Answer:
The percent yield for the reaction is 63.7%
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 %
