The percentage modulation of AM changes from 50% to 70%. Originally at 50% modulation, carrier power was 70 W. Now, determine th
1 answer:
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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
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
Which gives;
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;
- <u>HCl: 0.4 kmol/h</u>
- <u>Cl₂: 1.6 kmol/h</u>
- <u>H₂O: 1.6 kmol/h</u>
- <u>O₂: 0.5 kmol/h</u>
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