For the reaction $$\ce{AsCl5(g) <=> AsCl3(g) + Cl2(g)}$$ at $550$ K, the equilibrium constant ($K_p$) is $9.81$. Suppose that $3.150 \ g$ $\ce{AsCl5}$ is placed in an evacuated $600$ ml bulb, which is then heated to $550$K.
What is the partial pressure of $\ce{AsCl5}$ at equilibrium?
So at the moment, I understand that $$\ce{K_p=$$\frac{(P_{AsCl3})(P_{Cl_2})}{(P_{AsCl_5})}$$}$$ And that using a rearranged Ideal Gas Law I can get $$\ce{P=$$\frac{n(0.0125mol AsCl5)*R*T(550K)}{V (0.6L)}$$}$$ Giving me $$\ce{0.94 atm}$$ but I'm not sure where to go from here. Can somebody help by pointing me in the right direction?