chm152
posted by ami .
PCl3 + Cl2 = PCl5 Kc=26 @ 250 degree Celcius
1.0 mol of PCl3 , 1.0 mol of Cl2 , and 5.0 mol of PCl5 are placed in a 10.0L flask at 250 C and allowed to come to equilibrium. What is the concentration, in M,of cl2 of at equilibrium?
Choose one answer.
a. 3.4
b. 0.19
c. 0.16
d. 0.13
e. 0.20

PCl3 + Cl2 ==> PCl5
(PCl3) = 1.0 mol/10.0 L = xx M
(Cl2) = 1.0 mol/10.0 L = yy M
(PCl5) = 5.0 mol/10.0 L = zz M
First determine the reaction quotient, compare it with Keq, and determine which way the reaction will move to reach equilibrium. Then set up an ICE chart, substitute into the Keq expression, and solve for the unknown; use that to calculate each individual concn. I think the reaction will move to the left and I did a quickie calculation (but you should confirm it because I may have made a mistake in my quickie) and I'm looking at 0.13 M for the answer. 
If you will go to the original post, I saw there how you had worked the problem and I commented on where your error(s) were.