Chemistry
posted by Paul on .
At a certain temperature, 0.700 mol of SO3 is placed in a 3.50L container.
At equilibrium, 0.100 mol of O2 is present. Calculate Kc.
2SO3=2SO2 +)2

(SO3) = 0.700/3.50 = 0.2 mol
.........2SO3 ==> 2SO2 + O2
I........0.200.....0......0
C........2x......2x......x
E......0.2002x....2x.....x
At equil, x = 0.1 mol.
Calculate the concns of each and substitute into the Kc expression; calculate Kc. 
how do i find the concentrations?

The problem tells you that at equilibrium the amount of O2 = 0.1 mol.
That is in 3.50L; therefore, (O2) = 0.1mol/3.50L = x = 0.0286M
Then (SO3) = 0.2002x
(SO2) = 2x
(O2) = x = 0.0286 
Where do you substitute Kc into the expression?