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6 November, 03:53

C2H4+

O2

CO2 + H2O

A. What is my theoretical yield of H2O if I start with 9.35 g of Oz?

B. If I actually produced 2.63 g of H20, what is my percent yield?

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Answers (1)
  1. 6 November, 07:48
    0
    A. Theoretical yield = 3.51g

    B. %yield = 75%

    Explanation:

    The balanced equation for the reaction is given below:

    C2H4 + 3O2 - > 2CO2 + 2H2O

    Molar Mass of O2 = 16 x 2 = 32g/mol

    Mass of O2 from the balanced equation = 3 x 32 = 96

    Molar Mass of H20 = (2x1) + 16 = 18g/mol

    Mass of H2O from the balanced equation = 2 x 18 = 36g

    A. From the equation,

    96g of O2 produced 36g of H2O

    Therefore, 9.35g of O2 will produce = (9.35 x 36) / 96 = 3.51g of H2O

    Therefore, theoretical yield of water (H2O) = 3.51g

    B. Theoretical yield = 3.51

    Actual yield = 2.63g

    %yield = ?

    %yield = Actual yield/Theoretical yield x 100

    %yield = 2.63/3.51

    %yield = 75%
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