Chemistry problem about equilibrium concentrations

Question

A student ran the following reaction in the laboratory at 483 K: 

PCl₃(g) + Cl₂(g)     PCl₅(g) 

When she introduced 0.448 moles of PCl₃(g) and 0.477 moles of Cl₂(g) into a 1.00 liter container, she found the equilibrium concentration of PCl₅(g) to be 0.411 M. 

Calculate the equilibrium constant, K_c, she obtained for this reaction. 

K_c = 

Answer 1

PCl3(g) +  Cl2(g)  <==>   PCl5(g) 

0.448.......0.477.................0..........I

-x............-x......................+x.........C

0.448-x...0.477-x.............x...........E

We are told that at equilibrium PCl5 = 0.411 M so that is x

At equilibrium we have the following concentrations:

[PCl3] = 0.448 - 0.411 = 0.037

[Cl2] = 0.477 - 0.411 = 0.066

[PCl5] = 0.411

Kc = [PCl5] / [PCl3][Cl2] = (0.411) / (0.037)(0.066)

Kc = 168