Urea (CH_4N_2O) is a common fertilizer that can be synthesized by the reaction o

Question

Urea (CH_4N_2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH_3) with carbon dioxide as follows 2NH_3(aq) + CO_2(aq) rightarrow CH_4N_2O(aq) + H_2O(l)in an industrial synthesis of urea, a chemist combines 133 1 kg of ammonia with 211 4 kg of carbon dioxide and obtains 166 3 kg of urea Determine the limiting reactant Express your answer as a chemical formula Determine the theoretical yield of urea Express your answer using four significant figures Determine the percent yield for the reaction Express your answer using three significant figures

Explanation / Answer

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

133.1 kg of ammonia with 211.4 kg of carbon dioxide and obtains 166.3 kg of urea.

a. Determine the limiting reactant. NH3--------answer

b.Determine the theoretical yield of urea

Molar masses

NH4 =17.0337 g/mol

CO2 =44.0088 g/mol

CH4 N2O=60.061 g/mol

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133. 1 kg NH3 x (1000g NH3 /1 kg NH3)* (1 mol NH3 /17.0337g NH3) * (1 mol CH4N2O /2 mol NH3)*(60.061 g CH4N2O/ 1 mol CH4N2O)* (1 KG/1000 G)=

=234.655 Kg CH4 N2O-answer

211.4 kg co2 x (1000g co2 /1 kg co2)* (1 mol co2 /44.0088 g co2) * (1 mol CH4N2O /2 mol NH3)*(60.061 g CH4N2O/ 1 mol CH4N2O)* (1 KG/1000 G)=

=288.51 Kg CH4 N2O

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c.Determine the percent yield for the reaction.

% yield = Actual yield/ theoretical yield =(168.4 /234.65) *100%

=71.76 % answer

=

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