1. Write the balanced chemical equation:

The reaction between sulfur trioxide (SO₃) and oxygen (O₂) to form sulfur dioxide (SO₂) is:

2 SO₃ (g) + O₂ (g) → 2 SO₂ (g)

2. Calculate the moles of SO₃:

* Molar mass of SO₃ = 80.06 g/mol

* Moles of SO₃ = (5.00 g) / (80.06 g/mol) = 0.0625 mol

3. Determine the moles of O₂ required:

* From the balanced equation, 2 moles of SO₃ react with 1 mole of O₂.

* Moles of O₂ needed = (0.0625 mol SO₃) * (1 mol O₂ / 2 mol SO₃) = 0.03125 mol O₂

4. Apply the Ideal Gas Law to find the volume of O₂:

* Ideal Gas Law: PV = nRT

* P = Pressure = 5.25 atm

* V = Volume (what we want to find)

* n = Moles of O₂ = 0.03125 mol

* R = Ideal Gas Constant = 0.0821 L·atm/mol·K

* T = Temperature = 350 °C + 273.15 = 623.15 K

5. Solve for V:

* V = (nRT) / P

* V = (0.03125 mol * 0.0821 L·atm/mol·K * 623.15 K) / 5.25 atm

* V ≈ 0.308 L

Therefore, approximately 0.308 liters of oxygen gas are needed at 350°C and 5.25 atm to completely convert 5.00 grams of sulfur trioxide.