Is my work correct? Question: What is the percent yeild of the nicke(II) coordination? [Ni(H2O)6]Cl2 used = 6.140 g [Ni(en)3]Cl2 produced = 7.001 g Work mol of [Ni(H2O)6]Cl2 6.140 g/237.67 g/mol = 0.0258 mol [Ni(H2O)6]Cl2 mol of [Ni(en)3]Cl2 0.0258 mol [Ni(H2O)6]Cl2 x (1 mol [Ni(H2O)6]Cl2/1mol [Ni(en)3]Cl2) = 0.0258 mol grams of [Ni(en)3]Cl2 0.0258 mol x 309.89 g/mol = 8.00 g % yield 7.001 g/8.00 g = 87.5%
Is my work correct? Question: What is the percent yeild of the nicke(II) coordination? [Ni(H2O)6]Cl2 used = 6.140 g [Ni(en)3]Cl2 produced = 7.001 g Work mol of [Ni(H2O)6]Cl2 6.140 g/237.67 g/mol = 0.0258 mol [Ni(H2O)6]Cl2 mol of [Ni(en)3]Cl2 0.0258 mol [Ni(H2O)6]Cl2 x (1 mol [Ni(H2O)6]Cl2/1mol [Ni(en)3]Cl2) = 0.0258 mol grams of [Ni(en)3]Cl2 0.0258 mol x 309.89 g/mol = 8.00 g % yield 7.001 g/8.00 g = 87.5%
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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Is my work correct?
Question: What is the percent yeild of the nicke(II) coordination?
- [Ni(H2O)6]Cl2 used = 6.140 g
- [Ni(en)3]Cl2 produced = 7.001 g
Work
- mol of [Ni(H2O)6]Cl2
- 6.140 g/237.67 g/mol = 0.0258 mol [Ni(H2O)6]Cl2
- mol of [Ni(en)3]Cl2
- 0.0258 mol [Ni(H2O)6]Cl2 x (1 mol [Ni(H2O)6]Cl2/1mol [Ni(en)3]Cl2) = 0.0258 mol
- grams of [Ni(en)3]Cl2
- 0.0258 mol x 309.89 g/mol = 8.00 g
- % yield
- 7.001 g/8.00 g = 87.5%
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