

5. Calculate the mass of aluminum:

* Mass of Al = moles of Al * molar mass of Al

* Mass of Al = 8 moles * 26.98 g/mol = 215.84 g

Therefore, you need 215.84 grams of aluminum to react with 192 grams of oxygen gas.