During an investigation, a student burns magnesium to form magnesium oxide. The starting mass of magnesium is measured as 21.3 g.

2Mg (s) + 02 (8) - 2MgO (s)

If the student recovers 30.2 g of magnesium oxide, what is the percent yield? (atomic mass of magnesium = 24.3 amu)

Percentage yield = 85.2%

Explanation:

Given dа ta:

Mass of Mg = 21.3 g

Actual yield of MgO = 30.2 g

Percentage yield = ?

Solution:

Chemical equation:

2Mg + O₂ → 2MgO

Number of moles of Mg = mass/molar mass

Number of moles of Mg = 21.3 g / 24.3 g/mol

Number of moles of Mg = 0.88 mol

Now we will compare the moles of MgO with Mg.

Mg : MgO

2 : 2

0.88 : 0.88

Mass of MgO:

Mass of MgO = moles * molar mass

Mass of MgO = 0.88 mol * 40.3g/mol

Mass of MgO = 35.46 g

Actual yield of MgO = 30.2 g

Percentage yield:

Percentage yield = Actual yield/theoretical yield * 100

Percentage yield = 30.2 g / 35.46 g * 100

Percentage yield = 85.2%