50.0kg of N2[g] and 10.0 kg of H2 [g] are mixed to produce NH3[g] formed.Identify the limiting reagent in production of NH3 in this situaton

 

Given:

 

Mass of N₂ = 50 kg

 

                 = 50 × 10³ gm

 

No. of moles of  N₂

 

 

Mass of H₂ = 10 kg

                 = 10 × 10³ gm

 

No. of moles of  H₂

 

 

Formation of ammonia;

 

 

From reaction, 1 mole of N₂ requires 3 moles H₂.

 

So 1.78× 10³ mole of N₂ will require  (1.78×10³ ×3)  = 5.34 ×10³ moles of H₂.

 

Here 5×10³ moles of H₂ are given, therefore H₂ is the limiting reagent.

 

 

Now  5×10³ moles of H₂ will give,

 

 

Therfore the amount of NH₃ produced is 3.33×10³ mole.