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    an ester is hydrolysed as shown:
    RCOOR + h20 ----> RCOOH + ROH
    what is percentage yield of RCOOH when 0.5g of RCOOH (mr=100) is obtained from 1g of RCOOR (mr=150)?

    please can someone explain how to answer this
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    (Original post by hanakm)
    an ester is hydrolysed as shown:
    RCOOR + h20 ----> RCOOH + ROH
    what is percentage yield of RCOOH when 0.5g of RCOOH (mr=100) is obtained from 1g of RCOOR (mr=150)?

    please can someone explain how to answer this
    all you need t know is that

    n = m / Mr

    n is number of moles
    m is mass
    Mr is relative molecular mass

    The first thing to work out is the maximum amount of product that could be made.
    Moles of starting material is 1gram/ 150grams per mole = 0.00666 moles.

    the reaction converts 1 mole of starting material to 1 mole of products so 0.00666 moles is the maximum amount of product we could obtain.

    0.00666 moles of product is equivalent to 0.00666moles x 100 grams per mole = 0.666 grams

    We actually only make 0.5grams so the percentage yield is
    (0.5/0.666)x100 = 75% yield
 
 
 
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