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22 December, 05:19

Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide: CaO (s) + H2O (l) → Ca (OH) 2 (s) In a particular experiment, a 5.50-g sample of CaO is reacted with excess water and 6.77 g of Ca (OH) 2 is recovered. What is the percent yield in this experiment?

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  1. 22 December, 06:55
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    The % yield is 93.25 %

    Explanation:

    Step 1: Data given

    Mass of CaO = 5.50 grams

    Water is in excess

    Mass of Ca (OH) 2 produced = 6.77 grams

    Molar mass Ca (OH) 2 = 74.09 g/mol

    Step 2: The balanced equation

    CaO (s) + H2O (l) → Ca (OH) 2 (s)

    Step 3: Calculate moles CaO

    Moles CaO = mass CaO / molar mass CaO

    Moles CaO = 5.50 grams / 56.08 g/mol

    Moles CaO = 0.098 moles

    Step 4: Calculate moles Ca (OH) 2

    For 1 mol CaO we need 1 mol H2O to produce 1 mol Ca (OH) 2

    For 0.098 moles CaO we'll have 0.098 moles Ca (OH) 2

    Step 5: Calculate mass Ca (OH) 2

    Mass Ca (OH) 2 = moles * molar mass

    Mass Ca (OH) 2 = 0.098 moles * 74.09 g/mol

    Mass Ca (OH) 2 = 7.26 grams

    Step 6: Calculate % yield

    % yield = (actual yield / theoretical yield) * 100 %

    % yield = (6.77 grams / 7.26 grams) * 100%

    % yield = 93.25 %

    The % yield is 93.25 %
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