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2) 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:

The temperature change equals 5.06°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

This is how I calculated
5*5*5.06*6.50= 822.25/4.180= 196.7

You haven't used specific heat H2O anywhere.
q = mass H2O X specific heat H2O x delta T + Ccal*delta T.
What's the 5*5 bit?
100 mL solution = 100 g H2O
q = [100g x 4.180 x 5.06] + [6.50 x 5.06] = ?
q/n = dH/mol = ?/0.05 = x.

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