Question

When limestone (which is principally CaCO3) is heated, carbon dioxide and quicklime (CaO) are produced by the reaction CaCO3(s) →∆ CaO(s) + CO2(g) . If 16.3 g of CO2 was produced from the ther- mal decomposition of 41.48 g of CaCO3, what is the percentage yield of the reaction?

Answer 1

The balanced chemical equation for the thermal decomposition of calcium carbonate is:

CaCO3(s) → CaO(s) + CO2(g)

According to the equation, for every one mole of calcium carbonate decomposed, one mole of carbon dioxide is produced. We can use this information to calculate the theoretical yield of carbon dioxide:

Molar mass of CaCO3 = 100.09 g/mol
Molar mass of CO2 = 44.01 g/mol

Number of moles of CaCO3 = 41.48 g / 100.09 g/mol = 0.4144 mol
Theoretical yield of CO2 = 0.4144 mol × 1 mol CO2/1 mol CaCO3 × 44.01 g/mol = 18.23 g

Therefore, the theoretical yield of CO2 is 18.23 g.

The percentage yield can be calculated using the following formula:

Percentage yield = (actual yield / theoretical yield) × 100%

In this case, the actual yield of CO2 is given as 16.3 g. Substituting the values into the formula, we get:

Percentage yield = (16.3 g / 18.23 g) × 100%
Percentage yield = 89.5%

Therefore, the percentage yield of the reaction is 89.5%.