Flue-gas averaging: A flue gas contains 0.15 kmol CO2, 0.05 kmol O2, and 0.80 kmol N2 per kmol of mixture. What is the average molecular weight of the gas mixture?

Chemical Engineering Stoichiometry Difficulty: Medium
Choose an option
  • A
    28.6
  • B
    30.0
  • C
    30.6
  • D
    32.6

Answer

Correct Answer: 30.6

Explanation

Introduction / Context:Determining a gas mixture’s average molecular weight is a routine step in combustion calculations, gas density estimates, and compressor sizing. The average is a mole-fraction-weighted sum of component molecular weights.

Given Data / Assumptions:

  • Mole fractions: yCO2 = 0.15, yO2 = 0.05, yN2 = 0.80 (sum = 1.00).
  • Molecular weights: MCO2 = 44, MO2 = 32, MN2 = 28 kg/kmol.
  • Ideal mixing assumed.

Concept / Approach:The average molecular weight Mmix is given by Mmix = Σ yi * Mi. This uses mole fractions as weights because each kmol of mixture contains yi kmol of species i.

Step-by-Step Solution:

Step 1: Compute contributions: 0.15 * 44 = 6.60.Step 2: Add oxygen term: 0.05 * 32 = 1.60.Step 3: Add nitrogen term: 0.80 * 28 = 22.40.Step 4: Sum = 6.60 + 1.60 + 22.40 = 30.60 kg/kmol.

Verification / Alternative check:As a reasonableness check, the mixture is mostly N2, so the result should be close to 28 but pulled upward by CO2 (44) and slightly by O2 (32). A value around 30 is sensible; 30.6 matches the calculation.

Why Other Options Are Wrong:

  • 28.6: Too low; underweights CO2 contribution.
  • 30.0: Rounded down; not the precise weighted sum.
  • 32.6: Too high; would require more CO2/O2 than present.

Common Pitfalls:Using mass fractions by mistake; the correct weighting for Mmix is mole fraction. Also, forgetting to ensure that fractions sum to unity before applying the formula.

Final Answer:30.6

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