1)Determine the limiting reactant. Express your answer as a chemical formula. 2)
ID: 851750 • Letter: 1
Question
1)Determine the limiting reactant.
Express your answer as a chemical formula.
2) Determine the theoretical yield of urea
Express your answer using four significant figures.
3) Determine the percent yield for the reaction.
Express your answer using three significant figures.
Urea (CH4N2O) kg of urea. 1)Determine the limiting reactant. Express your answer as a chemical formula. 2) Determine the theoretical yield of urea Express your answer using four significant figures. 3) Determine the percent yield for the reaction. Express your answer using three significant figures. {rm kg} of carbon dioxide and obtains 161.8kg kg of ammonia with 211.4 kg 2;{rm{NH}}_3 (aq); + ;{rm{CO}}_2 (aq); to ;{rm{CH}}_4 {rm{N}}_2 {rm{O}}(aq); + ;{rm{H}}_2 {rm{O}}(l) In an industrial synthesis of urea, a chemist combines 140.5kg ({rm{NH}}_3 ) with carbon dioxide as follows: 2NH3(aq)+CO2(aq)?CH4N2O(aq)+H2O(l) ({rm{CH}}_4 {rm{N}}_2 {rm{O}}) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3)Explanation / Answer
1)
Molar mass of NH3,
MM = 1*MM(N) + 3*MM(H)
= 1*14.01 + 3*1.008
= 17.034 g/mol
mass(NH3)= 140500.0 g
use:
number of mol of NH3,
n = mass of NH3/molar mass of NH3
=(140500.0 g)/(17.03 g/mol)
= 8.248*10^3 mol
Molar mass of CO2,
MM = 1*MM(C) + 2*MM(O)
= 1*12.01 + 2*16.0
= 44.01 g/mol
mass(CO2)= 211400.0 g
use:
number of mol of CO2,
n = mass of CO2/molar mass of CO2
=(211400.0 g)/(44.01 g/mol)
= 4.803*10^3 mol
Balanced chemical equation is:
2 NH3 + CO2 ---> CH4NO2 + H2O
2 mol of NH3 reacts with 1 mol of CO2
for 8.248*10^3 mol of NH3, 4.124*10^3 mol of CO2 is required
But we have 4.803*10^3 mol of CO2
so, NH3 is limiting reagent
Answer: NH3
2)
we will use NH3 in further calculation
Molar mass of CH4NO2,
MM = 1*MM(C) + 4*MM(H) + 1*MM(N) + 2*MM(O)
= 1*12.01 + 4*1.008 + 1*14.01 + 2*16.0
= 62.052 g/mol
According to balanced equation
mol of CH4NO2 formed = (1/2)* moles of NH3
= (1/2)*8.248*10^3
= 4.124*10^3 mol
use:
mass of CH4NO2 = number of mol * molar mass
= 4.124*10^3*62.05
= 2.559*10^5 g
Answer: 2.559*10^2 Kg
3)
% yield = actual mass*100/theoretical mass
= 1.618*10^5*100/2.559*10^5
= 63.23%
Answer: 63.2%
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