Answer:
37.82kJ/mol is the molar heat of neutralization
Explanation:
The neutralization of a monoprotic acid, HX, with NaOH as example is:
HX + NaOH → H₂O + NaX + ΔH
Where ΔH is molar heat of neutralization (The heat that is released when 1 mole of HX is neutralized).
As neutralization of 156mmol = 0.156mol HX release 5.90kJ. When 1 mole is neutralized the released heat is:
ΔH = 5.90kJ / 0.156moles
ΔH = 37.82kJ/mol is the molar heat of neutralization
