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    i was hoping someone could help me with a Kc question i have come across that i cant seem to get right:

    0.204 mol CH3COOC2H5 and 0.645 mol H2O were introduced into a flask, acidified and left until a steady state was reached. a sample was then analysed, and 0.114 mol of CH3COOH was found. calculate Kc for the reaction.

    CH3COOC2H5(l) + H2O(l) -> CH3COOH(l) + C2H5OH(l)
    0.204 0.645 0.114 x
    i dont even know how to go about it, there is no volume of the flask givem. is the question just wrong, or am i missing something?

    the answer is given as 0.27

    any help would be appreciated!!!
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    is the anser given in the question, or did you get it from the markscheme?
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    Since 0.114 moles of CH3COOH is present in the equilibrium mixture, 0.114 moles of CH3COOC2H5 will have been used up. Hence:

    moles CH3COOC2H5 in equilibrium mixture = 0.204 - 0.114 = 0.0900 moles

    The same is true for water, hence:

    moles H2O in equlibrium mixture = 0.645 - 0.114 = 0.531 moles

    If 0.114 moles of CH3COOH is present in the equilibrium mixture, then 0.114 moles of C2H5OH must also be present in the equilibrium mixture.

    Therefore:

    Kc = 0.114 x 0.114 / 0.531 x 0.0900 = 0.272 (3sf)

    It's all to do with molar ratios, which in this case is 1:1.

    The volumes cancel out, so it's not required.
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    Okay...i think i get that. i just didnt get what the mols of C2H5OH would be, but i take it if you take amounts from the equation, you assume it is 0.114?

    excellent! thanks a lot!!!
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    Yeah, it's all about the equation. The equation basically says: for every mole of ethanoic acid produced, you will also get one mole of ethanol.
 
 
 
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