Question

Find the equilibrium moles for the following reaction with initially 0.051 mol for Ethyl acetate, 0.260 mol for water and 0.03045 mol of acetic acid at equilibrium.

I'm not sure how to get the ones at equilibrium for the left side or even how much you have to take away for the reaction.

CH3CH2COOCH3 + H2O ---> CH3CH2OH + CH3COOH

initial:
EthylAc: 0.051 mol
Water: 0.260 mol
EthylOH: --------
Acetic: --------

rxn:
EthylAc:
Water:
EthylOH:
Acetic:

eq:
EthylAc: ?
Water: ?
EthylOH: ?
Acetic: 0.03045 mol

Thank you

Answer 1

does it mean that because acetic acid has 0.03045 mol at equilibrium that x = 0.03045 mol? So then Ethanol would also have 0.03045 mol at eq and EtAc and H20 would both be their initial [conc] - x?

Answer 2

because it's a 1:1 ratio for all, then at rxn, we would have -x for both EtAc and H2O and + x on the right, so x = 0.03045.
Is this correct?

Answer 3

You have run your sentence together and I can't be sure about the concentrations. From the table, however, it appears initially you have EtOAc of 0.051 and H2O of 0.260 (so initially EtOH = 0 and acetic acid = 0).
Yes, if you have 0.03045 mols acetic acid at equilibrium you will have 0.03045 mols EtOH at equilibrium.
.......EtOAc + H2O ==> EtOH + HAc
I......0.051..0.260.....0......0
C.......-x.....-x.......x.......x
E.....0.051-x.0.260-x...x......x
So = 0.03045 which allows you to place numbers for all and calculate Keq.(at least I suppose that is what you are to do).

Answer 4

okay thanks Dr.Bob, that's exactly what I did too.