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 percent yield of oxygen is 87.84%.

Step-by-step explanation:

To calculate the percent yield of oxygen, we need to first calculate the theoretical yield of oxygen, which is the amount of oxygen that would be expected to be produced if the reaction went to completion. From the given information, we know that 3.00 g of mercury (II) oxide decomposes to form mercury and oxygen. Since the molar mass of HgO is 216.59 g/mol, the number of moles of HgO can be calculated by dividing the mass by the molar mass:

Number of moles of HgO = 3.00 g / 216.59 g/mol = 0.01385 mol of HgO

According to the balanced chemical equation, 2 moles of HgO produce 1 mole of O2. Therefore, the number of moles of O2 produced can be calculated by multiplying the number of moles of HgO by the stoichiometric coefficient:

Number of moles of O2 = 0.01385 mol HgO * (1 mol O2 / 2 mol HgO) = 0.00693 mol O2

Since the molar mass of O2 is 32.00 g/mol, we can calculate the theoretical yield of oxygen in grams:

Theoretical yield of O2 = 0.00693 mol O2 * 32.00 g/mol = 0.222 g O2

Finally, we can calculate the percent yield of oxygen by dividing the actual yield of oxygen (0.195 g) by the theoretical yield of oxygen (0.222 g) and multiplying by 100:

Percent yield of oxygen = (0.195 g O2 / 0.222 g O2) * 100 = 87.84%

Answer 2

The correct answer is 87.8%