Calorimetry question

Question

If 51.8 mL of 1.3 M NaOH react with excess HCl creating 2742.3 J of heat, what is the ΔH_{neutralization} in kJ/mol?

 

NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

Answer 1

NaOH(aq) + HCl(aq) ===> NaCl(aq) + H2O(l) ... balanced equation

moles of NaOH used = moles HCl used = 51.8 ml NaOH x 1 L/1000 ml x 1.3 mol/L = 0.06734 moles

This number of moles created 2742.3 J of heat. The ∆H_{neutralization} per mole would thus be...

2742.3 J / 0.06734 moles = 40,723 J/mole = 40.7 kJ/mole (to 3 significant figures)