Answer: 54.4 kJ/mol
Explanation:
First we have to calculate the moles of HCl and NaOH.


The balanced chemical reaction will be,

From the balanced reaction we conclude that,
As, 1 mole of HCl neutralizes by 1 mole of NaOH
So, 0.05 mole of HCl neutralizes by 0.05 mole of NaOH
Thus, the number of neutralized moles = 0.05 mole
Now we have to calculate the mass of water:
As we know that the density of water is 1 g/ml. So, the mass of water will be:
The volume of water = 

Now we have to calculate the heat absorbed during the reaction.

where,
q = heat absorbed = ?
= specific heat of water = 
m = mass of water = 100 g
= final temperature of water = 
= initial temperature of metal = 
Now put all the given values in the above formula, we get:


Thus, the heat released during the neutralization = 2.72 KJ
Now we have to calculate the enthalpy of neutralization per mole of
:
0.05 moles of
releases heat = 2.72 KJ
1 mole of
releases heat =
Thus the enthalpy change for the reaction in kJ per mol of HCl is 54.4 kJ