the number of moles of Aluminum in 1.3741 g was calculated to be 0.051 moles and that the chemistry student weighed out. For every mole of Aluminum that you start with, you can create one mole of Alum [KAl(SO4)2 ⋅12 H2O(s)]. What is the theoretical yield before drying the Alum precipitate?
the number of moles of Aluminum in 1.3741 g was calculated to be 0.051 moles and that the chemistry student weighed out. For every mole of Aluminum that you start with, you can create one mole of Alum [KAl(SO4)2 ⋅12 H2O(s)]. What is the theoretical yield before drying the Alum precipitate?
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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the number of moles of Aluminum in 1.3741 g was calculated to be 0.051 moles and that the chemistry student weighed out. For every mole of Aluminum that you start with, you can create one mole of Alum [KAl(SO4)2 ⋅12 H2O(s)]. What is the theoretical yield before drying the Alum precipitate?
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Step 1
The molar ratio between aluminum and alum is 1:1, meaning that one mole of aluminum should theoretically produce one mole of alum. Therefore, if we have 0.051 moles of aluminum, the theoretical yield of alum that can be produced is also 0.051 moles.
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