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My question is for the part circled in pencil, how do we get -12,970.4 and then -25,940.8 j/mole. Are we normalizing twice? If so, why?

My question is for the part circled in pencil, how do we get -12,970.4 and then -25,940.8 j/mole. Are we normalizing twice? If so, why?

Chemistry: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter14: Chemical Equilibrium
Section: Chapter Questions
Problem 14.97QE: According to the Resource Conservation and Recovery Act (RCRA), waste material is classified as...
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My question is for the part circled in pencil, how do we get -12,970.4 and then -25,940.8 j/mole. Are we normalizing twice? If so, why? 

Chemistry I Science Supplemental Problems from Chapter 7 CCC - Dr. Feudale 4. When 500 mL of 1.00 M barium nitrate solution at 25°C is mixed with 500 mL of 1.00 M sodium sulfate solution at 25°C, a white solid forms and the temperature of the mixture increases to 28.1°C. Calculate AH for this precipitation reaction per mole of BaSO4. Net Ionic reaction: Ba2+(aq) + SO, (aq) → BaSO4(s) Assume no heat lost to the surroundings: Heat released by the reaction = Heat absorbed by solution Assume Csoln = Cwater = 4.184 J/g-°C and dsoln = dwater = 1.00 g/mL Because the temperature of the solution increased, the reaction must be exothermic -AHreaction = qsolution = msolnCsolnATsoln -AHreaction = -12,970.4 J Moles of Ba2+ = Moles of SO42- = Moles of BaSO4 = (0.500 L)(1.00 M) = 0.500 moles %3D %3D %3D -AHreaction = -12,970.4 J/0.500 moles =-25,940.8 J/mole = -25.9 kJ/mole CCC - Dr. Feudale Chemistry I Science Supplemental Problems from Chapter 7 4. When 500 mL of 1.00M barium nitrate solution at 25°C is mixed with 500 mL of 1.00 M sodium sulfate solution at 25°C, a white solid forms and the temperature of the mixture increases to 28.1°C. Calculate AH for this precipitation reaction per mole of BaSO4. 1.00g 4e184y 3.1°C I mi 19°C 6485.2 -12,970.43=-AH %3D 500ml %3D 0.5
Transcribed Image Text:Chemistry I Science Supplemental Problems from Chapter 7 CCC - Dr. Feudale 4. When 500 mL of 1.00 M barium nitrate solution at 25°C is mixed with 500 mL of 1.00 M sodium sulfate solution at 25°C, a white solid forms and the temperature of the mixture increases to 28.1°C. Calculate AH for this precipitation reaction per mole of BaSO4. Net Ionic reaction: Ba2+(aq) + SO, (aq) → BaSO4(s) Assume no heat lost to the surroundings: Heat released by the reaction = Heat absorbed by solution Assume Csoln = Cwater = 4.184 J/g-°C and dsoln = dwater = 1.00 g/mL Because the temperature of the solution increased, the reaction must be exothermic -AHreaction = qsolution = msolnCsolnATsoln -AHreaction = -12,970.4 J Moles of Ba2+ = Moles of SO42- = Moles of BaSO4 = (0.500 L)(1.00 M) = 0.500 moles %3D %3D %3D -AHreaction = -12,970.4 J/0.500 moles =-25,940.8 J/mole = -25.9 kJ/mole CCC - Dr. Feudale Chemistry I Science Supplemental Problems from Chapter 7 4. When 500 mL of 1.00M barium nitrate solution at 25°C is mixed with 500 mL of 1.00 M sodium sulfate solution at 25°C, a white solid forms and the temperature of the mixture increases to 28.1°C. Calculate AH for this precipitation reaction per mole of BaSO4. 1.00g 4e184y 3.1°C I mi 19°C 6485.2 -12,970.43=-AH %3D 500ml %3D 0.5
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