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Zorluk: KolayComposition and Pollution of Air

A 250 cm3250\text{ cm}^3 sample of dry air is passed over excess heated phosphorus in a closed tube to remove all the oxygen gas present. Assuming oxygen constitutes 21%21\% by volume of dry air, what is the volume of the remaining gas mixture in cm3\text{cm}^3?

Cevap: 197.5 cm^3

Cevap

The volume of the remaining gas mixture is 197.5 cm3197.5\text{ cm}^3.
Because oxygen makes up 21%21\% by volume of dry air, a 250 cm3250\text{ cm}^3 sample contains 0.21×250 cm3=52.5 cm30.21 \times 250\text{ cm}^3 = 52.5\text{ cm}^3 of oxygen. Heated phosphorus reacts with all the oxygen to form solid phosphorus oxide, leaving behind 250 cm3−52.5 cm3=197.5 cm3250\text{ cm}^3 - 52.5\text{ cm}^3 = 197.5\text{ cm}^3 of unreacted gases.

Adım Adım Çözüm

1
Calculate the volume of oxygen gas in the initial sample
52.5 cm352.5\text{ cm}^3
Oxygen makes up 21%21\% by volume of dry air, so 0.21×250 cm3=52.5 cm30.21 \times 250\text{ cm}^3 = 52.5\text{ cm}^3.
2
Determine the remaining gas volume after complete removal of oxygen
197.5 cm3197.5\text{ cm}^3
Phosphorus reacts completely with oxygen, leaving the unreacted components of air: 250 cm3−52.5 cm3=197.5 cm3250\text{ cm}^3 - 52.5\text{ cm}^3 = 197.5\text{ cm}^3.

Anahtar Kavram

Percentage composition of air by volume
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Konular

Composition and Pollution of Air

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