7) 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: 8 H2S(g) + 4O2(g)  S8(l) + 8 H2O Under optimal conditions the Claus process gives 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 98.0%  yield?

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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7) 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:
8 H2S(g) + 4O2(g)  S8(l) + 8 H2O
Under optimal conditions the Claus process gives 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 98.0% 
yield?

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Why do you multipy the # of H2S mols by 4, when 4 corresponds to the reactory amount of O2?

The logic there is somewhat confusing to me.

I understand how/why you found out the mols of the equation but dividing the sample's MM by the equation confuses me.

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