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Enthalpy neutralisation question

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    • Thread Starter

    Just from a past paper this question I don't seem to understand the answer.

    The standard enthalpy change for the reaction of calcium hydroxide with
    hydrochloric acid was found by reacting 0.0100 mol of solid calcium hydroxide
    with 50.0 cm3 of a 1.00 mol dm–3 solution of hydrochloric acid (an excess), in a polystyrene cup. The temperature rose from 21.2 °C to 26.7 °C.

    Q) Calculate the standard enthalpy change, ΔH ○, for the reaction. Include a sign and units in your answer.

    I have established that the change in enthalpy is 1149.5J, but the mark scheme says this must be divided by 0.01 moles in order to get the standard enthalpy change. I infact divided it by 0.02 seeing as there is a 1:2 ratio between the calcium oxide and the water produced,

    Ca(OH)2(aq) + 2HCl(aq) → CaCl2(aq) + 2H2O(l)

    , which for neutralisations is what I think the energy change has to be divided by. But could someone help me out on this? Thanks!!

    Look up the definition for standard enthalpy change- the enthalpy change to form 1 mole of a water.

    Edit: Just realised it's even simpler.

    You divide the energy change by the number of moles right? Well you should use the compound which is limiting (not in excess). Therefore if you have 0.01mol of calcium hydroxide, you use that.

    The balanced equation involves two moles of HCl for every 1 mole of Ca(OH)2
    • Thread Starter

    Ah I see, silly mistake. Thanks to both of you!
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