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50.0 mL 2.00M HCI and 50.0 mL 2.00 M NaOH mixed: HCI(aq) + NaOH(aq) → H,0() + NaCI(aq) - Csoln = Cwater = 4.184 J/(g °C) dsoln = dwater = 1.00 g/mL - mgoln = (50 mL + 50 mL) = 100 mL = 100 g - ΔΤ, %3D %3D %3D %3D AT 13.8 °C (same mass water, twice the moles) soln Asoin = (100 g)(4.184 J/(g °C)(13.8 °C) = +5773.92 J - Moles HCI: (M*L) = (2.00 M)(0.050 L) = 0.100 moles - Moles NaOH = (2.00 M)(0.050 L) = 0.100 moles %3D %3D %3D %3D %3D 9soln = +5773.92 J/0.100 moles = +57,739.2 J J/mole AHn = -57.7 kJ/mole %3D rxn Pearson Education, Inc.

50.0 mL 2.00M HCI and 50.0 mL 2.00 M NaOH mixed: HCI(aq) + NaOH(aq) → H,0() + NaCI(aq) - Csoln = Cwater = 4.184 J/(g °C) dsoln = dwater = 1.00 g/mL - mgoln = (50 mL + 50 mL) = 100 mL = 100 g - ΔΤ, %3D %3D %3D %3D AT 13.8 °C (same mass water, twice the moles) soln Asoin = (100 g)(4.184 J/(g °C)(13.8 °C) = +5773.92 J - Moles HCI: (M*L) = (2.00 M)(0.050 L) = 0.100 moles - Moles NaOH = (2.00 M)(0.050 L) = 0.100 moles %3D %3D %3D %3D %3D 9soln = +5773.92 J/0.100 moles = +57,739.2 J J/mole AHn = -57.7 kJ/mole %3D rxn Pearson Education, Inc.

General Chemistry - Standalone book (MindTap Course List)
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter6: Thermochemisty
Section: Chapter Questions
Problem 6.111QP: In a calorimetric experiment, 6.48 g of lithium hydroxide, LiOH, was dissolved in water. The...
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For the part circled, what if the molarity is different and one has a molarity of 3 and one has M=2. When we normalize, are we dividing by the limiting reactant?

Calorimetry Example: Understanding AHn If 50.0 mL 2.00 M HCI and 50.0 mL 2.00 M NaOH mixed: HCI(aq) + NaOH(aq) → H20(1) + NaCl(aq) - Csoln = Cwater = 4.184 J/(g °C) – dooln = dwater = 1.00 g/mL - msoln = (50 mL + 50 mL) = 100 mL = 100 g %3D %3D - ΔΤ, ATSoln = 13.8 °C (same mass water, twice the moles) %3D 9soln = (100 g)(4.184 J/(g °C)(13.8 °C) = +5773.92 J Moles HCI: (M*L) = (2.00 M)(0.050 L) = 0.100 moles Moles NaOH = (2.00 M)(0.050 L) = 0.100 moles geoln = +5773.92 J /0.100 moles = +57,739.2 J J/mole %3D %3D %3D %3D %3D %3D AHn =-57.7 kJ/mole rxr © 2014 Pearson Education, Inc.
Transcribed Image Text:Calorimetry Example: Understanding AHn If 50.0 mL 2.00 M HCI and 50.0 mL 2.00 M NaOH mixed: HCI(aq) + NaOH(aq) → H20(1) + NaCl(aq) - Csoln = Cwater = 4.184 J/(g °C) – dooln = dwater = 1.00 g/mL - msoln = (50 mL + 50 mL) = 100 mL = 100 g %3D %3D - ΔΤ, ATSoln = 13.8 °C (same mass water, twice the moles) %3D 9soln = (100 g)(4.184 J/(g °C)(13.8 °C) = +5773.92 J Moles HCI: (M*L) = (2.00 M)(0.050 L) = 0.100 moles Moles NaOH = (2.00 M)(0.050 L) = 0.100 moles geoln = +5773.92 J /0.100 moles = +57,739.2 J J/mole %3D %3D %3D %3D %3D %3D AHn =-57.7 kJ/mole rxr © 2014 Pearson Education, Inc.
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