Calculate the Molar Enthalpy of Neutralization (ΔHn) in kJ/mol of the reaction between a monoprotic acid and a monoprotic base, given the following information:

Question

The temperature change equals 6.92°C,
50.0 mL of 1.00 M concentration of Acid
50.0 mL of 1.00 M concentration of Base
Heat capacity of the calorimeter is 6.50 J/°C.
The specific heat of water is 4.180 J/g°C

Answer 1

50 mL x 1.00M = 0.05 mols acid and 0.05 mols base.
q = delta H = dH = (mass H2O x specific Heat H2O x delta T) + (Ccal*delta T) where Ccal = heat capacity of the calorimeter.
Then q/n = dH/mol

Answer 2

57.8512kjml-

Answer 3

57851.2 jml-