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12. In an experiment, 28.0mL of 0.250M HNO3 and 53.0mL of 0.320M KOH are mixed. Calculate the amount of water formed (in grams), and the concentration of excess ion that is involved in the reaction. Hint: write both the balanced molecular and net ionic equations for this reaction. - H20(1) + KNO3(aq) KOH(aq) + HNO3(aq) K*(aq) + OH (aq) + H*(aq) + NO; (aq) → H2O(1) + K*(aq) + NO3 (aq) H*(aq) + OH (aq) Molecular: Total ionic: Net ionic: - H20(1) This is a limiting reagent problem so we need to find which reactant is present in smaller stoichiometric amount. Using the information provided and the balanced chemical equation, we find that HNO3 is limiting and the hydroxide ion, OH, is present in excess. Mass of H2O formed = 0.126 g Moles OH excess = moles OH start - moles OHreacted Moles OH excess = 0.01696-0.0070 = 0.00996 moles OH- [OH excess] = 0.00996 moles /0.0810 L = 0.123 M OH-

12. In an experiment, 28.0mL of 0.250M HNO3 and 53.0mL of 0.320M KOH are mixed. Calculate the amount of water formed (in grams), and the concentration of excess ion that is involved in the reaction. Hint: write both the balanced molecular and net ionic equations for this reaction. - H20(1) + KNO3(aq) KOH(aq) + HNO3(aq) K*(aq) + OH (aq) + H*(aq) + NO; (aq) → H2O(1) + K*(aq) + NO3 (aq) H*(aq) + OH (aq) Molecular: Total ionic: Net ionic: - H20(1) This is a limiting reagent problem so we need to find which reactant is present in smaller stoichiometric amount. Using the information provided and the balanced chemical equation, we find that HNO3 is limiting and the hydroxide ion, OH, is present in excess. Mass of H2O formed = 0.126 g Moles OH excess = moles OH start - moles OHreacted Moles OH excess = 0.01696-0.0070 = 0.00996 moles OH- [OH excess] = 0.00996 moles /0.0810 L = 0.123 M OH-

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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My question is, for the part underlined in purple, why are we looking for excess OH- and not excess KOH? Does this have to do with the net equation?

12. In an experiment, 28.0mL of 0.250M HNO3 and 53.0mL of 0.320M KOH are mixed. Calculate the amount of water formed (in grams), and the concentration of excess ion that is involved in the reaction. Hint: write both the balanced molecular and net ionic equations for this reaction. KOH(aq) + HNO3(aq) → H2O(l) + KNO3(aq) K*(aq) + OH (aq) + H*(aq) + NO; (aq) H*(aq) + OH (aq) → H2O(1) Molecular: → H2O(1) + K*(aq) + NO3 (aq) -> Total ionic: Net ionic: This is a limiting reagent problem so we need to find which reactant is present in smaller stoichiometric amount. Using the information provided and the balanced chemical equation, we find that HNO3 is limiting and the hydroxide ion, OH¯, is present in excess. Mass of H2O formed = 0.126 g %3D Moles OH excess = moles OH start- moles OH reacted %3D %3D Moles OH excess = 0.01696 - 0.0070 = 0.00996 moles OH- %3D %3D [OH excess] = 0.00996 moles / 0.0810 L = 0.123 M OH- %3D
Transcribed Image Text:12. In an experiment, 28.0mL of 0.250M HNO3 and 53.0mL of 0.320M KOH are mixed. Calculate the amount of water formed (in grams), and the concentration of excess ion that is involved in the reaction. Hint: write both the balanced molecular and net ionic equations for this reaction. KOH(aq) + HNO3(aq) → H2O(l) + KNO3(aq) K*(aq) + OH (aq) + H*(aq) + NO; (aq) H*(aq) + OH (aq) → H2O(1) Molecular: → H2O(1) + K*(aq) + NO3 (aq) -> Total ionic: Net ionic: This is a limiting reagent problem so we need to find which reactant is present in smaller stoichiometric amount. Using the information provided and the balanced chemical equation, we find that HNO3 is limiting and the hydroxide ion, OH¯, is present in excess. Mass of H2O formed = 0.126 g %3D Moles OH excess = moles OH start- moles OH reacted %3D %3D Moles OH excess = 0.01696 - 0.0070 = 0.00996 moles OH- %3D %3D [OH excess] = 0.00996 moles / 0.0810 L = 0.123 M OH- %3D
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