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 data:

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%

the answer to this question is 3.