What is the theoretical yield in moles!

Consider the following reaction:

2SO2(g) + O2(g) => 2SO3(g)

If 277.6 mL of SO2 is allowed to react with 151.9 mL of O2 (both measured at 327 K and 45.8 mmHg), find the limiting reactant. What is the theoretical yield of SO3 in moles?

*Make sure the answer is in moles for the units, as the program requires this.

Thanks!

To find the moles, use the ideal gas law.
Convert the V to L and T to kelvins to have units consistent with those on R (0.08206 L atm / K mol.
45.8 mmHg x (1 atm / 760 mmHg) = 0.0603 atm

PV = nRT
n = PV / RT = (0.0603 atm x 0.2776 L) / (0.08206 L atm / K mol x 327 K) = 0.000623 mol SO2
n = PV / RT = (0.0603 atm x 0.1519 L) / (0.08206 L atm / K mol x 327 K) = 0.000341 mol O2

2 mol SO2 react with 1 mol O2
0.000623 mol SO2 x (1 mol O2 / 2 mol SO2) = 0.000312 mol O2
There are more moles of O2 than needed (0.000341, but only 0.000312 needed) to use up all the SO2; SO2 is the limiting reagent and determines the yield.

2 mol SO2 produces 2 mol SO3
0.000623 mol SO2 x (2 mol SO3 / 2 mol SO2) = 0.000623 mol SO3 produced (theoretically)