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 122.5kg of ammonia with 211.4kg of carbon dioxide and obtains 170.3kg of urea. Determine the limiting reactant. (express your answer as a chemical formula) Determine the theoretical yield of urea. (express using four significant figures) m=___________kg Determine the percent yield for the reaction. (express your answer using three significant figures)

Question

20-11-2019
Chemistry

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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 122.5kg of ammonia with 211.4kg of carbon dioxide and obtains 170.3kg of urea. Determine the limiting reactant. (express your answer as a chemical formula) Determine the theoretical yield of urea. (express using four significant figures) m=___________kg Determine the percent yield for the reaction. (express your answer using three significant figures)

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thomasdecabooter thomasdecabooter · Answer 1 · 2019-11-20T16:36:08+01:00

Answer:

NH3 is the limiting reactant

The theoretical yield is 216.0 kg urea

The % for this reaction is 78.8 %

Explanation:

Step 1: 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

Step 2: The balanced equation

2NH3(aq) + CO2(aq) --> CH4N2O(aq) + H2O(l)

Step 3: Calculate moles of NH3

Number of moles = mass / Molar mass

Moles NH3 = 122500 grams / 17.031 g/mol

Moles NH3 = 7192.77 moles

Step 4: Calculate moles of CO2

Moles CO2 = 211400 / 44.01 g/mol

Moles CO2 = 4803.45 moles

Step 5: 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

Step 6: 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

Step 7: 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

Step 8: Calculate % yield

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

% yield = (170.3 / 216) *100% = 78.8%

The % for this reaction is 78.8 %