If the mask contains 0.150 mol KO2 and 0.100 mol water, how many moles of oxygen can be produced? KO2(s) + H2O(ℓ)→ KOH(s) + O2(g)

Question

Potassium superoxide, KO2, is used in re-breathing masks to generate oxygen according to the reaction below. If the mask contains 0.150 mol KO2 and 0.100 mol water, how many moles of oxygen can be produced? KO2(s) + H2O(ℓ)→ KOH(s) + O2(g)

Answer 1

KO₂(s) + H₂O(ℓ)→ KOH(s) + O₂(g) ... UNBALANCED equation. Must first balance the equation.

4KO₂(s) + 2H₂O(ℓ)→ 4KOH(s) + 3O₂(g) ... BALANCED EQUATION

Since we are given the moles of BOTH reactants, we must find which reactant is limiting. One easy way to do this is to divide the moles of each reactant by the corresponding coefficient in the balanced equation. Whichever value comes out less represents the limiting reactant.

For KO₂: 0.150 mols / 4 = 0.0375

For H₂O: 0.100 mol / 2 = 0.0500

Since 0.0375 is less than 0.0500, KO₂ is the limiting reactant.

Once we know the limiting reactant, we use the moles of that reactant and the stoichiometry of the balanced equation to find moles or product formed:

0.150 mols KO₂ x 3 mol O₂ / 4 mol KO₂ = 0.113 moles O₂ formed (3 sig. figs.)