In the reaction below, 22 g of H2S with excess O2 produced 5 g of sulfur. ? H2S + ? O2 → ? S + ? H2O . What is the percent

Question

In the reaction below, 22 g of H2S with excess
O2 produced 5 g of sulfur.
? H2S + ? O2 → ? S + ? H2O .
What is the percent yield of sulfur?
Answer in units of %.

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Nho 5 years 2021-08-30T00:40:16+00:00 1 Answers 23 views 0

Answers ( )

  1. giabao
    0
    2021-08-30T00:41:47+00:00
    Reply

    Answer:

    24.1 %

    Explanation:

    This is the reaction of oxygen between hydrogen sulfide.

    The equation is:

    2H₂S  +  O₂ →  2S  +  2H₂O

    As the oxygen is the excess, limting reagent is the H₂S.

    We convert the mass to moleS: 22g / 34.06 g/mol =

    0.646 moles

    Ratio is 2:2. 2 moles of sulfide can produce 2 moles of sulfur.

    Then, 0.646 moles of sulfide will produce 0.646 moles of S.

    We convert the moles to mass: 0.646 mol . 32.06 g/mol =

    20.71 g

    That’s the theoretical yield.

    % yield = (produced yield / theoretical yield) . 100

    % yield = (5 g/ 20.71g) . 100 =  24.1 %

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