<u>Answer:</u> The enthalpy of the reaction is 
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
....(1)
Given mass of metal = 0.295 g
Molar mass of metal = 57.78 g/mol
Putting values in above equation, we get:

- To calculate the moles of a solute, we use the equation:

We are given:
Volume of hydrochloric acid = 65 mL = 0.065 L (Conversion factor: 1 L = 1000 mL)
Molarity of the solution = 1 moles/ L
Putting values in above equation, we get:

- For the given chemical reaction:

By Stoichiometry of the reaction:
1 mole of metal reacts with 2 moles of hydrochloric acid..
So, 0.0051 moles of metal will react with =
of hydrochloric acid.
As, given amount of hydrochloric acid is more than the required amount. So, it is considered as an excess reagent.
Thus, metal is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
If 1 mole of metal absorbs 104 J of heat
Then, 0.0051 moles of metal will absorb =
of heat.
Converting joules to kilo joules, we use the conversion factor:

Converting the given value in kilo joules, we get:

Hence, the enthalpy of the reaction is 