i used a formula I derived myself from the equation i showed above:

Original:- qrxn = - nΔH°= CcalΔTcal + CvVacidΔTacid + CvVbaseΔTbase

Derived: - nΔH° = (Ccal + CvVacid)(Tf – Ti) + CvVbase(Tf-Ti)

I need to plug them into the original equation above (derived a eq. only because I got a larger number than friend):

ΔH° = 64163 J/mol

n = 0.05 mol

change in temperature for the acid and the calorimeter = 5.97°C

change in temperature of the base = 6.31°C

volume of both base and acid = 50 mL

heat capacity of water = 4.16 J/mL °C

Using derived: I got 109.6

Compared to a friend: She got 34.5

I need you or someone to calculate it for me and tell me if im right or wrong.

Thank you, that’s all the info I have.