Chemistry Problem

Question

Some indoor air-purification systems work by converting a little of the oxygen in air to ozone, which kills mold and mildew spores and other biological air pollutants. The chemical equation (unbalanced) for ozone generation is O₂(g)→O₃(g).

One such system claims to generate 4.0 g of ozone per hour from dry air passing through the purifier at a of flow of 5.0 L / min. If one L of indoor air contains 0.28 grams O₂, what is the percent yield of this process?

Answer 1

3O2(g) ===> 2O3(g)  balanced equation

0.28 gO2/L x 1mol/32 g = 0.00875 moles O2/L

5.0 L/min x 60 min/hour x 0.00875 moles/L = 2.625 moles O2/hour processed

2.625 moles O2 x 2 moles O2/3 moles O2 = 1.75 moles O3 prepared/hr

1.75 moles O3 x 48 g/mole = 84 g O3 prepared/hour

% yield = 4.0g/84 g (x100) = 4.8% yield