Answer:
NH3 is the limiting reactant
The theoretical yield is 216.0 kg urea
The % for this reaction is 78.8 %
Explanation:
<u>Step 1:</u> Data given
Mass of ammonia = 122.5 kg
Mass of carbon dioxide = 211.4 kg
Mass of urea produced = 170.3 kg
Molar mass of ammnoia = 17.031 g/mol
Molar mass of carbon dioxide = 44.01 g/mol
Moalr mass of urea = 60.06 g/mol
<u>Step 2:</u> The balanced equation
2NH3(aq) + CO2(aq) --> CH4N2O(aq) + H2O(l)
<u>Step 3:</u> Calculate moles of NH3
Number of moles = mass / Molar mass
Moles NH3 = 122500 grams / 17.031 g/mol
Moles NH3 = 7192.77 moles
<u>Step 4:</u> Calculate moles of CO2
Moles CO2 = 211400 / 44.01 g/mol
Moles CO2 = 4803.45 moles
<u>Step 5</u>: Calculate limiting reactant
For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O
NH3 is the limiting reactant. It will completely be consumed (7192.77 moles).
CO2 is in excess. There will be consumed 7192.77/2 = 3596.4 moles
There will remain 4803.45 - 3596.4 = 1207.05 moles of CO2
<u>Step 6:</u> Calculate moles of urea produced:
For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O
For 7192.77 moles of NH3, we have 3596.4 moles of urea produced
<u>Step 7: </u>Calculate mass of urea
Mass urea = moles urea * molar mass urea
Mass urea = 3596.4 moles * 60.06 g/mol
Mass urea = 216000 grams = 216 kg = theoretical yield
<u>Step 8</u>: Calculate % yield
% yield = (actual yield / theoretical yield)*100%
% yield = (170.3 / 216) *100% = 78.8%
The % for this reaction is 78.8 %
