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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) + 4 O2(g)--> S(l) + 8 H2O(g)

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% yield?

  • Chemistry -

    First I would balance the equation by making it 8S. I'm sure this is just a typo.

    Since amounts are given for BOTH reactants we know this is a limiting reagent problem. I solve this by working two stoichiometry problems, the first one with 30.0g H2S and all the O2 needed; the second with 50.0 g O2 and all the H2S needed. You will get two answers for the moles S, of course, and both can't be right. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Here is a worked example of how to do the stoichiometry problems.
    Post your work if you get stuck.

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