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4 July, 11:16

Consider the balanced equation for the following reaction:

3Ca (ClO3) 2 (aq) + 2Li3PO4 (aq) → Ca3 (PO4) 2 (s) + 6LiClO3 (aq)

Determine the theoretical yield of LiClO3 (aq) in grams if the percent yield of LiClO3 (aq) is 81.0% and 9.45 moles of LiClO3 (aq) forms.

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  1. 4 July, 12:53
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    1055.83g

    Explanation:

    Data obtained from the question include:

    Percentage yield of LiClO3 = 81.0%

    Actual yield of LiClO3 = 9.45 moles

    Now, let us convert 9.45 moles of LiClO3 to grams. This is illustrated below:

    Molar Mass of LiClO3 = 7 + 35.5 + (16x3) = 7 + 35.5 + 48 = 90.5g/mol

    Number of mole of LiClO3 = 9.45 moles

    Mass of LiClO3 = ?

    Mass = number of mole x molar Mass

    Mass of LiClO3 = 9.45 x 90.5

    Mass of LiClO3 = 855.225g

    Now we obtain the theoretical yield as follow:

    %yield = Actual yield/Theoretical yield x 100

    81% = 855.225 / Theoretical yield

    81/100 = 855.225 / Theoretical yield

    Cross multiply to express in linear form

    Theoretical yield x 81 = 100x855.225

    Divide both side by 81

    Theoretical yield = (100x855.225) / 81

    Theoretical yield = 1055.83g

    Therefore, the theoretical yield of LiClO3 is 1055.83g
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