A 14.549 g sample of CaCl2 was added to 12.344 g of K2CO3 and mixed in water. A 3.522 g yield of CaCO3 was obtained.

Question

What is the limiting reagent?

Calculate the percent yield of CaCO3

Answer 1

CaCl₂(aq) + K₂CO₃(aq) ==> CaCO₃(s) + 2KCl(aq) ... balanced equation

First, we'll find the limiting reactant:

moles CaCl₂ present = 14.549 g CaCl₂ x 1 mol/110.98 g = 0.1311 moles CaCl₂

moles K₂CO₃ present = 12.344 g K₂CO₃ x 1 mol/138.21 g = 0.08931 moles K₂CO₃

Since mol ratio of CaCl₂ : K₂CO₃ is 1:1, the K2CO3 is limiting

Percent yield will be dictated by the limiting reactant.

Theoretical yield of CaCO₃: 0.08931 mols K₂CO₃ x 1 mol CaCO_3/mol K₂CO₃ = 0.08931 mols CaCO₃

0.08931 mols CaCO₃ x 100.09 g CaCO₃/mol = 8.939 g CaCO₃ theoretical yield

Percent yield = actual yield / theoretical yield (x100%)

% yield = 3.522 g / 8.939 g (x100%) = 39.40%