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 129.2 kg of ammonia with 211.4 kg of carbon dioxide and obtains 180.1 kg of urea.

1).Determine the theoretical yield of urea

2).Determine the percent yield for the reaction.

Answer 1

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 129.2 kg of ammonia with 211.4 kg of carbon dioxide and obtains 180.1 kg of urea.

Solution :-

a)Thereotical yield calculation

Using the mole ratio of the both reactant lets calculate the mass of the urea produced

(129.2 kg NH3 * 1000 g / 1 kg) *(1 mol NH3/17.03 g) *(1 mol CH4N2O/2 mol NH3) = 3793.3 mol urea

(211.4 kg CO2*1000g /1 kg)*(1 mol CO2/44.01 g)*(1 mol urea / 1 mol CO2) =4803 mol urea

NH3 gives less moles of urea therefore NH3 is limiting reactant

So the maximum moles of urea that can be formed = 3793.3 mol

Now lets convert moles of urea to its mass

Mass of urea = moles * molar mass

                  = 3793.3 mol * 60.06 g per mol

                  = 227826 g urea

Lets convert to kg

227826 g * 1 kg /1000 g = 228 kg urea

So the threotical yield = 228 kg urea

b)Now lets calculate the percent yield

% yield = (actual yield / theoretical yield )*100%

             = (180.1 kg /228 kg)*100%

             = 79.00 %