Question

Urea [(NH₂)₂CO] is prepared by reacting ammonia with carbon dioxide.

2NH_3(g)  + CO_2(g)  → (NH₂)₂CO_(aq) + H₂O_(l)

In one process, 637.2 g of NH₃ are treated with 1142 g of CO₂.

i. Which of the two reactants is the limiting reagent?

ii. Calculate the mass of (NH₂)₂CO formed.

iii. How much excess reagent (in grams) is left at the end of the reaction?

Answer 1

i. If 637.2 g of NH₃ react completely, calculate the number of moles of (NH₂)₂CO, that could be produced, by the following relation. 

Mass of NH₃ → Moles of NH₃ → Moles of (NH₂)₂ CO

Moles of (NH₂)₂ CO =

= 18.71 moles of (NH₂)₂CO

If 1142 g of CO react completely, calculate the number of moles of (NH₂)₂CO, that could be produced, by the following relation. 

Mass of CO₂ → Moles of CO₂ → Moles of (NH₂)₂CO

Moles of (NH₂)₂CO = 

= 25.95 mol of (NH₂)₂CO

Since NH₃ produces smaller amount of (NH₂)₂CO, the limiting reagent is NH₃.

ii. Mass of (NH₂)₂CO = 

= 1124 g(NH₂)₂CO

iii. Starting with 18.71 moles of (NH₂)₂CO, we can determine the mass of CO₂ that reacted using the mole ratio from the balanced equation and the molar mass of CO₂ by the following relation.

Moles of (NH₂)₂ CO → Moles of CO₂ → Grams of CO₂

Mass of CO₂ reacted

= 823.4 g CO₂

The amount of CO₂ remaining = 1142 g – 823.4 g = 318.6 g ≈ 319 g CO₂ remaining

i. Limiting reagent =  NH₃

ii. Mass of (NH₂)₂CO produced = 1124 g

iii. Mass of CO₂ remaining = 319 g