Enriching air with pure oxygen: What is the required mole ratio of air to added O2 to obtain an oxygen-enriched stream containing 50% O2 by volume (assume dry air contains 21% O2)?

Difficulty: Medium

Correct Answer: 1.72

Explanation:


Introduction / Context:
Air enrichment by blending pure oxygen with air is used in combustion intensification and medical or environmental applications. Determining the correct blend ratio is a straightforward material-balance problem using mole fractions (equal to volume fractions for gases at common T and P).


Given Data / Assumptions:

  • Dry air contains 21% O2 by volume (mole basis).
  • Target mixture contains 50% O2 by volume.
  • Pure O2 is blended with air; inert nitrogen comes only from air.


Concept / Approach:
Let x = moles of air, y = moles of pure O2 added. Oxygen in the blend = 0.21x + y; total moles = x + y. The design requirement is (0.21x + y)/(x + y) = 0.50. Solve for y/x to get the mixing ratio, then invert to obtain air:oxygen.


Step-by-Step Solution:

Set (0.21x + y)/(x + y) = 0.50.Cross-multiply: 0.21x + y = 0.50x + 0.50y.Group y terms: y − 0.50y = 0.50x − 0.21x → 0.50y = 0.29x.Solve: y/x = 0.58 → air:O2 = x:y = 1 : 0.58 = 1.724 ≈ 1.72.


Verification / Alternative check:
Assume x = 1 mol air, y = 0.58 mol O2. O2 fraction = (0.21 + 0.58)/(1 + 0.58) = 0.79/1.58 ≈ 0.50, confirming the calculation.


Why Other Options Are Wrong:

  • 0.58 is the inverse ratio (O2 per air), not air per O2.
  • 0.50 and 0.20 are unrelated to the derived ratio.
  • 2.50 gives an O2 fraction far from 50%.


Common Pitfalls:
Using mass instead of mole/volume fractions for gas blending; forgetting that the inerts (chiefly N2) come only from air and dilute the final O2 fraction.


Final Answer:
1.72

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