Determine the percent yield for this reaction.?
2NH₃+CO₂ --> CN₂H₄O + H₂O
50.0g of ammonia reacted with 50.0g of carbon dioxide and 60.0g of urea was obtained.
I calculate the ammount of NH₃ and CO₂ in moles which I get 0.3408 and 0.8802 respectively. Then I determine which is the limiting reactant which is NH₃. (0.1704 mol CO₂ is required to react with all of the NH₃ and 1.7604 mol NH₃ is required to react with all of the CO₂). Then I calculate the theoretical yield and get 0.1704mol which is 10.24g of CN₂H₄O which is less than 60.07. What I did wrong? Thank you in advance :)
2NH₃+CO₂ --> CN₂H₄O + H₂O
50.0g of ammonia reacted with 50.0g of carbon dioxide and 60.0g of urea was obtained.
I calculate the ammount of NH₃ and CO₂ in moles which I get 0.3408 and 0.8802 respectively. Then I determine which is the limiting reactant which is NH₃. (0.1704 mol CO₂ is required to react with all of the NH₃ and 1.7604 mol NH₃ is required to react with all of the CO₂). Then I calculate the theoretical yield and get 0.1704mol which is 10.24g of CN₂H₄O which is less than 60.07. What I did wrong? Thank you in advance :)
1 Answer
See below.
Explanation:
Step 1: Calculate the number of moles of each reactant.
Step 2: Identify the limiting reagent.
We need 2 moles of
We have more than enough ammonia for the amount of
Step 3: Calculate how much
Since there is a 1:1 ratio between how many moles of
If for instance, 1 mole of
Step 4: Convert moles to grams.
This number is our theoretical yield in grams.
Step 5: Determine percent yield.
The equation for percent yield is:
We have all of the numbers necessary, so we can plug in:
I hope this helps!