If 15.0 g of nitrogen reacts with 15.0 g of hydrogen, 10.5 g of ammonia is produced. What is the percent yield?

Percent yield =  57%

Explanation:

Given data:

Mass of nitrogen = 15.0 g

Mass of hydrogen = 15.0 g

Mass of ammonia produced = 10.5 g

Percent yield = ?

Solution:

Chemical equation:

N₂+ 3H₂      →   2NH₃

Number of moles of hydrogen:

Number of moles = mass/molar mass

Number of moles = 15.0 g/ 2 g/mol

Number of moles = 7.5 mol

Number of moles of nitrogen:

Number of moles = mass/molar mass

Number of moles = 15.0 g/ 28 g/mol

Number of moles = 0.54 mol

Now we will compare the moles of ammonia with nitrogen and hydrogen from balance chemical equation.

              N₂       :        NH₃

               1         :          2

           0.54       :     2×0.54 = 1.08

             H₂         :           NH₃

              3           :            2

              7.5        :      2/3×7.5= 5 mol

Theoretical yield of ammonia:

Mass = number of moles × molar mass

Mass = 1.08 × 17 g/mol

Mass = 18.36 g

Percent yield:

Percent yield = (actual yield / theoretical yield) ×100

Percent yield = (10.5 g/ 18.36 g) ×100

Percent yield = 0.57 ×100

Percent yield =  57%

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