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?

Question

Chemistry

Laney Gardner

10 March, 16:44

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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Answers (1)

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Gunner · Answer 1

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%