Hydrogen sulfide is poisonous. it must be removed from crude oil. claus process allows for removing h2s from crude oil by converting it into elemental sulfur. a feed stream to the claus process consists of 10 mole % h2s and 90 mole % co2. one third of the stream is sent to a furnace where the h2s is burnt completely with stoichiometric amount of air fed at 1 atm and 25 oc. the combustion reaction is as follows: h2s+ 3 2 o2 so2+h2o. the produce gases from this reaction are them mixed with the remaining two-thirds of the feed stream and sent to a reactor in which the following reaction goes to completion: 2h2s+so23s+2h2o. the gases leave the reactor at 10.0 m3 /min, 320 oc, and 205 kpa absolute. assuming ideal gas behavior, determine the feed rate of air in kmol/min. how much sulfur is produced in kg/min?

Here is the link to the all information u need i mean answer key

Ok so co(g) is not pronoused that. the only clue is that youpronouse it the oppisote way. 

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.