Chemistry help please?

Question

Sulfuryl chloride, SO2Cl2, decomposes to SO2 and Cl2 at 648K with a Kc=0.045 by the following reaction:
SO2Cl2(g) « SO2(g) + Cl2(g)

You place 16.0 grams of SO2Cl2 into a 5.5 liter flask and heat the flask to 648K. What is the equilibrium concentration of SO2 in the flask? Enter your answer in scientific notation.
Answer:

You now carry out the reaction described above in a 1.00 liter flask. You begin with 18.5 grams of SO2Cl2 and include enough Cl2 gas to give a partial pressure of 1.2 atm at 648 K. What is the initial concentration of chlorine gas in this flask in moles/liter?
Answer:

Using the data from the preceeding question, determine the equilibrium concentration of SO2 in the 1.00 liter flask at 648 K. Enter your answer in scientific notation.
Answer:

Answer 1

The initial concentration of SO2Cl2 is 16 gram / [135 (molec. mass) * 5.5 (liter)] = 0.022 molar. This decomposes to x moles of SO2 and x moles of Cl2, reducing the initial concentration by x:

SO2Cl2 ----> SO2 + Cl2

0.022-x...->......x..........x

The equilibrium equation is

kc = [SO2]*[Cl2] / [SO2Cl2]

kc = x^2 / (0.022 - x)

x^2 + kc*x - 0.022*kc = 0 For kc = 0.045

x^2 + 0.045*x - 9.9*10^-4 = 0 solve for x

x=0.016M. [scientific notation 1.6*10^-2M] This is the concentration of SO2 in moles/liter. The molec mass of SO2 is 64, so the mass concentration is 1.03 g/liter

The no of moles of Cl2 gas is found from the gas law, PV=nRT. n = PV/RT P = 1.2 atm, V = 1 liter, T = 648 ºK and R = 8.314 joule/ºK*mole.

n = 0.023 mole in one liter, so [Cl2] = 0.023M [2.3*10^-2 M]

18.5 g of SO2Cl2 is 0.137 moles

SO2Cl2 ----> SO2 + Cl2
0.137-x...->....x.......0.023 *** [EDIT: corrected number to 0.137]

kc = [SO2]*[Cl2] / [SO2Cl2]

kc = 0.023*x / (0.137-x)

0.137*kc - x*kc = 0.023*x

x(0.023 + kc) = 0.137*kc

x = 0.137*kc / (0.023 + kc)

x = 0.091 M [9.1*10^-2 M]