CHEM
posted by Akle on .
I know I posted this question again but no one is really helping me. Can someone please help me as soon as possible.
In a constantpressure calorimeter, 60.0 mL of 0.340 M Ba(OH)2 was added to 60.0 mL of 0.680 M HCl. The reaction caused the temperature of the solution to rise from 24.07 °C to 28.70 °C. If the solution has the same density and specific heat as water, what is ΔH for this reaction (per mole of H2O produced)? Assume that the total volume is the sum of the individual volumes.

This is my work.....
(18g) (4.184 J)(28.7024.07)
I get 350.95
then i divide it by 0.0340 and get 10322.17. Then I divide it by a 1000 and get 10.322... It is wrong. Can you take a look at my work and tell me what i did wrong and confirm the right answer for me thanks! 
Where do you come up with 18g? That is 60 mL of one reagent plus 60 ml of the other. With a density of 1.00 g/mL, the water has a mass of 120 grams. So substitute 120 for the 18 and recalculate that part.
The second error is in the moles. I calculated 60.0 mL x 0.340M = 20.4 mmoles Ba(OH)2 and
60.0 mL x 0.680M = 40.8 mmoles HCl. Therefore mol H2O formed is
......Ba(OH)2 + 2HCl ==> BaCl2 + 2H2O
......0.0204...0.0408...0.0204.0.0408
So q from above divided by 0.0408 gives q/mole H2O. Watch the s.f., a common problem with these databases. 
Your post doesn't give if that is to be J/mol or kJ/mol. Watch that too.

I got an answer of 56.976. The answer is negative though since the reaction is exothermic. Other than that great explanation. Thank you so much. You are the best!!!! :]

yeah i converted it. It was suppossed to be kj :]