Question

1-Urea, which has the chemical formula (NH₂)₂CO, is a fertilizer that can be prepared by reacting ammonia (NH₃) with carbon dioxide (CO₂). Given the following chemical equation, what is the theoretical yield of urea (in grams) if 8.19 mol carbon dioxide is the limiting reactant?

2 NH₃(g) + CO₂(g) → (NH₂)₂CO(aq) + H₂O(l)

2- Suppose 4.77 g of NH3 are mixed with 7.52 g of CO2. Determine the limiting reagent and theoretical yield of urea

2 NH₃(g) + CO₂(g) → (NH₂)₂CO(aq) + H₂O(l)

What is the equation balanced also?

Answer 1

2 NH3 + CO2 (NH2)2CO + H2O

1. As per the balanced chemical equation 1 mol CO2 produces 1 mol of urea.

1 mol CO2 : 1 mol urea

8.19 mol CO2 : 8.19 mol urea

Now,

Mol of urea = Wt of urea / Molar mass

Wt of urea = 8.19 mol * 60 g/mol

= 491.4 g

Thus,

Theoretical yield = 491.4 g

2. Mol of NH3 = 4.77 g / 17 g/mol = 0.28 mol

Mol of CO2 = 7.52 g / 44 g/mol = 0.17 mol

As per the equation-

2 mol NH3 : 1 mol CO2

0.28 mol NH3 : 0.14 mol CO2

This means 0.28 mol NH3 will react with 0.17 mol of CO2.

Thus,

CO2 is in excess while NH3 is limiting reagent.

Now,

2 mol NH3 : 1 mol urea

0.28 mol NH3 : 0.14 mol urea

Mol of urea = 0.14 mol

Wt of urea = 0.14 mol * 60g/mol = 8.4 g

Thus,

Theoretical yield = 8.4 g