Caco3(s) ∆→cao(s) + co2(g).
if 13.2 g of co2 was produced from the thermal decomposition of 40.1 g of caco3, what
is the percentage yield of the reaction?

74.9%.

Explanation:

Relative atomic mass data from a modern periodic table:

What's the theoretical yield of this reaction?

In other words, what's the mass of the CO₂ that should come out of heating 40.1 grams of CaCO₃?

Molar mass of CaCO₃:

.

Number of moles of CaCO₃ available:

.

Look at the chemical equation. The coefficient in front of both CaCO₃ and CO₂ is one. Decomposing every mole of CaCO₃ should produce one mole of CO₂.

.

Molar mass of CO₂:

.

Mass of the 0.400655 moles of expected for the 40.1 grams of CaCO₃:

.

What's the percentage yield of this reaction?

.