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    • Thread Starter


    I cant understand all of the following question, mark scheme attached:

    (c) Write an equation for the reaction of propanoyl chloride with water.
    An excess of water is added to 1.48 g of propanoyl chloride. Aqueous sodium hydroxide
    is then added from a burette to the resulting solution.
    Calculate the volume of 0.42 mol dm–3 aqueous sodium hydroxide needed to react
    exactly with the mixture formed. (5 marks)

    M1 CH3CH2COCl + H2O → CH3CH2COOH + HCl
    (penalise wrong alkyl group once at first error)

    M2 Mr of CH3CH2COCl = 92.5
    (if Mr wrong, penalise M2 only)

    M3 moles of CH3CH2COCl = 1.48/92.5 = 0.016 1

    M4 moles NaOH = 2 × 0.016 = 0.032 1
    (allow for × 2 conseq to wrong no of moles)

    M5 volume of NaOH = 0.032/0.42 = 0.0762 dm3 or 76.2 cm3
    (if ×2 missed in M4 lose M5 also)

    I can understand up to M3 however I cannot see why the moles of NaOH have been multiplied by 2... could someone please explain this to me?:confused:

    Thanks for reading...
    • Thread Starter

    Oh damn.. i just read the mark scheme working again and I can see how 2 'separate moles' of acid are formed in M1 hence the consequential NaOH x 2...

    Sorry!... how do i delete this thread?
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