Calculating Equilibrium Partial Pressures Given Kp and Mass For the reaction
  $$\ce{AsCl5(g) &lt;=&gt; AsCl3(g) + Cl2(g)}$$
  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?
Say you start with $A$ moles of $\ce{AsCl5}$, and then you reach a temperature $T$, for which the equilibrium constant is $K_p$.

In order to achieve equilibrium, some amount of forward reaction ($\ce{AsCl5 -&gt; AcCl3 + Cl2}$) happens to consume $x$ moles of $\ce{AsCl5}$. Based on the stoichiometry of the equation, you know that exactly $x$ new moles of the products are formed.

$$K_p=\frac{P_{\ce{AsCl3}}P_{\ce{Cl2}}}{P_{\ce{AsCl5}}}$$
Since for $V$, $R$, $T$ are constant, we can replace the above equation by:
$$K_p=\frac{RT}{V}\frac{n_{\ce{AsCl3}}n_{\ce{Cl2}}}{n_{\ce{AsCl5}}}=\frac{RT}{V}\frac{x^2}{A-x}$$
Solve this for $x$, and substitute in the ideal gas equation to get the answer.