Answer: The enthalpy change of neutralization is 0.571 kJ/mol
Explanation:
The balanced chemical reaction for neutralization experiment using NaOH and HCl will be :
[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]
As NaOH is the limiting reagent, which is 1.254102 moles which will decide how much heat is released.
When 1.254102 moles of NaOH reacts, heat released will be = 716 J
Thus 1 mole moles of NaOH reacts, heat released will be = [tex]\frac{716}{1.254102}\times 1=571J=0.571kJ[/tex] (1kJ=1000J)
Thus the enthalpy change of neutralization is 0.571 kJ/mol
