A plant is removing a trace of Cl2 from a waste gas stream by passing it over a solid granular absorbed in a tubular packed bed.
At present, 63.2 % removal is being accomplished, but it is believed that greater removal could be achieved if the flow rate were increased by a factor of 4, the particle diameter were decreased by a factor of 3, and the packed tube length increased by 50%. What percentage of chlorine would be removed under the proposed scheme? ( The chlorine transferring to the absorbent is removed completely by a virtually instantaneous chemical reaction.)
1 answer:
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Answer:
3.375moles of O₂
Explanation:
The reaction expression is given as;
2KClO₃ → 2KCl + 3O₂
Number of moles of KClO₃ = 2.25moles
Now;
To find the number of moles of O₂ produced we use known number of moles.
So;
2 mole of KClO₃ will produce 3 moles of O₂
2.25moles of KClO₃ will produce = 3.375moles of O₂