Unit 4 - stupid Kc (equilibrium concentration) doubt

I posted this as a separate thread (unanswered).... might as well delete that one and ask here?

CH3 COOH(l) + CH3 CH2 OH(l) <reversible react.>CH3 COOCH2 CH3 (l) + H2 O(l)

Initial concentration
-------------------------------
CH3COOH - 0.4 mol
CH3CH2OH - 0.3 mol
CH3COOCH2CH3 - 0.00 mol
H20 - 0.15

Equilibrium concentration
----------------------------------
CH3COOH - 0.2 mol
CH3CH2OH - UNKNOWN
CH3COOCH2CH3 - UNKNOWN
H20 - UNKNOWN


Q) Find the unknown values

If you do not post with specific doubts people are less inclined to help you...

What are you having problems with?

Q) Find the unknown values - I don't know how to solve this and would like to know how to find the answers.

You know how much ethanoic acid has reacted at equilibrium (0.4-0.2 mol), so using the reaction stoichiometry you can find out how many moles of the other components have either been made or used up.

Then, by subtracting the reactant values from the initial moles and adding the product values to the initial moles you have the moles of all components.

Oh! I get it. (maybe)

0.4 becomes 0.2 (that's 0.4-0.2)

since all moles of other compounds are same...

CH3CH2OH -> 0.3 - 0.2 = 0.1 mol

and in the product side I do...

CH3COOCH2CH3 -> 0.00 + 0.2 = 0.2
H20 -> 0.15 + 0.2 = 0.35

correct? or am I thinking it wrong and got this by fluke?