A 14.570 g sample of CaCl2 was added to 12.081 g of K2CO3 and mixed in water. A 3.566 g yield of CaCO3 was obtained.

Question

A  14.570 g sample of  was added to 12.081 of  and mixed in water. A 3.566 g yield of  was obtained. Calculate the percent yield of 

Answer 1

This question involves Stoichiometry, limiting reagents, dimensional analysis, and a whole of keeping track of significant figures. First we must set up the chemical equation for the reaction that is occurring.

We are told there is a reaction between the two reagents in (effectively) an aqueous solution that produces CaCO₃. Using Stoichiometry we find the balanced chemical equation to be

CaCl_2(aq) + K₂CO_3(aq) -> CaCO_3(s) + 2KCl_(aq).

We know that the CaCO₃ is solid from solubility rules/tables. Since there is information about the masses of the two reagents and the mass of the product we should be tipped off that this is a limiting reagent problem. We must now use dimensional analysis to determine the theoretical yield for the given amount of each reagent. We will also need information about the masses of the compounds involved.

From the periodic table we find:

mass_CaCl2 = (40.078) + 2 * (35.453) = 110.984 g/mol CaCl₂

mass_K2CO3 = 2 * (39.098) + (12.011) + 3 * (15.999) = 138.204 g/mol K₂CO₃

mass_CaCO3 = (40.078) + (12.011) + 3 * (15.999) = 100.086 g/mol CaCO₃

Now we can use these values in dimensional analysis to arrive at our two theoretical yield results, namely:

(14.570 g CaCl₂) * (1 mol CaCl₂) / (110.984 g CaCl₂) * (1 mol CaCO₃) / (1 mol CaCl₂) * (100.086 g CaCO₃) / (1 mol CaCO₃) = 13.139 g CaCO₃

(12.081 g K₂CO₃) * (1 mol K₂CO₃) / (138.204 g K₂CO₃) * (1 mol CaCO₃) / (1 mol K₂CO₃) * (100.086 g CaCO₃) / (1 mol CaCO₃) = 8.7489 g CaCO₃.

Since 8.7489 g CaCO₃ < 13.139 g CaCO₃, the K₂CO₃ must be the limiting reagent, and the appropriate value to use for the theoretical yield is 8.7489 g CaCO₃. Now we must use the formula for the percent yield to find:

(3.566 g CaCO₃ actual yield) / (8.7489 g CaCO₃ theoretical yield) x 100% = 40.76%

Although it's not a great percent yield, the value seems reasonable. It is certainly between 0-100%.