Question

Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2 NH3 (aq) + CO2 (aq) → CH4N2O (aq) + H2O (1) In an industrial synthesis of urea, a chemist combines 124.2 kg of ammonia with 211.4 kg of carbon dioxide and obtains 169.8 kg of urea.

Question 1:

Determine the limiting reactant

Question 2:

Determine the theoretical yield of urea

Question 3:

Determine the percentage yield for the reaction

Answer 1

The balanced equation of reaction is

2NH3(aq) + CO2(aq) ------------------> CH4NO(aq) + H2O(l)

124.2x10³ /17 211.4x10³ /44 0 0 initial moles

Q1)

To calculate the limiting reagent   

=7.30x10³ = 4.8x10³

the ratio is

7.30x10³ /2=3.6x10³ 4.8x10³ /1 =4.8x10³

Thus NH3 has lower value thus NH3 is the limiting reagent

Q2)

The theoretical yield is obtianed using the limiting reagent.

  2NH3(aq) + CO2(aq) ------------------> CH4NO(aq) + H2O(l)

124.2x10³ /17 211.4x10³ /44 0 0 initial moles

=7.30x10³ = 4.8x10³

0 4.8x10³ -3.6 x10³ 3.6x10³ - after reaction

Thus moles of urea formed = 3.6x10³ moles

mass of urea formed = moles x molar mass

= 3.6x10³ mol x 60g/mol

= 216 x10³ g

= 216 kg

Thus the theoretical yield of urea = 216 kg

Q3)

actual yield of urea = 169.8 kg

% yield = actual yield x 100/theoretical yield

= 169.8x100/216

= 78.61%

The % yield = 78.61% of urea