Asked by Jennie
Ok, Need some more help... This is what I have so far.
4. Urea (NH2)2CO is prepared by reacting ammonia with carbon dioxide. The byproduct is water. 637.2g of ammonia are reacted with 787.3 g of carbon dioxide.
molar mass:NH3=17 CO2=44 CO(NH2)2=60.1
a. (10 points) Write a balanced chemical equation.
2NH3+CO2--->CO(NH2)2 +H20
Is this right?
b. (5 points) Determine which of the two reactants (ammonia or carbon dioxide) is a limiting reactant. Show
calculations to support your choice. No credit will be given if your supporting calculations are wrong.
2 moles NH3 is require 1 mole of CO2 to produce one molecule of CO(NH2)2. 2 moles NH3 is 637.2/17=37.5 moles =37.5/2=18.74 moles in 1 NH3?
Available CO2 = 787.3 = 787.3/44=17.89 moles CO2 So since CO2 is smaller, wouldn't it be the limiting agent?
c. (10 points) Which of the two reactants should you use to calculate the theoretical yield (= the maximum amount
possible) of urea? Explain why. UMMM I would calculate the NH3 because it is the larger one? why? I don't have a clue help?
4. Urea (NH2)2CO is prepared by reacting ammonia with carbon dioxide. The byproduct is water. 637.2g of ammonia are reacted with 787.3 g of carbon dioxide.
molar mass:NH3=17 CO2=44 CO(NH2)2=60.1
a. (10 points) Write a balanced chemical equation.
2NH3+CO2--->CO(NH2)2 +H20
Is this right?
b. (5 points) Determine which of the two reactants (ammonia or carbon dioxide) is a limiting reactant. Show
calculations to support your choice. No credit will be given if your supporting calculations are wrong.
2 moles NH3 is require 1 mole of CO2 to produce one molecule of CO(NH2)2. 2 moles NH3 is 637.2/17=37.5 moles =37.5/2=18.74 moles in 1 NH3?
Available CO2 = 787.3 = 787.3/44=17.89 moles CO2 So since CO2 is smaller, wouldn't it be the limiting agent?
c. (10 points) Which of the two reactants should you use to calculate the theoretical yield (= the maximum amount
possible) of urea? Explain why. UMMM I would calculate the NH3 because it is the larger one? why? I don't have a clue help?
Answers
Answered by
DrBob222
Answered above.
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