posted by Jarrod on .
Consider the following reaction:
2 SO2 (g) + O2 (g)----> 2 SO3 (g)
If 285.3 mL of SO2 is allowed to react with 158.9 mL of O2 (both measured at 315 K and 50.0 mmHg), what is the limiting reactant and theoretical yield of SO3 in moles? If 187.2 mL of SO3 is collected (measured at 315 K and 50.0 mm Hg), what is the percent yield for the reaction?
I am having troubles starting this problem, anyone know where to begin? I know we need to get from SO2 to SO3 and likewise O2 to SO3 but starting out with Ml I am not sure how to get a molar mass.
Use PV = nRT, substitute for P, V, R, and T, and solve for n = mols. Go from there.
I tried doing it that way already but it just doesn't look correct, because I set up PV=nRT to n=PV/RT. Entering the numbers I would get
((0.0658 atm)(0.2853))/((0.08206 L*atm/K*mol)(315K)) and I would get 0.000726 moles. After that would I just go straight to the number of moles of SO2 and SO3 so set up like so: 0.000726 mol SO2* 2 mol SO3/ 2 mol SO2* 80.07 g/ 1 mol SO3 ?
7.26E-4 is correct for moles SO2.
And your next step is ok but you don't need to go all the way to grams (yet).
7.26E-4 x (2 moles SO3/2 moles SO2) = 7.26E-4 mols SO3 (if all of the SO2 reacts).
Now do the same thing with oxygen, see how many moles oxygen you have initially, and convert that to moles SO3.
Those two numbers you get for moles SO3 probably will not be the same which means one of them is wrong. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Use that smaller value, convert that to grams SO3, and that is the theoretical yield.
Now you have the limiting reagent.
Now you have the theoretical yield.
Now yu can calculate the percent yield if you obtain the amount given in the problem.
Okay I believe I got it, thank you so much for the help! :)
You're very welcome. Come again. Thank you for showing your work It helps us diagnose the problem.