Ammonia is oxidized to nitric oxide in the following reaction:
4NH3 + 5O2 --------> 4NO + 6H2O
a. Calculate the ratio (lb-mole O2 react/lb-mole NO formed).
b. If ammonia is fed to a continuous reactor at a rate of 100.0 kmol NH3/h, what oxygen feed rate (kmol/h) would correspond to 40.0% excess O2?
c. If 50.0 kg of ammonia and 100.0 kg of oxygen are fed to a batch reactor, determine the limiting reactant, the percentage by which the other reactant is in excess, and the extent of reaction and mass of NO produced (kg) if the reaction proceeds to completion.

Question

sunarafet84621 sunarafet84621

05-02-2020
Chemistry

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Ammonia is oxidized to nitric oxide in the following reaction:
4NH3 + 5O2 --------> 4NO + 6H2O
a. Calculate the ratio (lb-mole O2 react/lb-mole NO formed).
b. If ammonia is fed to a continuous reactor at a rate of 100.0 kmol NH3/h, what oxygen feed rate (kmol/h) would correspond to 40.0% excess O2?
c. If 50.0 kg of ammonia and 100.0 kg of oxygen are fed to a batch reactor, determine the limiting reactant, the percentage by which the other reactant is in excess, and the extent of reaction and mass of NO produced (kg) if the reaction proceeds to completion.

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baraltoa baraltoa · Answer 1 · 2020-02-06T14:47:42+01:00

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