Consider the reaction of PCl3(g) + Cl2(g) -> PCl5(g) EQUILIBRIUM PROBLEM!!

Consider the reaction of PCl3(g) + Cl2(g) -> PCl5(g)

At a certain temperature, 0.500 mol of PCl5 was placed into a 0.250 L container and allowed to react. At equilibrium, the container held 0.100 mol of PCL5. What is the value of K for this reaction?

It's either :
a)0.100
b)0.500
c)0.156
d)0.625
e)0.833

pleas show your work because i want to see how you got the answer. and if you can use an ice table

Initial:
[PCl5] = 0.500 mol / 0.250 L = 2 M
[PCl3] = 0
[Cl2] = 0

Change:
Final moles of PCl5 = 0.1, so change is -0.40 mol/ 0.25L = 1.6 M
Since 1 mol PCl3 and 1 mol Cl2 are formed for every mole of PCl5. So,

moles PCl3 = 0.400
moles Cl2 = 0.400

Final
[PCl5] = 0.100 mol/ 0.25L = 0.4 M
[PCl3] = 0.400 mol / 0.250 L = 1.6 M
[Cl2] = 0.400 mol / 0.250 L = 1.6 M

K = [PCl5] /[PCl3][Cl2] = 0.4 / 1.6(1.6) = 0.156

So c) is the correct answer