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31 December, 08:38

Consider the chemical equation for the production of water: 2 H2+O2>2 H2O. If 100 grams of oxygen gas are used, what would the percent yield be if 75 g of H2O was produced?

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  1. 31 December, 08:53
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    66.7%

    Explanation:

    The balanced equation for the reaction is given below:

    2H2 + O2 - > 2H2O

    Next, we shall determine the mass of O2 that reacted and the mass of H2O produced from the balanced equation. This is illustrated below:

    Molar mass of O2 = 16x2 = 32g/mol

    Mass of O2 from the balanced equation = 1 x 32 = 32g

    Molar mass of H2O = (2x1) + 16 = 18g/mol

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

    Summary:

    From the balanced equation above,

    32g of O2 reacted to produce 36g of H2O.

    Next, we shall determine the theoretical yield of H2O. This can be obtained as follow:

    From the balanced equation above,

    32g of O2 reacted to produce 36g of H2O.

    Therefore, 100g of O2 will react to produce = (100 x 36) / 32 = 112.5g of H2O.

    Therefore, the theoretical yield of H2O is 112.5g

    Finally, we shall determine the percentage yield of H2O as follow:

    Actual yield = 75g

    Theoretical yield = 112.5g

    Percentage yield = ?

    Percentage yield = Actual yield / Theoretical yield x 100

    Percentage yield = 75/112.5 x 100

    Percentage yield = 66.7%

    Therefore, the percentage yield of H2O is 66.7%
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