If a gas mask contained 35.55 g of KO2 and 5.4 g of water, what amount of oxygen (in grams) could be produced?

Question

Potassium superoxide KO₂ is used in rebreathing gas masks to produce oxygen.

4KO₂ + 2H₂O à 4KOH + 3O₂

If a gas mask contained 35.55 g of KO₂ and 5.4 g of water, what amount of oxygen (in grams) could be produced?

Answer 1

4KO₂ + 2H₂O ==> 4KOH + 3O₂ .. balanced equation

Whenever given the amounts of both reactants, as in this case, we must first determine which reactant is present in limiting supply. An easy way to do this is to simply divided the moles of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less then represents the limiting reactant:

For KO₂: 35.55 g KO₂ x 1 mol KO₂ / 71.097 g = 0.5000 moles (÷4->0.125)

For H₂O: 5.4 g H₂O x 1 mol H₂O / 18.02 g = 0.2997 moles (÷2->0.15)

Since 0.125 is less than 0.15, KO₂ is the limiting reactant. We will now use moles of KO₂ (0.500) and the stoichiometry of the balanced equation to determine amount of oxygen (O₂) produced.

0.500 mols KO₂ x 3 mols O₂ / 4 mols KO₂ x 32 g O₂ / mol = 12.0 g O₂