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21 June, 05:28

Paige heated 3.00 g mercury (II) oxide (HgO, 216.59 g/mol) to form mercury (Hg, 200.59 g/mol) and oxygen (O2, 32.00 g/mol). She collected 0.195 g oxygen. What was the percent yield of oxygen?

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  1. 21 June, 06:36
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    We first write the chemical equation:

    2HgO → 2Hg + O₂

    Next, we calculate the moles of HgO present:

    moles = 3 / 216.59

    moles = 0.014

    Each mole of oxygen gas needs 2 moles of HgO to be produced.

    Theoretical moles of oxygen gas produced = 0.014 / 2

    = 0.007

    Theoretical mass of oxygen = 0.007 x 32 = 0.224 grams

    Percentage yield = actual yield / theoretical yield x 100

    = 0.195 / 0.224 x 100

    = 87.0%
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