Check xample #1) 25mL of 1.0M HCI is mixed with 40mL of 0.80M NaOHHCI + NaOH → H₂O + NaClFirst, let's calculate the literature enthalpy of reaction from the values in the table above:SubstanceH₂OHC1NaOHNaClAHreaction=AHproducts-AHreactantsAHreaction = (AHH20+ AHNaCl) - (AHHCI + AHNaOH)AHreaction = (-286 kJ/mol + -407 kJ/mol) - (-164 kJ/mol + -469 kJ/mol)AHreaction = -60 kJ/molLet's calculate the limiting reagent to determine how many mols of water will be formed:Volume of reactant ÷ 1000 x Molarity of reactant x mole-to-mole ratio to waterHere's the first reactant: 25mL of 1.0M HCI141000 mL1.0 mol HCl 1 mol H20XX-1 mol HClKLet's do this for the second reactant, 40mL of 0.80M NaOH1 mol H201 mol NaOH25mL of HCI x40mL of NaOH x1 L 0.80 mol NaOHX1000 mL1LX-AHr in kJ/mol-286-164-469-407= 0.025 mol H₂O= 0.032 mol H₂OThe limiting reagent is the compound that produced the fewest number of moles. In this case,the 25mL of 1.0M HCI only produced 0.025 mol H₂O, so that is the limiting reagent.Next, multiply the limiting reagent (the moles) by the AH reaction calculated above0.025 mol H₂O x -60 kJ/mol = -1.5 kJAnd that's your answer! The reaction should release -1.5 kJ of energy. xample #2) 10mL of 2.5M HCI is mixed with 30mL of 1.0M NaOHHCI + NaOH → H₂O + NaClAs we ease into these, it's the same reaction as above!SubstanceH₂OHC1NaOHNaClAHreaction=AHproducts- AHreactantsAHreaction = (AHH20+ AHNaCl) - (AHHCI+ AHNaOH)AHreaction = (-286 kJ/mol + -407 kJ/mol) - (-164 kJ/mol + -469 kJ/mol)AH reaction = -60 kJ/molLet'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: 10mL of 2.5M HCI10*2.5X= 0.025 mol HCl1000Let's do this for the second reactant, 30mL of 1.0M NaOH3010000.03 mol NaOHThe limiting reagent is the compound that produced the fewest number of moles. If you didyour calculations right, it should have been the HCI! Multiply that number of moles buy theAHreaction that was calculated above(CO0.025545 mol H₂0) x -60 kJ/mol = 1.5AHf in kJ/mol-286-164-469-4074_kJ

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Substance H₂O HC1 NaOH NaCl = reaction=AHproducts-AHreactants reaction (AHH20 + AHNaci) - (AHHCI + AHNaOH) = reaction (-286 kJ/mol + -407 kJ/mol) - (-164 kJ/mol + -469 kJ/mol) Hreaction = -60 kJ/mol t's calculate the limiting reagent to determine how many mols of water will be formed: emember: Volume of reactant ÷ 1000 x Molarity of reactant x mole-to-mole ratio to water ere's the first reactant: 10mL of 2.5M HCI 10 *2.5 = 0.025 mol HCl 1000 YOU t's do this for the second reactant, 30mL of 1.0M NaOH 30 2 0.03 mol NaOH 1000 -1.5 AHf in kJ/mol -286 -164 -469 -407 me limiting reagent is the compound that produced the fewest number of moles. If you did our calculations right, it should have been the HCI! Multiply that number of moles buy the Hreaction that was calculated above 0.0255mol H₂O) x -60 kJ/mol = kJ

Substance H₂O HC1 NaOH NaCl = reaction=AHproducts-AHreactants reaction (AHH20 + AHNaci) - (AHHCI + AHNaOH) = reaction (-286 kJ/mol + -407 kJ/mol) - (-164 kJ/mol + -469 kJ/mol) Hreaction = -60 kJ/mol t's calculate the limiting reagent to determine how many mols of water will be formed: emember: Volume of reactant ÷ 1000 x Molarity of reactant x mole-to-mole ratio to water ere's the first reactant: 10mL of 2.5M HCI 10 *2.5 = 0.025 mol HCl 1000 YOU t's do this for the second reactant, 30mL of 1.0M NaOH 30 2 0.03 mol NaOH 1000 -1.5 AHf in kJ/mol -286 -164 -469 -407 me limiting reagent is the compound that produced the fewest number of moles. If you did our calculations right, it should have been the HCI! Multiply that number of moles buy the Hreaction that was calculated above 0.0255mol H₂O) x -60 kJ/mol = kJ

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter8: Thermochemistry

Section: Chapter Questions

Problem 12QAP: The heat of neutralization, Hneut, can be defined as the amount of heat released (or absorbed), q,...

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