Interpretation −
Concept introduction − Percent yield is the ration of actual yield to theoretical yield.
Answer to Problem 35SSC
Percent yield is 87.9%
Explanation of Solution
Mass of Mg
= Mass of Mg and crucible − Mass of empty crucible
=
=
Step 1: Balance the chemical equation
Make mass to mole conversion
For conversion from mass to mole, we divide mass by molecular mass, so in this case moles of Mg are:
Step 3: Make moles to mole conversion
As we know from the balanced chemical equation that 2 moles Mg produces 2 moles MgO. Therefore, moles of MgO produced by 0.0983 moles of Mg:
Step 4: Makes mole to mass conversion
For conversion of moles to mass, we multiply mole by molecular mass, so in this case mass of Mg is:
Thus, percent yield is 87.9%
Chapter 11 Solutions
Glencoe Chemistry: Matter and Change, Student Edition
Additional Science Textbook Solutions
Organic Chemistry
Chemistry: Structure and Properties
Chemistry: A Molecular Approach
General Chemistry: Principles and Modern Applications (11th Edition)
Organic Chemistry (8th Edition)
Chemistry: The Central Science (13th Edition)
- ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill EducationPrinciples of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
- Organic ChemistryChemistryISBN:9780078021558Author:Janice Gorzynski Smith Dr.Publisher:McGraw-Hill EducationChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningElementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY