Mass of KOH....

This is a limiting reactant problem. Compute the amount of product with each reactant. The lesser amount is the theoretical yield, and tells the limiting reactant.

4KO2(s) + 2H2O(l) --> 4KOH(s) + 3O2(g)

27.1g ........ 22.1g ............?g

27.1g KO2 x (1 mol KO2 / 71.1g KO2) x (1 mol KOH / 1 mol KO2) x (56.1g KOH) / 1 mol KOH) = 21.4g KOH

22.1g H2O x (1 mol H2O / 18.0g H2O) x (2 mol KOH / 1 mol H2O) x (56.1g KOH / 1 mol KOH) = 138g KOH

Therefore, 21.4g KOH is produced and KO2 is the limiting reactant.