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11. 6.55 g of ammonia (NH3) react completely according to the following reaction:2NH3 + CO2 > CN2OH4 + H2OWhat is the theoretical yield of urea (CN2OH4) for this reaction?-If in the lab, 8.75g of urea was recovered, what was the percent yield of the lab?

User Michel Milezzi
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1 Answer

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26 votes

What we're going to do is to use the stoichiometry of the problem:

First, we need to pass all amounts in grams to moles. (Just dividing by the molecular weight)

Now that we have the moles, we are going to use the proportion of the chemical equation:

Now,

The theoretical yield of urea, is 11.5g.

Finally, to find the percent yield, what we need to do is to divide the actual yield (The experimental value) and the theoretical yield:

So that's the answer.

11. 6.55 g of ammonia (NH3) react completely according to the following reaction:2NH-example-1
11. 6.55 g of ammonia (NH3) react completely according to the following reaction:2NH-example-2
11. 6.55 g of ammonia (NH3) react completely according to the following reaction:2NH-example-3
11. 6.55 g of ammonia (NH3) react completely according to the following reaction:2NH-example-4
User Ankit Batra
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3.1k points