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!!