4 KO2 + 2CO2 rightwards arrow 2 K2CO3 + 3 O2

Question

How many moles of O2 can be produced from 12 moles of KO2 and 10 moles of CO2? If you were to produce 100.0 grams of O2 what would be the percent yield of the reaction?

Answer 1

This is a limiting reagent (LR) problem. You know that because amounts are given for BOTH reactants.

Using the coefficients in the balancaed equation, convert 12 mols KO2 to mols O2.
Do the same to convert mols CO2 to mols O2.
It is likely the two values will not be the same which means one of them is not right; the correct value in LR problems is ALWAYS the smaller value and the reagent responsible for the smaller number is the LR.

Using the smaller number, convert to grams O2. grams O2 = mols O2 x molar mass O2. This is the theoretical yield for O2.

Then percent yield = (g O2 produced/theoretical yield)*100 = ?

Answer 2

First find limiting reactant (KO2). The answer for the first question is 9 mols of oxygen.

For the second question 100/(9*16) * 100% = 69.4%