The equilibrium constant ...

Just started a new topic on the equilibrium constant and am attempting the first piece of work set on it. Unsure if i'm doing this correctly, and would rather rectify any mistakes earlier on than later ... confirmations or corrections most appreciated :smile:

1) When 1.0 mol each of ethanol and ethanoic acid are mixed together at a fixed temperature, the reaction mixture at equilibrium is found to contain 0.66 mol of ethyl ethanoate.

a) Write an equation for the reaction between ethanol and ethanoic acid (1)

CH₃COOH + CH₃CH₂OH (reversible reaction arrow sign) CH₃COOC₂H₅ + H₂O

(Am i right in thinking water is produced as well?)

b) The total volume of the reaction mixture is 0.10 dm³. Calculate the equilibrium concentrations of ethyl ethanoate and ethanol in the reaction mixture (2)

Equilibrium concentration of ethyl ethanoate = 0.66 / 0.1 = 6.6

Equilibrium concentration of ethanol = (1-0.66) / 0.1 = 0.34 / 0.1 = 3.4

c) Write an expression for the equilibrium constant, K_c, for this reaction (2)

K_c = {[Ethyl ethanoate]*[water]} / {[Ethanol]*[Ethanoic acid]}

d) Calculate the value of the equilibrium constant for this reaction at this fixed temperature (2)

K_c = {6.6 * 6.6} / {3.4*3.4}
= 3.768 no units

Would the concentration of the water also be 0.66, or have I done that wrong?