What is the percent yield?

Question

Quicklime (CaO) can be prepared by roasting limestone (CaCO3) according to the reaction

CaCO3(s) ∆−→CaO(s) + CO2(g).

When 3.4 × 103g of CaCO3 are heated, the actual yield of CaO is 1.03 × 103g. What is the percent yield?

Answer in units of %.

Answer 1

I'm going to have to make some assumptions with your question here: I assume by 3.4x103g you mean 3.4 x 10^3 g, and similar for the other value. Worse case scenario the problem can be set up with these assumptions, and if they are wrong, the correct values can be used to "fix" it.

The first thing we need here is the molar masses of CaCO3 and CaO. For simplicity, we will do some rounding. CaCO3 has a molar mass of roughly 100 g/mol, and CaO has a molar mass of 56 g/mol. Since our given values are only to 2 significant figures, it is likely any refinement to these values (based on the periodic table) will be lost to rounding at the end of the calculation.

The problem starts with 3.4x10^3g of reactant. With a molar mass of 100 g/mol, dividing the mass of reactant by the molar mass gives 34 mol of reactant. By stoichiometry, there should also be 34 moles of CaO produced; we can worry about the mass of this if we want, or we can concern ourselves with the number of moles produced! Either technique is fine. Since I already know i should get 34 moles of product, I will personally leave it at that, and find out how many moles of product I *actually* received.

Since the mass of the produce was determined to be 1.03 x 10^3 g, we can find this out. We know the molar mass to be roughly 56 g/mol; dividing 1.03 x 10^3 g by the molar mass, we arrive at ~18.4 moles of product. Since we should have received 34 moles, we lost a lot!

To determine the percent yield, we divide the actual yield (the 18.4 moles) by the theoretical yield (the 34 moles), multiply by 100, and call it a percentage. Therefore, our percent yield is:

18.4 moles / 34 moles * 100 = 54%

So the percent yield is 54%.