A 250\text{ cm}^3 sample of dry air is passed over excess heated phosphorus in a closed tu… — JAMB UTME

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

Answer: 197.5 cm^3

Answer

The volume of the remaining gas mixture is

197.5\text{ cm}^3

Because oxygen makes up

21\%

by volume of dry air, a

250\text{ cm}^3

sample contains

0.21 \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\text{ cm}^3 - 52.5\text{ cm}^3 = 197.5\text{ cm}^3

of unreacted gases.

Step-by-Step Solution

Calculate the volume of oxygen gas in the initial sample

52.5\text{ cm}^3

Oxygen makes up

21\%

by volume of dry air, so

0.21 \times 250\text{ cm}^3 = 52.5\text{ cm}^3

Determine the remaining gas volume after complete removal of oxygen

197.5\text{ cm}^3

Phosphorus reacts completely with oxygen, leaving the unreacted components of air:

250\text{ cm}^3 - 52.5\text{ cm}^3 = 197.5\text{ cm}^3

Key Concept

Percentage composition of air by volume

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic