A common laboratory reaction is the neutralization of an acid with a base. When 40.3 mL of 0.500 M HCl at 25.0°C is added to 75 mL of 0.500 M NaOH at 25.0°C in a coffee cup calorimeter (with a negligible heat capacity), the temperature of the mixture rises to 28.2°C. What is the heat of reaction per mole of NaCl (in kJ/mol)? Assume the mixture has a specific heat capacity of 4.18 J/(g•K) and that the densities of the reactant solutions are both 1.07 g/mL. Enter your answer to three significant figures in units of kJ/mol.

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
Chapter5: Thermochemistry
Section: Chapter Questions
Problem 5.62QE: A 50-mL solution of a dilute AgNO3 solution is added to 100 mL of a base solution in a coffee-cup...
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A common laboratory reaction is the neutralization of an acid with a base. When 40.3 mL of 0.500 M
HCl at 25.0°C is added to 75 mL of 0.500 M NAOH at 25.0°C in a coffee cup calorimeter (with a
negligible heat capacity), the temperature of the mixture rises to 28.2°C. What is the heat of
reaction per mole of NaCl (in kJ/mol)? Assume the mixture has a specific heat capacity of 4.18
J/(g-K) and that the densities of the reactant solutions are both 1.07 g/mL.
Enter your answer to three significant figures in units of kJ/mol.
Transcribed Image Text:A common laboratory reaction is the neutralization of an acid with a base. When 40.3 mL of 0.500 M HCl at 25.0°C is added to 75 mL of 0.500 M NAOH at 25.0°C in a coffee cup calorimeter (with a negligible heat capacity), the temperature of the mixture rises to 28.2°C. What is the heat of reaction per mole of NaCl (in kJ/mol)? Assume the mixture has a specific heat capacity of 4.18 J/(g-K) and that the densities of the reactant solutions are both 1.07 g/mL. Enter your answer to three significant figures in units of kJ/mol.
Expert Solution
Step 1

The volume of HCl = 40.3 mL OR 0.0403 L

The molarity of HCl = 0.5 M

The volume of NaOH = 75 mL or 0.075 L

The molarity of NaOH = 0.5 M

Moles=Molarity×VolumeMoles of HCl =0.5 M×0.0403 L=0.021 moleMoles of NaOH =0.5 M×0.0.075 L=0.038 mole

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