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: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter3: Chemical Reactions
Section: Chapter Questions
Problem 92QRT: You mix 25.0 mL of 0.234-M FeCl3 solution with 42.5 mL of 0.453-M NaOH.
Calculate the maximum mass,...
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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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