Question

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?

Answer 1

the balanced chemical equation for decomposition of HgO is as follows

2HgO --> 2Hg + O₂

stoichiometry of HgO to O₂ is 2:1

number of HgO moles heated are - 3.00 g / 216.59 g/mol = 0.0139 mol

according to stoichiometry of reaction -

number of O₂ moles formed = 0.0139 mol/ 2 = 0.00695 mol

mass of O₂ to be formed - 0.00695 mol x 32.00 g/mol = 0.2224 g

but the actual yield = 0.195 g

percent yield = actual yield / theoretical yield x 100 %

percent yield = 0.195 g / 0.2224 g x 100 % = 87.7 %

answer is 87.7 %

Answer 2

Answer:

The correct answer is 87.8%