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)< br />
In an industrial synthesis of urea, a chemist combines 128.9 kg of ammonia with 211.4 kg of carbon dioxide and obtains 166.3 kg of urea.

Determine the theoretical yield of urea

The theoretical yield of urea : = 227.4 kg

Given

Reaction

2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l)

128.9 kg of ammonia

211.4 kg of carbon dioxide

166.3 kg of urea.

Required

The theoretical yield of urea

Solution

mol Ammonia (MW=17 g/mol)

=128.9 : 17

= 7.58 kmol

mol CO₂(MW=44 g/mol) :

= 211.4 : 44

= 4.805 kmol

Mol : coefficient of reactant , NH₃ : CO₂ :

= 7.58/2 : 4.805/1

=3.79 : 4.805

Ammonia as limiting reactant(smaller ratio)

Mol urea based on mol Ammonia :

=1/2 x 7.58

=3.79 kmol

Mass urea :

=3.79 kmol x 60 g/mol

= 227.4 kg

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :