For the part underlined in orange, how are we getting moles of OH-? What are the steps. Solve use dimensional analysis please, this is my first chem class and I don't know the other ways well. Thank you.
For the part underlined in orange, how are we getting moles of OH-? What are the steps. Solve use dimensional analysis please, this is my first chem class and I don't know the other ways well. Thank you.
Living By Chemistry: First Edition Textbook
1st Edition
ISBN:9781559539418
Author:Angelica Stacy
Publisher:Angelica Stacy
ChapterU4: Toxins: Stoichiometry, Solution Chemistry, And Acids And Bases
SectionU4.15: Holey Moley: Preparing Solutions
Problem 8E
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For the part underlined in orange, how are we getting moles of OH-? What are the steps. Solve use dimensional analysis please, this is my first chem class and I don't know the other ways well. Thank you.
![12. In an experiment, 28.0mL of (0.250M HNO1 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.
Molecular: KOH(aq) + HNO3(aq) → H2O(I) + KN03(aq)
Total ionic:
K*(aq) + OH (aq) + H*(aq) + NO3 (aq) – H2O(I) + K*(aq)
H*(aq) + OH (aq) → H2O(I)
+ NO, (aq)
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 H20 formed = 0.126 g
%3D
Moles OH excess = moles OH start-moles OH reacted
1
Moles OH excess = 0.01696 - 0.0070 = 0.00996 moles OH-
%3D
%3D
%3D
[OH excess] = 0.00996 moles / 0.0810 L = 0.123 M OH-
%3D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F61b75115-d70f-4fe2-af93-2076876ad69a%2F3b8aa7e4-841b-4704-b6f4-34a770809930%2Fklkrfvk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:12. In an experiment, 28.0mL of (0.250M HNO1 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.
Molecular: KOH(aq) + HNO3(aq) → H2O(I) + KN03(aq)
Total ionic:
K*(aq) + OH (aq) + H*(aq) + NO3 (aq) – H2O(I) + K*(aq)
H*(aq) + OH (aq) → H2O(I)
+ NO, (aq)
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 H20 formed = 0.126 g
%3D
Moles OH excess = moles OH start-moles OH reacted
1
Moles OH excess = 0.01696 - 0.0070 = 0.00996 moles OH-
%3D
%3D
%3D
[OH excess] = 0.00996 moles / 0.0810 L = 0.123 M OH-
%3D
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