Calculate the equilibrium constant, Kc, for the ethanol-acetic acid reaction based on this experiment.?

Consider the titration between the 1/100 of the equilibrium mixture and Ba(OH)₂.
2CH₃COOH(aq) + Ba(OH)₂(aq) → Ba(CH₃COO)₂(aq) + 2H₂O(l)
Mole ratio CH₃COOH : Ba(OH)₂ = 2 : 1

No. of moles of Ba(OH)₂ needed = (0.1020 mol/L) × (28.95/1000 mL) = 0.002953 mol
No. of moles of CH₃COOH in 1/100 of the reaction mixture = (0.002953 mol) × 2 = 0.005906 mol
No. of moles of CH₃COOH in the reaction mixture = (0.005906 mol) × 100 = 0.5906 mol

Consider the esterification :
____________ C₂H₅OH ___ + ___ CH₃COOH ___ ⇌ ___ CH₃COOC₂H₅ ___ + ___ H₂O …... Kc
Initial: _____ 0.5000 mol ______ 1.000 mol ___________ 0 mol __________ 0 mol
Change: ______ -y mol __________ -y mol ___________ +y mol __________ +y mol
At eqm: __ (0.5000 - y) mol ___ (1.000 - y) mol ________ y mol ___________ y mol

At eqm:
1.000 - y = 0.5906
y = 0.4094
[C₂H₅OH] = (0.5000 - 0.4094) M = 0.0906 M
[CH₃COOH] = 0.5906 M
[CH₃COOC₂H₅] = [H₂O] = 0.4094 M

Let V L be the volume of the reaction mixture.

Kc = [CH₃COOC₂H₅] [H₂O] / {[C₂H₅OH] [CH₃COOH]}
= (0.4094/V M)² / {(0.0906/V M) × (0.5906/V M)
= 0.4094² / {0.0906 × 0.5906)
= 3.13

Here's a hint: You know the concentration and volume of BaOH2 so you can calculate the moles reacted, multiply that by 2 and it will give you the moles of acetic acid. Next you need to multiply that by 100 to give the total acetic acid in the equilibrium mixture. Then you write out the equation for the equilibrium constant, but you need to find the concentrations of the other compounds still. Do that by taking the initial amount and subtracting the amount reacted. (For the reactants) or by just using the amount reacted for the products. Remember water doesn't go in the mass action expression