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#3 acid via the oxidation of ammonia is: The first step in the reaction sequence for the production of nitric 4NH3 + 502 2 4NO + 6H2O 75% conversion is achieved with an equimolar mixture of ammonia and oxygen fed at the rate of 100 mol/h. Determine the outlet compositions. (Hint: determine the limiting reactant).

#3 acid via the oxidation of ammonia is: The first step in the reaction sequence for the production of nitric 4NH3 + 502 2 4NO + 6H2O 75% conversion is achieved with an equimolar mixture of ammonia and oxygen fed at the rate of 100 mol/h. Determine the outlet compositions. (Hint: determine the limiting reactant).

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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4 mol NHH3 produce 6mol HzO I mol NH3 goes to lomol Hhe o 4 A37.5 mol Nt3 produees → (375 ) = S6.25 MolHzO at outlet 12.5 mol unreacted NH3 37.5 mol NO T6.25 mol HzO will be obtained 10l6.25 y. NHs o (12.5/106.25) (l00%) Z NO a (375/10625 |(100%) Il.76% ミ 35.30% Outlet こ COMP 7. H20> (5le.25/106.25 )(100.) = 52.94
Transcribed Image Text:4 mol NHH3 produce 6mol HzO I mol NH3 goes to lomol Hhe o 4 A37.5 mol Nt3 produees → (375 ) = S6.25 MolHzO at outlet 12.5 mol unreacted NH3 37.5 mol NO T6.25 mol HzO will be obtained 10l6.25 y. NHs o (12.5/106.25) (l00%) Z NO a (375/10625 |(100%) Il.76% ミ 35.30% Outlet こ COMP 7. H20> (5le.25/106.25 )(100.) = 52.94
#3 acid via the oxidation of ammonia is: The first step in the reaction sequence for the production of nitric 4NH3 + 502 2 4NO + 6H2O 75% conversion is achieved with an equimolar mixture of ammonia and oxygen fed at the rate of 100 mol/h. Determine the outlet compositions. (Hint: determine the limiting reactant). #3 4NHg +5024NO + C6tzO NHs Reactor NO R NHyo So mol hr equimolar SO MOI hr 75Y. conversion achieved. NHs reacted in rxN > SO Mol (0.75)=37.5 MOl NHs unreacte d > 50-37.5 = 12.5 4MoI NHl3 5mol Oz neeoded 5 Moi 02 AMo Nti3 ( 37.5 reacted N) : 46.86 mol Oz hour 4 Mol NH3 - 4mol NO 4 MOI NO (34 5 mal Nto receted) =375 Mo) NO hour 4 mol NHs produce Comol HzO Imol NHs goes to lomol He O 4
Transcribed Image Text:#3 acid via the oxidation of ammonia is: The first step in the reaction sequence for the production of nitric 4NH3 + 502 2 4NO + 6H2O 75% conversion is achieved with an equimolar mixture of ammonia and oxygen fed at the rate of 100 mol/h. Determine the outlet compositions. (Hint: determine the limiting reactant). #3 4NHg +5024NO + C6tzO NHs Reactor NO R NHyo So mol hr equimolar SO MOI hr 75Y. conversion achieved. NHs reacted in rxN > SO Mol (0.75)=37.5 MOl NHs unreacte d > 50-37.5 = 12.5 4MoI NHl3 5mol Oz neeoded 5 Moi 02 AMo Nti3 ( 37.5 reacted N) : 46.86 mol Oz hour 4 Mol NH3 - 4mol NO 4 MOI NO (34 5 mal Nto receted) =375 Mo) NO hour 4 mol NHs produce Comol HzO Imol NHs goes to lomol He O 4
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