Experiment 11 lab report - Experiment 12 Heat of Neutralization CHE 1211L-0 Mr Matthew Edwin Thomas Miller Minseong Park 1 The balanced equation of

# Experiment 11 lab report - Experiment 12 Heat of...

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Experiment 12 : Heat of Neutralization CHE 1211L-0 Mr. Matthew Edwin Thomas Miller Minseong Park
1. The balanced equation of neutralization of HCl and NaOH shows a 1:1 molar ratio of the two reactants. Therefore both six moles of HCl and six moles of NaOH are the limiting reactants and will yield the same enthalpy change. a. Enthalpy of neutralization of HCl and NaOH: 56.4 J/mmol b. 6 moles of HCl = 6 moles of NaOH = 6 * 56.4 J/mmol = 338.4 J/mmol 2. a. 1 Joule = 0.239 calories b. The heat of neutralization of acetic acid i. 38.5 J/mmol * 0.239 calories / 1 Joule = 9.20 cal/mmol 3. Polystyrene cups are used in the experiment because Polystyrene is a good insulator, which traps the heat so that it helps the results to be more accurate. If two of Polystyrene cups were used, insulation will increase and will trap

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Unformatted text preview: the heat better. As a result, it will give us more accurate results. 4. Delta H naught is the delta H of one mole of a compound at 25 degrees C and 1 atm. The experiment was not performed under such a controlled environment. 5. a. Clean your 600mL beaker. Then obtain 150mL of the NaOH solution in the 600mL beaker. - 3 minutes b. Using your pipets, measure 20.0mL of the NaOH solution into a polystyrene cup, and start record the temperature with digital temperature. - 2 minutes c. Add the acid to the NaOH solution in the polystyrene cup. Stir the mixture with the probe and continue collecting data for five minutes. -6 minutes d. Repeat the procedure with same acid. - 11 minutes e. Repeat the procedure twice for each of acids - 44 minutes...
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