Posted by Alexis Moran on Monday, October 10, 2011 at 9:13pm.
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

CHEMISTRY  DrBob222, Tuesday, October 11, 2011 at 1:33am
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.

CHEMISTRY  Margaret, Wednesday, February 13, 2013 at 11:40pm
I did what you said but I keep getting the wrong answer no matter what I do.
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