Ask Question
24 January, 12:52

Solid urea, CN2OH4, (MM=60.06 g) and liquid water are produced by heating ammonia

(MM=17.03 g) gas and carbon dioxide (MM=44.01 g) gas according to the equation

below. This reaction is known to proceed with a percent yield of 85.5%. What is the

theoretical yield if 13.74 g of urea is actually collected?

2NH3 (g) + CO2 (g) →CN2OH4 (s) + H2O (1)

+3
Answers (1)
  1. 24 January, 12:54
    0
    16.1 g

    Explanation:

    Percent yield = (actual / theoretical) x 100

    85.5 = (13.74 / theoretical) x 100

    0.855 = (13.74 / theoretical)

    0.855 (theoretical) = 13.74

    theoretical = 13.74 / 0.855

    Theoretical = 16.070 = 16.1 g (3 sig figs)
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Solid urea, CN2OH4, (MM=60.06 g) and liquid water are produced by heating ammonia (MM=17.03 g) gas and carbon dioxide (MM=44.01 g) gas ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers