Since the pandemic has hit, more people are growing their own gardens and processing the fruits of their work. One thing that is flying off the shelves are spices for canning, such as alum. From this lab, you have seen how easy it is to prepare an alum, and the ingredients to make this are not difficult to obtain. You are going to make a batch of pickles and the recipe requires 6.05g of alum for each quart jar of pickles and you plan on making 50 jar of pickles. The % yield of making alum is only 84.4%.  a. What is the actual yield of alum that you need to prepare to make your pickles?  b. What is the theoretical yield of alum required to prepare the pickles?  c. What mass of aluminum do you need to start with to make the required amount of alum for the preparation of the pickles?

Chemistry: The Molecular Science
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Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter13: The Chemistry Of Solutes And Solutions
Section13.7: Colligative Properties Of Solutions
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Since the pandemic has hit, more people are growing their own gardens and processing the fruits of their work. One thing that is flying off the shelves are spices for canning, such as alum. From this lab, you have seen how easy it is to prepare an alum, and the ingredients to make this are not difficult to obtain. You are going to make a batch of pickles and the recipe requires 6.05g of alum for each quart jar of pickles and you plan on making 50 jar of pickles. The % yield of making alum is only 84.4%. 

a. What is the actual yield of alum that you need to prepare to make your pickles? 

b. What is the theoretical yield of alum required to prepare the pickles? 

c. What mass of aluminum do you need to start with to make the required amount of alum for the preparation of the pickles?

Expert Solution
Step 1: Actual yield of alum required to make pickles

Each quart jar requires = 6.05g alum

So, 50 quart jars would require = 6.05*50 g = 302.5g alum

% yield of making alum = 84.4%

It means, 100 g of sample contains 84.4 g of alum.

So, 302.5 g of alum would be present in = 100*302.584.4g=358.41g of sample of alum.

                                                   

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