Chemistry question help! Consider the following reaction: 2SO2(g)+O2(g)→2SO3(g) What is the theoretical yield of SO3?

At STP, 1 mole of any gas occupies 22400 mL of volume.
Initial no. of moles of SO₂ = (272.5 mL) / (22400 mL/mol) = 1.22 × 10⁻² mol
Initial no. of moles of O₂ = (167.9 mL) / (22400 mL/mol) = 7.50 × 10⁻³ mol

Equation for the reaction :
2SO₂(g) + O₂(g) → 2SO₃(g)
Mole ratio SO₂ : O₂ : SO₃ = 2 : 1

If 1.22 × 10⁻² mol SO₂ completely reacts, O₂ needed = (1.22 × 10⁻² mol) × (1/2) = 6.1 × 10⁻³ mol < 7.50 × 10⁻³ mol
O₂ is in excess, and thus SO₂ is the limiting reactant.

According to the above equation, mole ratio SO₂ : SO₃ = 2 : 2
Theoretical yield of SO₃ = (1.22 × 10⁻² mol) × (2/2) = 1.22 × 10⁻² mol

I always encourage my students to consider the answer that they calculate: Does this answer appear logical? You have a small volume of SO2 and O2. This is measured in mL. You know that at STP , 1 mol of gas has volume = 22.4L In order to get a theoretical yield of 4.25 mol you must start with at minimum that amount of reactants. You have no way near 22.4L of reactants , let alone 4.25*22.4 = 95.2 L of reactants. The answer 4.25 mol SO3 is clearly impossible.

You have shown that SO2 is the limiting reactant.
The balanced equation tells you that 1 mol SO2 will produce 1 mol SO3
How many mol of SO2 do you have, if you have 272.5 mL of SO2 at STP
272.5mL = 0.2725L
1mol SO2 at STP = 22.4L
Mol SO2 in 0.2725L = 0.2725L/22.4L/mol = 1.22*10^-2 mol SO2
Because 1 mol SO2 produces 1 mol SO3 ,
Mol SO3 produced = 1.22*10^-2 mol - theoretical yield.