What is the percent yield of ammonia if the yield of ammonia is 15 g and you had
14 g of nitrogen to start with?
N2 + 3H2 → 2NH3

Percent yield = 88.23%

Explanation:

Given data:

Percent yield of ammonia = ?

Actual yield = 15 g

Mass of nitrogen = 14 g

Solution:

Chemical equation:

N₂ + 3H₂      →     2NH₃

Number of moles of nitrogen:

Number of moles = mass/molar mass

Number of moles = 14 g/ 28 g/mol

Number of moles = 0.5 mol

Now we will compare the moles of ammonia and nitrogen.

         N₂          :         NH₃

          1            :         2

         0.5        :        2/1×0.5 = 1 mol

Mass of ammonia/Theoretical yield

Mass = number of moles × molar mass

Mass = 1 mol × 17 g/mol

Mass = 17 g

Percent yield:

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

Percent yield =  (15 g/ 17 g)× 100

Percent yield = 88.23%