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?

.

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.