Part A Determine the limiting reactant Urea (CH4 N2O) is a common fertilizer tha

Question

Part A Determine the limiting reactant Urea (CH4 N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NHs) with carbon dioxode as follows 2 NHs (aq) +CO (aq) CH,N2(aH2O(1) In an industrial synthesis of urea, a chemist combines 131.6 kg of ammonia with 211.4 kg of carbon dioxide and obtains 167 5 kg of urea Express your answer as a chemical formula. Submit Request Answer v Part B Determine the theoretical yield of urea Express your answer using four significant figures. kg Submit Part C Determine the percent yield for the reaction Express your answer using three significant figures

Explanation / Answer

Part A

molar mass of NH3 = 17.031 g/mol

molar mass of CO2 = 44.01 g/mol

131.6 kg NH3 = 131600 gm NH3

211.4 kg CO2 = 211400 gm CO2

no.of mole = mass of compound in gm / molar mass of compound

no. of mole of NH3 = 121600 / 17.031 = 7139.92 mole

no. of mole of CO2 = 211400 / 44.01 = 4803.45 mole

According to reaction 1 mole CO2 react with 2 mole of NH3 molar retio between CO2 to NH3 = 1:2 therefore to react with 4803.45 mole of CO2 required NH3 = 4803.45 X 2 = 9606.90 mole but NH3 given only 7139.92 mole therefore NH3 is limiting reactant

Limiting reactant = NH3

Part B

According to reaction 2 mole HN3 produce 1 mole urea molar retio between NH3 to urea is 2:1 therefore 7139.92 mole of NH3 produce urea = 7139.92 / 2 = 3569.96 mole

urea produced = 3569.96 mole

molar mass of urea = 60.06 g/mol

gm of substance = no. of mole X molar mass

gm of urea produced = 3569.96 X 60.06 = 214411.8 gm = 214.4 kg

theoretical yield of urea = 214.4 kg

Part C

% yield = actual yield X 100 / theoretical yield

% yield of urea = 167.5 X 100 / 214.4 = 78.125 %

% yield of urea = 78.125 %

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