At a certain temperature, 0.660 mol of so3 is placed in a 3.00-l container.
2so3(g) < > 2so2 + o2(g)
at equilibrium, 0.160 mol of o2 is present. calculate kc.

When this equilibrium shifts to the right, forming "X" mol of O2
2SO3(g) ---> 2SO2(g) + O2(g) 0.660 - 2X --> --> 2X & X since X = 0.130 this becomes 0.400 <=> 0.260 & 0.130 find moles / litre: (0.400) / 4.5L --> --> 0.260 / 4.5L & 0.130 / 4.5 L which gives these molarities: 2SO3(g) ---> 2SO2(g) + 1 O2(g) [0.08889] --> [0.05778] & [0.02889] Kc = [SO2]^2 [O2] / [SO3]^2 Kc = [0.05778]^2 [0.02889] / [0.08889]^2 Kc = (0.003338) [0.02889] / [(0.007901) Kc = 0.0122

The correct answer would be 1. precipitation