For the given reaction -
                 Â
H2(g) + I2 (g) ---> 2 HI (g)
I(M)Â Â Â Â Â Â Â Â Â Â Â
0.6Â Â Â Â Â Â
0.6Â Â Â Â Â Â Â Â Â Â Â Â Â
0
CÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
-x      Â
-x             Â
+2x
Eq          Â
(0.6-x)Â Â Â
(0.6-x)Â Â Â Â Â Â Â Â 2x
Therefore -
                                    Â
Kc = 53.3 = (2x)2 / (0.6-x)2
By solving the above equation, the value of \"x\" is 0.470 M
Thus, at equilibrium, [HI] = 2(0.470 M) = 0.94
M