An equilibrium mixture of SO2,SO3 and O2 gases is maintained in a 11.5 L flask at a temp at which Kc=55.2 for the rxn:

Question

2SO2 + O2 <> 2SO3.

If the # of moles of SO2 and SO3 are =, how many moles of O2 are present?

I let SO2 and SO3 =x and O2=y. Since SO2 and SO3 will cancel in the equil. eqn, I got 1.81 X 10-2 mol O2 but book says 0.208???

Answer 1

Thanks for sharing your work and the answer. It helps tremendously. There are a number of problems and not all are yours.
2SO2 + O2 ==> 2SO3

(SO2) = (2x/11.5)M
(SO3) = (2x/11.5)M
O2 = (x/11.5)M

Kc = 55.2 = (SO3)^2/(SO2)^2(O2)
55.2 = (2x/11.5)^2/(2x/11.5)^2(x/11.5) = 55.2
The book answer is for moles = 0.208, which is correct. If you solve it as I indicated above, x = moles = 0.208.
If you solve it as you indicated, your x = molarity (since molarity is what is substituted into Kc), so your answer of 0.0181 is for molarity. To find moles, M x L = moles = 0.0181 x 11.5L = 0.208 moles.