At a certain temperature, 0.620 mol of so3 is placed in a 1.50-l container. 2so3< --> 2so2+o2. at equilibrium, 0.140 mol of o2 is present. calculate kc

2SO3 <> 2SO2 + O2 
initially... 
0.76 <> 0 + 0 

at eq. 
0.76-2x <> 2x + x 

given x = 0.16 

so no. of moles of SO2 at eq = 2x = 2X0.16 = 0.32 
no. of moles of SO3 = 0.76-2x = 0.76-0.32 = 0.44 

[SO3] = eq. conc. of SO3 = 0.44/3.5 = 0.126 M 
[SO2] = 0.32/3.5 = 0.092 M 
[O2] = 0.16/3.5 = 0.046 M 

Kc = [SO2]^2 [O2] / [SO3]^2 

Kc = 0.092^2 X 0.046 / 0.126^2 

Kc = 8.464 X 10^-3 X 0.046 / 0.016 

Kc = 3.893 X 10^-4 / 0.016 = 0.024

Ok where is the image ?

Density and pressure