4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g) Determine the mass (in g) of KOH formed if 34.6 g of KO2 reacts with 34.6 g of H2O.

Answer 1

KO₂ (potassium superoxide) is not a typical compound encountered in introductory chemistry. I'll assume it to be correct and proceed.

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

WORK IN MOLES FIRST, THEN CONVERT TO GRAMS

moles of KO₂ present = 34.6 g KO₂ x 1 mole/71.10 g = 0.4866 moles

moles of H2O present = 34.6 g H₂O x 1 mole/18 g = 1.922 moles

Which reactant is limiting? Easiest way is to divide each by it's coefficient and see which is smaller.

0.4866/4 = 0.122 and 1.922/2 = 0.961 so KO₂ IS LIMITING and determines KOH formation

moles of KOH theoretically formed = 0.4866 moles KO₂ x 4 moles KOH/4 moles KO₂ = 0.4866 moles KOH

mass of KOH = 0.4866 moles KOH x 56.1 g/mole = 27.29 g = 27.3 g (to 3 sig. figs.)