In the Deacon process for the manufacture of chlorine, HCl and O2 react to form Cl2 and H2O. Sufficient air (21 mole%O2, 79%N2) is fed to provide 35% excess oxygen, and the fractional conversion of HCl is 85%. (a) Calculate the mole fractions of the product stream components, using atomic species balances in your calculation. (b) Again calculate the mole fractions of the product stream components, only this time use the extent of reaction in the calculation. (c) An alternative to using air as the oxygen source would be to feed pure oxygen to the reactor. Running with oxygen imposes a significant extra process cost relative to running with air, but also offers the potential for considerable savings. Speculate on what the cost and savings might be. What would determine which way the process should be run?

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The answer is nitrogen-containing bases

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.

Answerchemical;