Question

Determine the theoretical yield and the percent yield if 21.8 g of K2CO3 is produced from reacting 27.9 g KO2 with 29.0 L of CO2 (at STP).   The molar mass of KO2 = 71.10 g/mol and K2CO3 = 138.21 g/mol.

Answer 1

Answers: (Theoretical yield = 27.1 g) ; (Percent yield = 80.4%)

1) We start by writing the balanced chemical equation, making sure there is the same amount of atoms on both sides of the arrow.

2) We calculate the amount of reagents in moles because the balanced equation is written per mole, so it will be easier to perform the next calculations. For KO₂ we use the given molar mass. For CO₂ we use the Standard Molar Volume (the volume occupied by one mole of any gas at STP), or alternatively we can use the ideal gas law.

3) We find out who's the limiting reagent. We calculate the hypothetical amount of K₂CO₃ that could be produced per each reagent separately. We choose the smaller value as the limiting reagent. This means that in this case we have more CO₂ than we need, some of it will be left over after the reaction is completed. The numbers in blue are taken from the balanced equation.

4) We find the theoretical yield by simply turning moles of K2CO3 into grams using the given molar mass.

5) We find the percent yield defined as the current or actual yield over the theoretical yield and expressed in % (times 100).