Hydrogen sulfide is an impurity in natural gas that must be removed. one common removal method is called the claus process, which relies on the reaction 8h2s(g)+4o2(g)→s8(l)+8h2o(g)under optimal conditions the claus process give 98% yield of s8 from h2s. if you started with 30.0 grams of h2s and 50.0 grams of o2, how many grams of s8 would be produced, assuming 98% yield

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AsiaWoerner AsiaWoerner · Answer 1 · 2018-09-20T17:14:42+02:00

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