Given :
[CO]=0.1550 M and [Cl2] = 0.178 MÂ Â
      at T = 1000 K.
Kc = 255
Lets use ICE chart.
CO (g)
+Â Â Â Â Â Â Â Â Â Â Cl2
(g) ⇌ COCl2 (g)
IÂ Â Â Â Â Â Â Â Â Â
0.1550
                      Â
0.178
            Â
0
CÂ Â Â Â Â Â Â Â Â
-x       Â
          Â
-x                   Â
+x
EÂ Â Â Â Â Â Â Â Â
(0.1550-x)Â Â Â Â Â Â
(0.178-x)Â Â Â Â Â Â Â Â x
Kc = [COCl2]/[CO][Cl2]
255 = x / ( 0.1550-x) ( 0.178-x)
255 = x/ (x2 – 0.333x + 0.02759 )
255 x2 – 84.915x + 7.035 = x
255 x2 – 85.915x + 7.035 = 0
We use quadratic formula to solve this problem
By using quadratic formula we get
x = 0.14 = [COCl2] = 0.14 M
[CO]= 0.1550 – 0.140 = 0.015 M
[Cl2]= 0.178-0.140 = 0.038 M
