# Help with chem homework? Calorimetry?

Here is my problem:

50.0 mL of 0.1 M HCl are combined with 50.0 mL of 0.1 M NaOH. The initial temperature

of both solutions is 25.4oC. The temperature of the final mixture is 26.9oC. Calculate the

heat of neutralization (DHreaction) in kJ/mole of water formed. Assume that volumes are

additive, the density of the solutions is the same as that of water (1 g/mL) and the specific

heat of the solutions is also the same as water, 4.184 J/g K. (Answer: 125.4 kJ/mol of H2O)

So far I calculated the heat of the solution, any help would be great, thanks!

### 1 Answer

- FernLv 71 year agoFavourite answer
100 grams x (26.9-25.4) x 4.184Joules/gC = 627 Joules

NaOH + HCl ===> NaCl + H2O

0.050 liters of NaOH x 0.10 moles NaOH/liter = 0.005 moles NaOH

0.005 moles of NaOH will react with 0.005 moles of HCl to produce 0.0050 moles of H2O

reaction is exothermic, so delta H is negative.

-627 Joules/0.005moles H2O = -125,400 Joules/mole H2O

-125,400 Joules x 1 kilojoule/1000 Joules = -125.4 kJ/mole H2O