Question

A 0.266 g piece of solid magnesium reacts with gaseous oxygen from the atmosphere to form solid magnesium oxide. In the laboratory a student weighs the mass of the magnesium oxide collected from this reaction as 0.244 g. 

 What is the percent yield of this reaction?

Answer 1

1) Balanced Chemical Equation

2Mg(s) +O₂ (g) -----> 2MgO (s)

Lab data:

Magnesium piece = 0.266 grams

Magnesium Oxide= 0.244 grams

2) Using the initial mass of magnesium, calculate the moles of magnesium-

Molar Mass of Mg = 24.31g/mol

Moles of Mg = 0.266 grams/24.31g/mol = 0.01094 mol

Mg reacts with atmospheric O₂, so O₂ is not a limiting reagent.

3) calculate the moles of magnesium oxide produced-

2 moles of Mg produces 2 moles of MgO

0.01094 moles of Mg produce 0.01094 moles of MgO

4) Calculate the mass of magnesium oxide.

Molar Mass of MgO = 40.31g/mol

Mass of MgO = 40.31g/mol x 0.01094 mol = 0.441g = Theoretical yield

5) %Yield = (Actual yield/ Theoretical yield) x100

             = (0.244 g/0.441g) x 100 = 55.32

Percent Yield = 55.32%