A 10.0 g sample of calcium carbonate (CaCO₃) is heated strongly and decomposes according to the reaction: CaCO3 ( 𝑠 ) → CaO ( 𝑠 ) + CO2 ( 𝑔 ) , what is the theoretical yield of CO2 (in grams)?

Question

a) What is the theoretical yield of CO₂ (in grams)?

b) If only 3.21 g of CO₂ is collected, what is the percent yield?

c) Explain the role of limiting reactants in this type of decomposition reaction.

Answer 1

The reaction shown, CaCO₃(s) ==> CaO(s) + CO₂(g), is already balanced. So we can now use this equation to answer the questions.

(a) Theoretical yield of CO₂: Since there are 10.0 g of CaCO₃ to start with, and assuming 100% conversion to CO₂, we use the moles of CaCO₃ to find moles of CO₂ and then convert this value to g of CO₂. We will do this using the molar mass of CaCO₃ (100. g/mol) and the molar mass of CO₂ (44.0 g/mol) along with dimensional analysis:

10.0 g CaCO₃ x 1 mol CaCO₃ / 100. g x 1 mol CO₂ / 1 mol CaCO₃ x 44.0 g / mol CO2 = 4.40 g CO₂

Thus the theoretical yield of CO₂ is 4.40 g (to 3 sig.figs.)

(b) % yield if only 3.21 g CO2 is collected: The % yield is the actual yield / theoretical yield (x100%)

3.21 g / 4.40 g (x100%) = 73.0 % (3 sig.figs.)

(c) The limiting reactant in any reaction is the one that is present in the least stoichiometric quantity. That is, the one that will run out first. In a typical decomposition reaction, there is only one reactant, so it is the limiting reactant by default. That being the case it limits the amount of product that can be formed.

Answer 2