Some indoor air-purification systems work by converting a little of the oxygen i

Question

Some indoor air-purification systems work by converting a little of the oxygen in the air to ozone, which oxidizes mold and mildew spores and other biological air pollutants. The chemical equation for the ozone generation reaction is O_29g) Right arrow 2 O_3(g) It is claimed that one such system generates 4.00 g O_3 per hour from dry air passing through the purifier at a flow of 5.00 L/min. if one liter of indoor air contains 0.280 g of O_2. What mole fraction of O_2 is converted to O_3 by the air purifier? What is the percent yield to the ozone generation reaction?

Explanation / Answer

flow rate = 5.00 L/min x 60 min/hr = 300 L/hr

mass of O2 per / hr = 300 L x 0.280 g O2/L= 84 g O2/hr

moles O2: 84 g/ 32 g/mole O2 = 2.625 moles O2/hr

moles O3: 4.00 / 48= 0.083 moles O3/hr

mole fraction of O2 is converted to O3 = 0.083 / 2.625 = 0.032

mole fraction of O2 is converted to O3 = 0.032

from the given equation :

3 mol of O2 ----------------> 2 mole of O3

2.625 mol O2 -------------> ?? mol of O3

moles of O3 = 1.75

mass of O3 = moles x molar mass

= 1.75 x 48

= 84 g

% yield = actual x 100 / theoretical

= 4 x 100 / 84

= 4.76 %

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