Question

Urea (CH4N20 ) is a common fertilizer that can be synthesized
by the reaction of ammonia (NH3) with carbon dioxide:
In an industrial synthesis of urea, a chemist combines 136.4 kg
of ammonia with 2 l l .4 kg of carbon dioxide and obtains l 68.4
kg of urea. Determine the limiting reactant, the theoretical yield of
urea, and percent yield for the reaction.

Chemical formula: 2NH3 + CO2 = CH4N2O + H2O

Answer 1

2NH3 + CO2 = CH4N2O + H2O
moles NH3 = 136400 g/17 = about 8024.
moles CO2 = 211400 g/44 = 4804
If NH3 is the limiting reagent(LR) we will need 8024/2 = 4012 moles CO2. Do we have that much CO2. Yes, therefore, NH3 is the LR and CO2 is the excess reagent (ER).
moles urea produced = 8024 mols NH3 x (1 mol urea/2 mols NH3) = 4012 moles urea. Then grams urea = mols x molar mass = ? g. This is the theoretical yield (TY). From the problem the actual yield (AY) is 168400 grams.
% yield = (AY/TY)*100 = ?
Post your work if you get stuck. I've rounded the numbers above so you should confirm and round to your specifications.