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17 April, 02:53

Magnesium carbonate decomposes on heating to form magnesium oxide and carbon dioxide. Suppose that 50.0 g of magnesium carbonate was heated, and that 20.0 g of magnesium oxide was subsequently collected. What was the percent yield of this reaction?

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  1. 17 April, 03:11
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    The percentage yield is 80%

    Explanation:

    First let us generate a balanced equation for the reaction. This is illustrated below:

    MgCO3 - > MgO + CO2

    Now, we proceed with our calculations as follows:

    Molar Mass of MgCO3 = 24 + 12 + (16x3) = 24 + 12 + 48 = 84g/mol

    Molar Mass of MgO = 24 + 16 = 40g/mol

    From the equation,

    84g of MgCO3 produced 40g of MgO.

    Therefore, 50g of MgCO3 will produce = (50 x 40) / 84 = 25g of MgO.

    Now we can obtain the %yield as follows:

    Experimental yield = 20g

    Theoretical yield = 25g

    %yield = Experimental yield / Theoretical yield x100

    %yield = 20/25 x 100

    %yield = 80%
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