If 4.55 moles of hydrogen were reacted with excess nitrogenin the equations N2 + 3 H2 = 2 NH3 and 48.7 g of ammonia product was recovered what would be the percent yield of the reaction?

Percent yield = 94.5%

Explanation:

Given data:

Number of moles of hydrogen = 4.55 mol

Mass of ammonia recovered = 48.7 g

Percent yield of ammonia = ?

Solution:

Chemical equation:

N₂+ 3H₂    →       2NH₃

Now we will compare the moles of ammonia and hydrogen.

               H₂        :         NH₃

                3         :          2

                4.55   :         2/3×4.55 = 3.03 mol

Theoretical yield of ammonia:

Mass = number of moles × molar mass

Mass = 3.03 mol × 17 g/mol

Mass = 51.51 g

Percent yield:

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

Percent yield = (48.7 g/ 51.51 g)  × 100

Percent yield = 0.945 × 100

Percent yield = 94.5%