CHEMISTRY
posted by Alexis Moran .
At a certain temperature , 0.960 mol of SO3 is placed in a 3.50 L container.
2SO3 > 2SO2 + O2
At equilibrium , 0.190 mol of O2 is present. Calculate kc.
i know that it starts off by
2SO3 > 2SO2 + O2
I 0.960 0 0
C 2x +2x x
_____________________
E 0.58 3.8 0.190
but then it says be sure to convert from moles to molar mass so i divided each equilibrium by 3.50 l . I did all of this but i still do not get the right answer !!! Someone help me please thank you

0.960/3.5L = 0.274M
0.190/3.5L = 0.0543M
............2SO3 ==> 2SO2 + O2
initial....0.274......0.....0
change.....2x........2x.....x
equil.....0.2742x....2x....0.0543
So you know x must be 0.0543 which will allow you to calculate 2x and from there SO3 and SO2 at equilibrium. Then substitute into Kc expression and solv for Kc. 
I did what you said but I keep getting the wrong answer no matter what I do.