Question

9. 6.59 g of ammonia (NH3, 17.031 g/mol) react completely in the presence of excess carbon dioxide (CO2, 44.01 g/mol) according to the following balanced equation,

2NH3(g) + CO2(g) → CH4N2O(s) + H2O(l) If 7.40 g of urea (CH4N2O, 60.056 g/mol) are produced,
calculate the percent yield for the reaction.

Answer 1

The percent yield for the reaction is 63.7%

Step-by-step explanation:

This problem can be solved, from different rules of three.

The reaction is:

2NH₃(g) + CO₂(g) → CH₄N₂O(s) + H₂O(l)

2 moles of ammonia, produce 1 mol of urea.

Mass / molar mass = moles

6.59g / 17.031 g/m = 0.387 moles

As ratio is 2:1, I will produce half a mole of urea.

(0.387 m . 1m)/ 2m = 0.193 m

Let's find out the mass.

Mass of urea = Moles of urea . molar mass of urea.

Mass = 0.193m . 60.056 g/m = 11.6 g

If the percent yield for the reaction was 100 %, we make 11.6 g, but I only obtained 7.40 g so let's calculate the new percent yield.

11.6 g ____ 100 %

7.40 g ___ (7.40g / 11.6g) .100 = 63.7 %