Using the following thermochemical data, what is the change in enthalpy for the following reaction? ca(oh)2(aq) + hcl(aq) cacl2(aq) + 2h2o(l)cao(s) + 2hcl(aq) cacl2(aq) + h2o(l),   dh = -186 kj cao(s) + h2o(l) d ca(oh)2(s), h = -65.1 kj

Ca(OH)2(aq) + 2HCl(aq)> CaCl2(aq) + 2H2O(l) ΔH-?

CaO(s) + 2HCl(aq)> CaCl2(aq) + H2O(l),  ΔH = -186 kJ

CaO(s) + H2O(l) > Ca(OH)2(s), ΔH = -65.1 kJ

1) Ca(OH)2 should be  reactant, so
CaO(s) + H2O(l) > Ca(OH)2(s) 
we are going to take as 
 Ca(OH)2(s)>CaO(s) + H2O(l), and ΔH = 65.1 kJ

2) Add 2 following equations
Ca(OH)2(s)>CaO(s) + H2O(l),                    and ΔH = 65.1 kJ
CaO(s) + 2HCl(aq)> CaCl2(aq) + H2O(l), and ΔH = -186 kJ

Ca(OH)2(s)+CaO(s) + 2HCl(aq)--->CaO(s) + H2O(l)+CaCl2(aq) + H2O(l)

Ca(OH)2(s)+ 2HCl(aq)---> H2O(l)+CaCl2(aq) + H2O(l)
By addig these 2 equation, we got the equation that we are needed,
so to find enthalpy of the reaction, we need to add  enthalpies of reactions we added.
ΔH=65.1 - 186 ≈ -121 kJ