posted by jesse .
A 50.0 mL solution of 1.2 M HCl at 24.1C is mixed with 50.0 mL of 1.3 M NaOH, also at 24.1C, in a coffee-cup calorimeter. After the reaction occurs, the temperature of the resulting mixture is 29.8C. The density of the final solution is 1.05 g/mL. Calculate the molar heat of neutralization. Assume the specific heat of the solution is 4.184 J/gC. The heat capacity of the calorimeter is 32.5 J/C
So you reacted 1.2M x 0.050L = 0.06 mols HCl with 0.06 mols NaOH in 100 mL solution.
mass solution = 100 mL x 1.05 g/mL = 105 grams.
delta T = dT = 29.8-24.1 = 5.7 degrees.
q = (mass soln x specific heat x dT) + (Ccal x dT)
q = (105 x 4.184 x 5.7) + (32.5 x 5.7) = ? J for 0.06 mols H2O formed.
Convert to J/1 mol and change that to kJ/mol.