Question

Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l) In an industrial synthesis of urea, a chemist combines 120.6 kg of ammonia with 211.4 kg of carbon dioxide and obtains 181.4 kg of urea.

Part A) Determine the limiting reactant.

Part B)Determine the theoretical yield of urea.

Part C) Determine the percent yield for the reaction.

Answer 1

A)

Molar mass of NH3,

MM = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

mass(NH3)= 120600.0 g

use:

number of mol of NH3,

n = mass of NH3/molar mass of NH3

=(1.206*10^5 g)/(17.03 g/mol)

= 7.08*10^3 mol

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

mass(CO2)= 211400.0 g

use:

number of mol of CO2,

n = mass of CO2/molar mass of CO2

=(2.114*10^5 g)/(44.01 g/mol)

= 4.803*10^3 mol

Balanced chemical equation is:

2 NH3 + CO2 ---> CH4N2O + H2O

2 mol of NH3 reacts with 1 mol of CO2

for 7.08*10^3 mol of NH3, 3.54*10^3 mol of CO2 is required

But we have 4.803*10^3 mol of CO2

so, NH3 is limiting reagent

B)

we will use NH3 in further calculation

Molar mass of CH4N2O,

MM = 1*MM(C) + 4*MM(H) + 2*MM(N) + 1*MM(O)

= 1*12.01 + 4*1.008 + 2*14.01 + 1*16.0

= 60.062 g/mol

According to balanced equation

mol of CH4N2O formed = (1/2)* moles of NH3

= (1/2)*7.08*10^3

= 3.54*10^3 mol

use:

mass of CH4N2O = number of mol * molar mass

= 3.54*10^3*60.06

= 2.126*10^5 g

= 212.6 Kg

Answer: 212.6 Kg

C)

% yield = actual mass*100/theoretical mass

= 181.4*100/212.6

= 85.32 %

Answer: 85.32 %