At equilibrium, 0.170 mol of o2 is present. calculate kc.

(missing in your question):

0.66 mol of SO3 is placed in 4.5 L container according to this reaction

2SO3 (g) ↔ 2SO2 (g) + O2 (g)

So the answer:

for the equilibrium reaction so:

Kc = [SO2]^2 * [O2] / [ SO3]^2

when concentration (c) = n / V when n = no. of moles / v (volume per L)

∴ [ SO3]° = 0.66 / 4.5 = 0.147 M

by using Ice table we can determine the concentrations:

2SO3 (g) ↔ 2SO2 (g) + O2 (g)

0.147-2X 2x X

∴ [ O2] = 0.17 M / 4.5 L = 0.0378 M

So X = 0.0378

∴ [SO3] = 0.147 - 2 * 0.0378 = 0.0714 M

and [ SO2] = 2 * 0.0378 = 0.0756 M

∴ by substitution in Kc formula

∴Kc = [0.0756]^2 * [0.0378] / [0.0714] = 0.003