Question

Sulfur trioxide, ammonia gas, and water will react to form ammonium sulfate. Determine the % yield when 3.78g of ammonium sulfate forms from the reaction of 2.51g of sulfur trioxide, 1.55g of ammonia, and 0.804g of water.

Answer 1

    SO3 +        2 NH3       +   H2O       ----------->    (NH4)2SO4

1 mol        2 mol                  1mol

    80 g       2x 17g = 34 g       18 g                            132 g

   2.51 g     1.55 g                  0.804 g                          3.78 g

First , we have to identify the limiting reagent.

moles = mass / molar mass in the balanced equation

moles of SO3 = 2.51 g/ 80 g/mol = 0.031 mol

moles of NH3 = 1.55 g/ 34 g /mol = 0.045 mol

moles of H2O = 0.804 g/ 18 g/mol = 0.044 mol

Hence, SO3 is present in lesser number of moles. Therefore, SO3 is the limiting reagent.

Yield is calculated based on SO3.

SO3 +        2 NH3       +   H2O       ----------->    (NH4)2SO4

1 mol     

    80 g                                                               132 g

   2.51 g                                                                 ?

?   = (2.51 g/ 80g) x 132 g ammonium sulfate

     = 4.14 g ammonium sulfate

This is theoretical yield of ammonium sulfate.

But the actual yield of ammonium sulfate = 3.78 g

Hence,

% yield = (actual yield / theoretical yield) x 100

              = ( 3.78 g/ 4.14 g) x 100

               = 91.3 %

Therefore,

% yield of ammonium sulfate = 91.3 %