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 tor the ozone generation reaction is 3 O_2 (g) rightarrow 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 of the ozone generation reaction?

Explanation / Answer

1.

5.00 L/min x 60 min/hr= 300 L/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
0.083/2.625=0.032

2.

Percent yield

1 litre contain 0.280 gm Oxygen

Flow rate is 5 lit/min, for 60 min

60 x 5 = 300 lit, it contain 300 x 0.280 = 84 gm of oxygen.

So, 84 gm oxygen generates 4.0 g O3.

Theoretically 3O2 gives 2O3 i.e. 48 gm should give 48 gm and that is 100%.

We are getting 4.0 g

4 x 100/48 = 8.33 %

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