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?

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 %