My question is for the part outlined in a box where it says theoretical yield. What is going on here? Thank you.

Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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My question is for the part outlined in a box where it says theoretical yield. What is going on here? Thank you.

17. Consider the production of urea from ammonia and carbon dioxide:
2NH3(g) + CO2(g) → CO(NH2)2(s) + H2O(1)
If the percent yield of the reaction is 75.0%, what mass of NH3, together with an excess of
CO2, is needed to obtain 50.3 grams of CO(NH2)2?
We need more NH3 than predicted by stoichiometry since the % yield for the reaction is only 75.0%
50.3 grams
Theoretical Yield
x100 = 75.0%
%3D
X grams
X = 67.07 g CO(NH2)2
Using 67.07 g of urea, we can obtain the mass of NH3 through stoichiometry: 38.0 g NH3
Page 6 of 7
Transcribed Image Text:17. Consider the production of urea from ammonia and carbon dioxide: 2NH3(g) + CO2(g) → CO(NH2)2(s) + H2O(1) If the percent yield of the reaction is 75.0%, what mass of NH3, together with an excess of CO2, is needed to obtain 50.3 grams of CO(NH2)2? We need more NH3 than predicted by stoichiometry since the % yield for the reaction is only 75.0% 50.3 grams Theoretical Yield x100 = 75.0% %3D X grams X = 67.07 g CO(NH2)2 Using 67.07 g of urea, we can obtain the mass of NH3 through stoichiometry: 38.0 g NH3 Page 6 of 7
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