Answer:
74.9%.
Explanation:
Relative atomic mass data from a modern periodic table:
- Ca: 40.078;
- C: 12.011;
- O: 15.999.
What's the <em>theoretical</em> 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 <em>percentage</em> yield of this reaction?
.
