At a certain temperature, 0.920 mol of SO3 is placed in a 4.00-L container.?

________ 2 SO₃(g) ⇌ 2 SO₂(g) + O₂(g)
Initial ___ 0.920 ______ 0 ______ 0 ___ mol
Change ___ -2y _____ +2y ____ +y ___ mol
At eqm _ 0.920-2y ____ 2y _____ y ___ mol

It is given that y = 0.200

Hence, at eqm :
[SO₃] = (0.920 - 2y) / 4 mol/L = (0.920 - 2 × 0.200) / 4 mol/L = 0.130 mol/L
[SO₂] = 2y/4 mol/L = 2 × 0.200 / 4 mol/L = 0.100 mol/L
[O₂] = 0.200 / 4 mol/L = 0.0500 mol/L

Kc = [SO₂]² [O₂] / [SO₃]² = 0.100² × 0.0500 / 0.130² = 0.0296 (mol/L)