For the reaction POCl3(g) <-> POCl(g) + Cl2(g) Kc = .30.?

Question

An initial .38 moles of POCl3 are placed in a 3.3 L container with initial concentrations of POCl and Cl2 equal to zero. What is the final concentration of POCl3? ( you must solve a quadratic equation)

1. final concentration = .219 M
2. final concnetration = .0526192 M
3. final concentration = .089 M
4. final concentration = .334152 M
5. final concentration = .0263096 M

Answer 1

.38 moles of POCl3 are placed in a 3.3 L, [POCl3] = 0.11515 M. Let the equilibrium concentration [Cl2] = X:
POCl3(g) <-> POCl(g) + Cl2(g)
0.11515 - X......X.............X
Kc = .30 = X^2 /(0.11515 - X)
X^2 + 0.3*X - 0.3*0.11515 = 0
(positive) X = 0.0888
the final concentration of POCl3 is 0.11515 - 0.0888 = 0.0263 M
Tell you what? I do not think you can keep 6 significant digits as in (5).