Question

Sulfur dioxide reacts with oxygen to form sulfur trioxide according to the equation

2SO2(g) + O2(g) <-> 2SO3(g)

Samples of sulfur dioxide, oxygen, and sulfur trioxide were added to a flask of volume 1.40 dm^3 and allowed to reach equilibrium at a given temperature. The flask contained 0.0550 mol of sulfur dioxide and 0.0720 mol of sulfur trioxide at equilibrium.

Kc has the numerical value of 27.9 under these conditions.
Calculate the amount, in moles, of oxygen gas in this equilibrium mixture.

Not really sure how to get the answer, as the only questions I've had up to this point made it easy to get the concentration, but here it's a gas

The only thing I've figured out is that
Kc = [SO3]^2 / [SO2]^2 x [O2]
So 27.9 = [SO3]^2 / [SO2]^2 x [O2]

Thank you so much!

Answer 1

You know from your Kc equation that you need concentrations. You have mols and volume; therefore, you know at equilibrium that [SO2] = 0.0550 mol/1.40 dm^3 and [SO3] = 0.0720 mol/1.40 dm^3. Therefore, [SO2] = approx 0.04 M and [SO3] = approx 0.05. You will need to recalculate both because I've only estimated. Now plug these into the equation.
........2SO2 + O2 ==> 2SO3
I............................
C............................
E........0.04...x.....0.05

So plug in the [SO2] and [SO3] values into Kc expression and solve for [O2] and the answer will be in units of of mols/dm^3 = M. The problem asks for mols. You know the volume and you know M, solve for mols. Post your work if you get stuck.

Answer 2

This helped a lot, thanks for the easy to understand explanations ^_^

Answer 3

You're welcome.

Answer 4

however many litters of sulphur trioxide are formed when 4800cm of sulphur dioxide is burned in air?