Question

Calculate the molar heat of neutralization in kJ/mol of the reaction between HA and BOH given the following information:
The temperature change equals 7C,
50 mL of 1 M concentration of Acid
50 mL of 1 M concentration of Base
Heat capacity of the calorimeter is 6.5 J/C.
Remember that the specific heat of water is 4.18 J/gC

Answer 1

q = (mass H2O x specific heat H2O x delta T) + (heat cap calorimeter x delta T).
q = (100 g x 4.18 x 7) + (6.5 x 7)= ?? joules.

Change to kJ. You want kJ/mol, you have kJ/0.05 mol.