Hydrogen sulfide has the distinctive unpleasant odor associated with rotten eggs, and it is poisonous. It often must be removed from crude natural gas and is therefore a product of refining natural gas. In such instances, the Claus process provides a means of converting H2S to elemental sulfur. Consider a feed stream to a Claus process that consists of 10.0 mole% H2S and 90.0% CO2. One-third of the stream is sent to a furnace where the H2S is burned completely with a stoichiometric amount of air fed at 1 atm and 25°C. The combustion reaction is H 2 S + 3 2 O 2 → S O 2 + H 2 O The product gases from this reaction are then mixed with the remaining two-thirds of the feed stream and sent to a reactor in which the following reaction goes to completion: 2 H 2 S + S O 2 → 3 S + 2 H 2 O The gases leave the reactor at 10.0 m 3 / min , 320°C, and 205 kPa absolute. Assuming ideal-gas behavior, determine the feed rate of air in kmol/min. Provide a single balanced chemical equation reflecting the overall process stoichiometry. How much sulfur is produced in kg/min?

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21kplatter 21kplatter · Answer 1 · 2019-10-22T19:22:09+02:00

Answer:

Given the yield is 98%

The reaction is

8H2S(g)+4O2(g)→ S8(l)+8H2O(g)

As per balanced equation eight moles of H2S will react with four moles of O2 to give one mole of S8

Let us calculate the moles of each reactant from given masses

30.0 grams of h2s

Moles = Mass / Molar mass = 30 / 34 = 0.88 moles

They need 0.44 moles of O2 to react

As 50.0 grams of o2 are given so

Moles = Mass / Molar mass = 50 / 16 = 3.125 moles

Thus H2S is the limiting reagent

8 moles of H2S will give one mole of S8

So 0.88 moles of H2S will give = 1 X 0.88 / 8 moles of S8 = 0.11 moles

However the yield is 98% so the actual moles of S8 formed = 0.98 X 0.11 = 0.1078 moles of S8

Mass of S8 produced = Moles X molar mass = 0.1078 X 256

= 27.6 grams of S8

Explanation: