Example #3) 15mL of 2.0M HCI is mixed with 30mL of 0.75M KOHHCI + KOH → H₂O + KCIAHreaction =SubstanceH₂OHC1KC1KOHAHreaction=AHproducts-AHreactantsAHreaction = (AHH20+ AHKCI) - (AHHCI + AHKOH)AHreaction = (-286+ -419- 1kg1Let's calculate the limiting reagent to determine how many mols of water will be formed:Remember: Volume of reactant ÷ 1000 x Molarity of reactant x mole-to-mole ratio to waterHere's the first reactant: 15mL of 2.0M HCIIS1500* 2.00.03 mol HCl301000Lof LOM KOHLet's do this for the second reactant, 30mL of 0.75M KOH0.750.0225 mol KOH=-)-(-164-61mol H₂O) xAHf in kJ/mol-286-164-419-480The limiting reagent is the compound that produced the fewest number of moles. Multiply thatnumber of moles buy the AH reaction that was calculated abovekJ/mol = -/-3725 kJ-+_ - 48010.0225CalcuFrom here on out, you are on your own! Calculate the enthalpy of reaction, and then thelimiting reagent. Next multiply the enthalpy of reaction by the limiting reagent. Good luck! Example #4) 50mL of 1.5M HCI is mixed with 60mL of 1.0M KOHHCI + KOH → H₂O + KCIAHf in kJ/mol-286-164-419-480Note: this is the same reaction as example #3, so the AH reaction should be the sameSubstanceH₂OHCIKCIKOHAHreaction=AHproducts- AHreactantsAHreaction =-bikiLet's calculate the limiting reagent to determine how many mols of water will be formed:Here's the first reactant: 50mL of 1.5M HCI50 * 1.5.10001.5 = 0.075 mol HelLet's do this for the second reactant, 60mL of 1.0M KOH601000* 1 = 0.obmolkottCalculate the amount of energy released:Call 0.06 * -61 = -3.66 kg

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Example #3) 15mL of 2.0M HCl is mixed with 30mL of 0.75M KOH

Introduction to General, Organic and Biochemistry

11th Edition

ISBN:9781285869759

Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres

Publisher:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres

Chapter6: Solutions And Colloids

Section: Chapter Questions

Problem 6.89P

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