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    Ethanoic acid reacts with ethanol in a reversible reaction represented by the equation below.
    In an experiment 3.0 mol of ethanoic acid were mixed with 1.0 mol of ethanol and when the reaction had reached equilibrium 0.9 mol of water had been formed.


    CH3COOH(l) + C2H5OH(l) <--> CH3COOC2H5(l) + H2O(l)

    The percentage of ethanoic acid converted into the ester CH3COOC2H5 in this reaction is

    A 22.5%

    B 30%

    C 43%

    D 90%

    Can someone send a worked solution to get the answer B? I'm getting C each time.
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    Tell you what, show your worked answer that gives C.
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    (0.9/3)*100 = 30%
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    Won't let me upload to TSR
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    Won't let me upload to TSR. I've posted the link to my solution, but it's getting approved first...
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    (Original post by TutorsChemistry)
    Tell you what, show your worked answer that gives C.
    Put the link together all as one,
    i.imgur
    .com/
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    (Original post by jacky33041)
    Put the link together all as one,
    i.imgur
    .com/
    BO1z0Dz.jpg


    The problem is that you are dividing by 2.1, the equilibrium quantity of ethanoic acid.
    The question asks how much has been converted, so you must divide by the initial quantity - divide by 3 rather than 2.1.
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    Thank you.
 
 
 
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Updated: March 12, 2018
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