Answer : The enthalpy of neutralization is, -113.9 KJ/mole
Explanation :
First we have to calculate the moles of
and
.


The balanced chemical reaction will be,

From the balanced reaction we conclude that,
As, 2 mole of
neutralizes by 1 mole of 
As, 0.0294 mole of
neutralizes by
mole of 
Thus, the number of neutralized moles = 0.0147 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 = 140.0 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 = -1.675 KJ
Now we have to calculate the enthalpy of neutralization.

where,
= enthalpy of neutralization = ?
q = heat released = -1.675 KJ
n = number of moles used in neutralization = 0.0147 mole

The negative sign indicate the heat released during the reaction.
Therefore, the enthalpy of neutralization is, -113.9 KJ/mole