The composition of gas in the feed, the percentage conversion and the
theoretical yield are combined to give the product stream composition.
Response:
The composition of gas in the product stream are;
- HCl: 0.4 kmol/h, Cl₂: 1.6 kmol/h, H₂O: 1.6 kmol/h, O₂: 0.5 kmol/h
<h3>How can percentage conversion give the contents of the product stream?</h3>
The amount of oxygen used = 30% exceeding the theoretical amount
Number of moles of hydrochloric acid = 4 kmol/h
Percentage conversion = 80%
Required:
The composition of the gas in the product feed.
Solution;
The given reaction is; 4HCl + O₂
2Cl₂ + 2H₂O
![Percentage \ conversion = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{Moles \ of \ limiting \ reactant \ supplied \ in \ the \, feed}}](https://tex.z-dn.net/?f=Percentage%20%5C%20conversion%20%3D%20%5Cmathbf%7B%20%5Cdfrac%7BMoles%20%5C%20of%20%5C%20limiting%20%5C%20reactant%20%5C%20reacted%7D%7BMoles%20%5C%20%20of%20%5C%20limiting%20%5C%20reactant%20%5C%20supplied%20%5C%20in%20%5C%20the%20%5C%2C%20feed%7D%7D)
Which gives;
![80 \% = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{4 \, kmol/h}}](https://tex.z-dn.net/?f=80%20%5C%25%20%3D%20%5Cmathbf%7B%20%5Cdfrac%7BMoles%20%5C%20of%20%5C%20limiting%20%5C%20reactant%20%5C%20reacted%7D%7B4%20%5C%2C%20kmol%2Fh%7D%7D)
Moles of limiting reactant reacted = 4 kmol/h × 0.80 = 3.6 kmol/h
Which gives;
Number of moles of HCl in the stream = 4 kmol/h - 3.6 kmol/h = 0.4 kmol/h
Number of moles of Cl₂ produced = 2 kmol/h × 0.8 = 1.6 kmol/h
Similarly;
Number of moles of H₂O produced = 2 kmol/h × 0.8 = 1.6 kmol/h
Number of moles of O₂ in the product stream = 30% × 1 kmol/h + 20% × 1 kmol/h = 0.5 kmol/h
The composition of the production stream is therefore;
Learn more about theoretical and actual yield here:
brainly.com/question/14668990
brainly.com/question/82989