Question

1.86 g H2 is allowed to react with 9.75 g N2 , producing 2.87g NH3.

Answer 1

Answer:

                     (a)  Theoretical Yield  =  10.50 g

                      (b)   %age yield  = 27.33 %

Explanation:

Answer-Part-(a)

                 The balance chemical equation for the synthesis of Ammonia is as follow;

                                          N₂ + 3 H₂ → 2 NH₃

Step 1: Calculating moles of N₂ as;

                   Moles = Mass / M/Mass

                   Moles = 9.75 g / 28.01 g/mol

                   Moles = 0.348 moles of N₂

Step 2: Calculating moles of H₂ as;

                   Moles = Mass / M/Mass

                   Moles = 1.86 g / 2.01 g/mol

                   Moles = 0.925 moles

Step 3: Finding Limiting reagent as;

According to equation,

                1 mole of N₂ reacts with  =  3 moles of H₂

So,

             0.348 moles of N₂ will react with  =  X moles of H₂

Solving for X,

                     X = 3 mol × 0.348 mol / 1 mol

                     X = 1.044 mol of H₂

It shows that to consume 0.348 moles of N₂ completely we require 1.044 mol of Hydrogen while, as given in statement we are only provided with 0.925 moles of H₂ hence, hydrogen  is limiting reagent. Therefore, H₂ will control the final yield.

Step 4: Calculating moles of Ammonia as,

According to equation,

                3 mole of H₂ produces  =  2 moles of NH₃

So,

             0.925 moles of H₂ will produce  =  X moles of NH₃

Solving for X,

                     X = 2 mol × 0.925 mol / 3 mol

                     X = 0.616 mol of NH₃

Step 5: Calculating theoretical yield of Ammonia as,

                     Theoretical Yield  =  Moles × M.Mass

                     Theoretical Yield  =  0.616 mol  × 17.03 g/mol

                     Theoretical Yield  =  10.50 g

Answer-Part-(b)

                    %age yield  = Actual Yield / Theoretical Yield × 100

                    %age yield  = 2.87 g / 10.50 g × 100

                    %age yield  = 27.33 %