CHEMISTRY

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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 -

    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.....0.274-2x....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 -

    I did what you said but I keep getting the wrong answer no matter what I do.

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