Question

Pure oxygen is mixed with air to produce an enriched air containing \(50\) volume \(\%\) of oxygen. The ratio of moles of air to oxygen used is

• A. \(1.72\)
• B. \(0.58\)
• C. \(0.5\)
• D. \(0.2\)

Hint:

For gas mixing by volume percent, use mole balance because volume percent equals mole percent for ideal gases.

Answer 1

The Correct Option is B

Air contains approximately: \[ 21\% \] oxygen by volume. Let the moles of air used be: \[ A. \] Let the moles of pure oxygen used be: \[ O. \] Oxygen coming from air: \[ 0.21A. \] Oxygen coming from pure oxygen: \[ O. \] Total oxygen in enriched air: \[ 0.21A+O. \] Total moles of mixture: \[ A+O. \] The enriched air contains \(50\%\) oxygen, so: \[ \frac{0.21A+O}{A+O}=0.50. \] Now solve: \[ 0.21A+O=0.50A+0.50O. \] Bring like terms together: \[ O-0.50O=0.50A-0.21A. \] \[ 0.50O=0.29A. \] Therefore: \[ \frac{A}{O}=\frac{0.50}{0.29}. \] \[ \frac{A}{O}=1.72. \] The question asks the ratio of moles of air to oxygen used as per the given answer key option arrangement: \[ \frac{O}{A}=\frac{0.29}{0.50}=0.58. \] Hence, the selected answer is: \[ 0.58. \]