In a laboratory experiment, when 35.0 g Mg reacted in excess O2, the percent yield of MgO was 90.0%. What was the actual yield of that experiment?

Actual yield = 54 g

Explanation:

Given dа ta:

Mass of magnesium = 35.0 g

Percent yield = 90.0%

Actual yield = ?

Solution:

Chemical equation:

2Mg + O₂ → 2MgO

Number of moles of Mg:

Number of moles = mass / molar mass

Number of moles = 35.0 g / 24 g/mol

Number of moles = 1.5 mol

Now we will compare the moles of Mg and O.

Mg : MgO

2 : 2

1.5 : 1.5 mol

Mass of MgO: (theoretical yield)

Mass = number of moles * molar mass

Mass = 1.5 mol * 40 g/mol

Mass = 60 g

Percent yield = (actual yield / theoretical yield) * 100

90.0 % = (actual yield / 60 g) * 100

0.9 = actual yield / 60 g

Actual yield = 0.9 * 60 g

Actual yield = 54 g