If 22.8 g of nitrogen with 4.93 g of hydrogen are combined in a closed system, what mass of ammonia will be produced?

             27.73 g of NH₃

Solution:

The Balance Chemical Reaction is as follow,

                                       N₂  +   3 H₂    →    2 NH₃  

Step 1: Find out the limiting reagent as;

According to Equation ,

               28 g (1 mole) N₂ reacts with  =  6.04 g (3 moles) of H₂

So,

                  22.8 g of N₂ will react with  =  X g of H₂

Solving for X,

                      X  =  (22.8 g × 6.04 g) ÷ 28 g

                      X  =  4.91 g of H₂

It means for total utilization of 22.8 g of N₂ we require 4.91 g of H₂, but we are provided with 4.93 g of H₂. Therefore, N₂ is the limiting reagent and will control the yield of product.

Step 2: Calculate Amount of NH₃ produced as;

According to Equation ,

                     28 g (1 mole) N₂ produces  =  34.06 g (2 moles) of NH₃

So,

                       22.8 g of N₂ will produce  =  X g of NH₃

Solving for X,

                      X =  (22.8 g × 34.06 g) ÷ 28 g

                      X  =  27.73 g of NH₃

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