Question

Urea is prepared by the reaction between ammonia and carbon dioxide. 

In one process, 637.2 g of NH₃ are allowed to react with 1142 g of CO₂ . 

(a) Which of the two reactants is the limiting reagent? 

(b) Calculate the mass of(NH₄)₂ CO formed. 

(c) how much of the excess reagent in grams is left at the end of the reaction?

Answer 1

No. of moles of ammonia = 637.2/17 = 37.45 mole 

No. of moles of CO₂ = 1142/44 = 25.95 moles 

As per the balanced equation, one mole of CO₂ requires 2 moles of ammonia.

∴ No. of moles of NH₃ required to react with 25.95 moles of CO₂ is = 2/1 x 25.95 = 51.90 moles. 

∴ 37.45 moles of NH₃ is not enough to completely react with CO₂ (25.95 moles).

Hence, NH₃ must be the limiting reagent, and CO₂ is excess reagent.

(b) 2 moles of ammonia produce 1 mole of urea. 

∴ Limiting reagent 37.45 moles of NH₃ can produce 1/2 x 37.45 moles of urea. 

= 18.725 moles of urea. 

∴ The mass of 18.725 moles of urea = (No. of moles) x (Molar mass) 

= 18.725 x 60 

= 1123.5 g of urea.

(c) 2 moles of ammonia requires 1 mole of CO₂ . 

∴ Limiting reagent 37.45 moles of NH₃ will require 1/2 x 37.45 moles of CO₂ 

= 18.725 moles of CO₂ 

∴ No. of moles of the excess reagent (CO₂) left = 25.95 – 18.725 = 7.225 

The mass of the excess reagent (CO₂) left = 7.225 x 44 = 317.9 g CO₂ .