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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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