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    I did this question of June 05 but theres one thing about the answer I dont understand.
    I worked out the moles of propanoyl chloride but apparently the moles of NaOh is double this. Why?

    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
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    (Original post by Tomato_Soup1992)
    I did this question of June 05 but theres one thing about the answer I dont understand.
    I worked out the moles of propanoyl chloride but apparently the moles of NaOh is double this. Why?

    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
    Acid chlorides are hydrolysed by water to carboxylic acids and HCl, both of which react with NaOH.

    RCOCl + H2O --> RCOOH + HCl

    RCOOH + NaOH --> RCOONa + H2O
    HCl + NaOH --> NaCl + H2O

    Hence 1 mole of an acid chloride produces two moles of acid on hydrolysis, which in turn needs 2 moles of NaOH for neutralisation
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    Didnt think about the HCl, thanks
 
 
 
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