3h2(g)+n2(g)→2nh3(g) what is the theoretical yield of ammonia, in kilograms, that we can synthesize from 5.25 kg of h2 and 32.7 kg of n2?

             29.75 Kg of NH₃

Solution:

              In order to calculate the theoretical yield, first we will identify the limiting reactant.
According to equation,

                      6 g (3 moles) H₂ requires  =  28 g (1 mole) N₂
So,
                      5250 g H₂ will require  =  X g of N₂

Solving for X,
                      X  =  (5250 g × 28 g) ÷ 6 g

                      X  =  25433 g of N₂

Hence, it is found that H₂ is the limiting reactant because N₂ is provided in excess (32700 g). Therefore,
As,
                   6 g (3 mole) H₂ produced  =  34 g ( 2 moles) of NH₃
So,
                       5250 g H₂ will produce  =  X g of NH₃

Solving for X,
                     X  =  (5250 g × 34 g) ÷ 6 g

                     X  =  29750 g of NH₃
Or,
                     X  =  29.75 Kg of NH₃

Adozen is 1221/12 = 1.751.75 x 12 = 21therefore, 1.75 is the answer.