Answered

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Calculate the ratio (lb-mole o2 react/lb-mole no formed). 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 40.0xcess o2?

Sagot :

175 Kmol/h is oxygen the fed rate that corresponds to 40% excess O₂.

Given chemical reaction is

4NH₃ + 5O₂ → 4NO + 6H₂O

The molecular weight of NH₃ is 17.03, O₂ is 32, NO is 30, and, H₂O  is 18.

To find the ratio,

Ratio = lb-mole O₂ react / lb-mole NO formed

Or, ratio = 5/4

Or, ratio = 1.25

NH₃ fed rate = 100 Kmol/h

We have to find out whether the Oxygen fed rate corresponds to 40% excess O₂.

For each 4 Kmol of NH₃, 5 Kmol of O₂ is required.

4 Kmol of NH₃ ≡ 5 Kmol of O₂

100 Kmol of NH₃ ≡ 125 Kmol of O₂

So, 125 Kmol of O₂  is required for 100 Kmol of NH₃.

For 40% excess of O₂, the required Kmol is = 125 + 125 × 40%

The required Kmol of oxygen is 175 Kmol/h for excess of O₂.

To learn more about ammonia(NH₃) and oxygen(O₂), visit: https://brainly.com/question/20923890

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