If 15.0g of nitrogen reacts with 15.0g of hydrogen, 10.5g of ammonia os produced. what is the percent yield of this reaction.

Percent yield = 57.2%

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 nitrogen:

Number of moles = mass/ molar mass

Number of moles = 15 g / 28 g/mol

Number of moles = 0.54 mol

Number of moles of hydrogen:

Number of moles = mass/ molar mass

Number of moles = 15 g / 2 g/mol

Number of moles = 7.5 mol

Now we compare the moles ammonia with hydrogen and nitrogen

                               N₂              :      NH₃

                                 1               :        2

                                 0.54         :      2/1 ×0.54 = 1.08 mol                                 

                                  H₂           :      NH₃

                                    3            :        2

                                   7.5          :      2/3×7.5= 5 mol

The number of moles of ammonia produced by nitrogen are less so it will limiting reactant.

Theoretical yield:

Mass = number of moles × molar mass

Mass = 1.08 mol × 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 = 57.2%

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