Mass balance on oxidation of ammonia for production of nitric acid

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SUMMARY

The mass balance on the oxidation of ammonia for nitric acid production involves the reaction 4NH3 + 5O2 ⇔ 4NO + 6H2O, with an 80% conversion rate of an equimolar mixture of NH3 and O2 fed at 100 mol/h. To determine output rates, calculate the moles of each species based on the initial flow rates and the conversion percentage. Specifically, for 100 moles of reactants, 80 moles of NH3 will react, producing 64 moles of NO and 120 moles of H2O, while 20 moles of NH3 and 20 moles of O2 will remain unreacted.

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Mangoes
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Homework Statement



The following chemical reaction takes place inside a reactor where 80% conversion is obtained with an equimolar mixture of NH3 and O2 fed at the flow rate of 100 mol/h.

4NH3 + 5O2 ⇔ 4NO + 6H2O

What are the output rates of all species?

There's a diagram shown where ammonia and oxygen are pictured as feed, output are the products, nitrogen monoxide and water, and the unused reactants, ammonia and oxygen.

The Attempt at a Solution



The problem wants me to do a mass balance on molecular species.

I've just recently started looking at processes involving reactions however and I'm not too sure on how to write balances when reactions are present. If we looked at NH3 for example,

input = consumption + output

I can denote different variables to represent input and ouput for NH3, but I'm not sure how I'd quantify the consumption of NH3 per unit time. Previously, I would be given the flow rate of one of the outputs of a species and then from stoichiometry I'd be able to get the consumption/generation of any of the species, but I'm not being given it this time.

I assume I have to use the 80% conversion somewhere, but that's for equimolar mixtures and I'm not sure how I'd account limiting reactants into the equation to get the appropriate consumption term.
 
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Mangoes said:

Homework Statement



The following chemical reaction takes place inside a reactor where 80% conversion is obtained with an equimolar mixture of NH3 and O2 fed at the flow rate of 100 mol/h.

4NH3 + 5O2 ⇔ 4NO + 6H2O

What are the output rates of all species?

There's a diagram shown where ammonia and oxygen are pictured as feed, output are the products, nitrogen monoxide and water, and the unused reactants, ammonia and oxygen.

The Attempt at a Solution



The problem wants me to do a mass balance on molecular species.

I've just recently started looking at processes involving reactions however and I'm not too sure on how to write balances when reactions are present. If we looked at NH3 for example,

input = consumption + output

I can denote different variables to represent input and ouput for NH3, but I'm not sure how I'd quantify the consumption of NH3 per unit time. Previously, I would be given the flow rate of one of the outputs of a species and then from stoichiometry I'd be able to get the consumption/generation of any of the species, but I'm not being given it this time.

I assume I have to use the 80% conversion somewhere, but that's for equimolar mixtures and I'm not sure how I'd account limiting reactants into the equation to get the appropriate consumption term.

For a few seconds I asked myself 80% conversion of what? - but if you look at the products it is clear they will be interested in the % conversion of the ammonia into NO.

Start with the NH3 conversion into NO. One mole of one converts into how many moles of the other?

And 80% conversion of anything converts a mole into how many?

The answer for two of the substances needs very little chemistry just some common sense.
For the other two a very little chemistry as well is also required.
 
Last edited:
Take as a basis 100 moles of reactants entering the reactor.

How many moles of NH3 enter?
How many moles of O2 enter?
For 80% conversion, how many moles of NH3 react?
How many moles of NH3 are left?
How many moles of O2 react?
How many moles of O2 are left?
How many moles of NO are produced?
How many moles of H2O are produced?
How many moles of each species exit the reactor?

Chet
 

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