TBR Gen Chem Stoichiometry Passage XI Question 72

[dennis-brodmann]

Someone asked about this question regarding why percent yield was in moles and not grams, but I have a somewhat different question.

I see that you can use moles, but what I don't understand is why TBR is taking the ratio for two different substances to find percent yield.

They gave the following reaction and question:

Ca(OH)2 (aq) + CO2 (g) –> CaCO3 (s) + H2O (l)
If 10.00 grams of Ca(OH)2 (s) produces 5.00 grams of CaCO3 (s), what is the percent yield for the reaction?

I know that %Yield= (Actual Yield/Theoretical Yield)*100%.
What I tried to do was use 10.00 grams of Ca(OH)2 to find the moles of CaCO3 (s) produced since that is the compound that is being produced (i.e. yielded).

What TBR does in the explanations, however, is taking 5 g/MW of CaCO3 as the actual yield divided by 10.00 grams of Ca(OH)2/MW of Ca(OH)2 as the theoretical yield.

I understand that from the balanced reaction, there is a 1:1 mole ratio for all the reactants and products, but for %yield, I thought that the ratio of Actual Yield to Theoretical Yield
had to be for the same substance.

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[Singhp03]

easiest way to solve these problems is via Dimensional analysis to get theoretical yield, TBR explains it in a weird way but once you use DA its all crystal clear

 

[LoLCareerGoals]

That's fine. Use the same substance. Do things the way you understand them. Find theoretical yield based on reactant initial amount and reaction stochiometry and divide actual by this theoretical amount.

 

[JLSant]

I know this thread is old, but I have the same question and I was wondering if someone could give a more in-depth explanation related to dimensional analysis? I have TBR right in front of me, but the explanation is rather brief.