[bpeck8]

Question.  

                 When 2.00 mol of SO2Cl2 is placed in a 2.00L flask at 303K, 56% of the SO2Cl2 decomposes to SO2 and Cl2:   SO2Cl2 (g) <=> SO2 (g) + Cl2 (g)
                 Calculate Kc for this reaction at this temperature.

Here is my work:
                        Kc = [SO2] [Cl2]
                                  [SO2Cl2]

                         2.00 mol SO2Cl2 = 1 M SO2Cl2             56% = .56 SO2Cl2 E
                             2.00 L                                           100

                             SO2Cl2               SO2             Cl2
                        I     1M                    0M               0M
                        C    1 - X                0 + X            0 + X
                        E    .56 M                 .44 M          .44M


                          Kc = (.44)(.44) = .34
                                    .56

I think we have to do something with the 56%, though i am not sure how to use it, so i did this. This feels wrong to me.

[Borek]

Not bad, although you got it reversed. If 56% of SO₂Cl₂ decomposed, how much was left - 56%, or 44%?

[bpeck8]

awww. ok i see what your saying. The .56 is the change and the .44 would be the equilibrium. which means that .56 would be the equilibrium for the products. right?

[EmperorTeepo]

Exactly. The .56 is what DECOMPOSED, so that is what your products' concentration will be.

2 mol SO2CL2/2L= 1 M, correct. .56 decomposed, so .44 is what is left. Since your products are all 1 mole ratio, .56 M is what is produced for each. So (.56)^2/ .44= Kc.