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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Thu Hương 5 years 2021-08-30T00:43:37+00:00 1 Answers 19 views 0

Answers ( )

  1. Xavia
    0
    2021-08-30T00:44:42+00:00
    Reply

    Answer:

    24%

    Explanation:

    From the question, the limiting reactant is H2S.

    The reaction equation is;

    2H2S + O2 → 2S + 2H2O

    Number of moles of H2S reacted = 22g/34 g/mol = 0.647 moles

    According to the reaction equation;

    2 moles of H2S yields 2 moles of sulphur

    0.647 moles of H2S yields 0.647 moles of sulphur

    So;

    Theoretical yield of sulphur = 0.647 moles * 32 g/mol = 20.7 g

    Actual yield = 5 g

    % yield = actual yield/theoretical yield * 100

    % yield =5 g/20.7 g * 100

    % yield = 24%

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )