!! LONG ANSWER !!
The theoretical yield of sulfur trioxide, or #SO_3#, will be 285.0 mL, and the percent yield will be 61.16%.
Start with the balanced chemical equation
#color(red)(2)SO_(2(g)) + O_(2(g)) -> color(red)(2)SO_(3(g))#
Notice that you have a #color(red)(2):1# mole ratio between sulfur dioxide, #SO_2#, and oxygen; this will be very important when comparing the number of moles of each gas that react, since it will tell you if one of the two compounds will act as a limiting reagent.
Your tool of choice will be the ideal gas law equation. Start by determining the number of moles of sulfur dioxide and oxygen that you start with
#PV = nrt => n = (PV)/(RT)#
#n_(O_2) = (48.5/760cancel("atm") * 159 * 10^(-3)cancel("L"))/(0.082 (cancel("L") * cancel("atm"))/("mol" * cancel("K")) * 330cancel("K")) = "0.0003750 moles "# #O_2#
#n_(SO_2) = (48.5/760cancel("atm") * 284.9 * 10^(-3)cancel("L"))/(0.082 (cancel("L") * cancel("atm"))/("mol" * cancel("K")) * 330cancel("K")) = "0.0006719 moles "# #SO_2#
According to the mole ratio, that many moles of oxygen would have required
#0.0003750cancel("moles "O_2) * (color(red)(2)"moles "SO_2)/(1cancel("mole "SO_2)) = "0.000750 moles "# #SO_2#
Since you've got fewer moles than required, sulfur dioxide will act as a limiting reagent, i.e. it will determine how many moles of oxygen will actually react
#0.0006719cancel("moles "SO_2) * ("1 mole "O_2)/(color(red)(2)cancel("moles "SO_2)) = "0.0003360 moles "O_2#
The number of moles of sulfur trioxide this reaction will produce is
#0.0006719cancel("moles "SO_2) * ("1 mole "SO_3)/(1cancel("mole "SO_2)) = "0.0006719 moles "SO_3#
This many moles will produce a volume of
#V = (nRT)/P = (0.0006719cancel("moles") * 0.082(("L" * cancel("atm"))/(cancel("mol") * cancel("K")) * 330cancel("K")))/(48.5/760cancel("atm"))#
#V = "0.285 L" = color(green)("285.0 mL")#
This represents the theoretical yield of #SO_3# #-># this is how much sulfur trioxide your reaction would produce if all (100%) of the sulfur dioxide reacts.
SInce your reaction produced 174.3 mL, the percent yield of the reaction will be smaller than 100%. Calculate percent yield by
#"% yield" = "actual yield"/"theoretical yield" * 100#
#"% yield" = 174.3/285.0 * 100 = color(green)("61.16%")#
