I've been stuck with this problem:

The following reaction has Kc = 2.8 · 10

^{2} at T=1000 K:

C

_{2}H

_{4} + H

_{2} C

_{2}H

_{6}A reactor with volume V = 10 L initially contains 0.32 mol C

_{2}H

_{4}, 0.16 mol H

_{2} and 0.68 mol C

_{2}H

_{6}.

Calculate the number of C

_{2}H

_{6} moles after:

a) the addition of 1.00 mol of C

_{2}H

_{6} in the reactor.

b) the volume of the reactor is increased by 20 L.

I know the answer to a) must be 1.515 moles, while b) must be 0.645 moles.

I can see these numbers make sense by LeChatelier principle. By adding 1 mole to 0.68 mol you get 1.68 mol of C

_{2}H

_{6}, but a part of it gets converted back to its reactants; by adding 20 L to the volume of the recipient, you shift the equilibrium position towards the reactants (because they are gaseous, so they are entropically favoured)

I tried to solve this exercise with ICE tables, resulting in

Numerator: 1.68 + x

Denominator: (0.32 - x) × (0.16 -x)

But when I solve for Numerator/Denominator = Kc, I do not get the answers above. I also do not know how to include volume in my calculation. I've tried using the ideal gas law and converting into partial pressures, but I still do not get the correct answer.