Answer:
0.28 M
Explanation:
Concentration of [COFâ‚‚] = 2.00 M
Considering the ICE table for the equilibrium as:
           2COF₂ (g)    ⇔      CO₂ (g) +    CF₄ (g)
t = o         2.00
t = eq         -2x               x            x
--------------------------------------------- --------------------------
Moles at eq:   2.00-2x             x            x
   Â
The expression for the equilibrium constant is:
[tex]K_c=\frac {[CO_2][CF_4]}{[COF_2]^2}=8.80[/tex] Â
So,
[tex]\frac{x^2}{(2.00-2x)^2}=8.80[/tex] Â
Solving for x, we get that
x = 0.86 M
Equilibrium concentrations :
[COâ‚‚] = [CFâ‚„] = 0.86 M
[COFâ‚‚] = 2.00 - 2*0.86 = 0.28 M