Question

Imagine you conduct a decomposition experiment of magnesium carbonate, MgCO3. If 25.0g of MgCO3 is heated to form 9.7 g MgO, what is the percent yield? Show calculations to support your answer.

Answer 1

MgCO₃ decompose to form MgO and CO₂.

MgCO₃ MgO + CO₂

Molar mass of MgCO₃ = 84.3139 g/mol

Molar mass of MgO = 40.3044 g/mol

Theoretically, 84.3139 g MgCO₃ decompose to form 40.3044 g of MgO

Therefore, theoretically, 25 g MgCO₃ decompose to form (40.3044 x 25)/84.3139 = 11.95 g of MgO

The actual yield of MgO = 9.7 g

Percent yield = Actual yield x 100/Theoretical yield

                      = (9.7 x 100)/11.95

                      = 81.17 %

Therefore the percent yield of MgO = 81.17 %