Your answer would be 0.00285 moles.
The answer of this answer is given in the attached file
Explanation:
The reaction equation will be as follows.

Hence, moles of Na = moles of electron used
Therefore, calculate the number of moles of sodium as follows.
No. of moles = 
=
(as 1 kg = 1000 g)
= 195.65 mol
As, Q =
where F = Faraday's constant
= 
=
mol C
Relation between electrical energy and Q is as follows.
E = 
Hence, putting the given values into the above formula and then calculate the value of electricity as follows.
E = 
= 
= 
As 1 J =
kWh
Hence,
kWh
= 3.39 kWh
Thus, we can conclude that 3.39 kilowatt-hours of electricity is required in the given situation.
Ca(OH)2(aq) + 2HCl(aq)------> CaCl2(aq) + 2H2O(l) ΔH-?
CaO(s) + 2HCl(aq)-----> CaCl2(aq) + H2O(l), Δ<span>H = -186 kJ
</span>
CaO(s) + H2O(l) -----> Ca(OH)2(s), Δ<span>H = -65.1 kJ
</span>
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
<span><u>CaO(s) + 2HCl(aq)-----> CaCl2(aq) + H2O(l), and ΔH = -186 kJ</u>
</span>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