Answer:
(a) 1.25 lb-mol O₂/ lb- mol NO
(b) 175 kmol/h
(c) NH₃ is the limiting reagent
70.3 % O₂ in excess
100 %
87.9 kg NO
Explanation:
(a) From the stoichiometry of the reaction we know that 5 mol O₂ react to produce 4 mol NO.
5 mol O₂ / 4 mol NO = 1.25 molO₂/molNO
This ratio is the same when using lb-mol, since lb-mol is used by chemical engineers and is equal to 454 mol.
Therefore the ratio is 1.25 lb-mol O₂/ lb- mol NO
(b) The stoichiometry of the reaction is still the same when working in continous reactios, so from the balanced chemical reaction we know:
5 mol O₂ / 4 mol NH₃ x 100 kmol/ h = 125 kmol / h
Now we want a 40.0 excess, then the flow rate iof O₂ is:
125 kmol/h x 1.40 = 175 kmol/h
(c) First lets calculate the number of moles of ammonia and oxygen and perform the calculations required utilizing the stoichiometry of the reaction:
mol NH₃ = 50.0 kg / 17.03 kg/kmol = 2.93 kmol NH₃
mol O₂ = 100.0 kg / 15.99 kg/kmol =6.25 kmol O₂
kmol O₂ required to react with 2.93 kmol NH₃ :
2.93 kmol NH₃ x 5 kmol O₂ / 4 kmol NH₃ = 3.67 kmol O₂
We have 6.25 kmol O₂ and require 3.67 kmol to react with 2.93 kmol NH₃, therefore, the limiting reagent is NH₃.
To calculate the % the other reactant is in excess:
( 6.25 kmol - 3.67 kmol ) / 3.67 kmol x 100 = 70.3 %
If the reaction proceeds to completion the extent of reaction is 100 % respect NH₃.
To determine the mass NO produced, we know from the stoichiometry of the reaction that the reaction is 1:1 respect NH₃ ( 4 mol NH₃ produces 4 mol NO ). So, the mass of NO produced in kg is :
kmol NO produced = 2.93 km
Molar mass NO = 30.01 kg/kmol
mass NO = 2.93 kmol x 30.01 kg/kmol = 87.9 kg
