Mole fraction from mixed masses:\nA gaseous mixture contains 14 kg of N2, 16 kg of O2, and 17 kg of NH3. What is the mole fraction of O2?

Introduction / Context:
Engineering problems frequently present mixture compositions by mass, yet gas-phase calculations require mole fractions. This item reinforces accurate conversion from mass to moles and correct normalization to obtain a mole fraction.

Given Data / Assumptions:

• Molecular weights: N2 = 28 kg/kmol, O2 = 32 kg/kmol, NH3 = 17 kg/kmol.
• Masses: 14 kg N2, 16 kg O2, 17 kg NH3.
• Ideal-gas mixture; temperature/pressure do not affect mole fraction.

Concept / Approach:
Convert each mass to kmoles, sum to get total kmoles, then divide O2 kmoles by total to get the mole fraction y_O2.

Step-by-Step Solution:
n_N2 = 14 / 28 = 0.5 kmol.n_O2 = 16 / 32 = 0.5 kmol.n_NH3 = 17 / 17 = 1.0 kmol.n_total = 0.5 + 0.5 + 1.0 = 2.0 kmol.y_O2 = n_O2 / n_total = 0.5 / 2.0 = 0.25.

Verification / Alternative check:
Compute other mole fractions: y_N2 = 0.25 and y_NH3 = 0.50; the sum 0.25 + 0.25 + 0.50 = 1.00 confirms correctness.

Why Other Options Are Wrong:
0.16 and 0.33 do not follow from the mass-to-mole conversions; 0.47 and 0.66 exceed the computed 0.25 and do not respect normalization with the given data.

Common Pitfalls:
Using mass fractions directly as mole fractions or misusing molecular weights; always convert masses to moles first.

Final Answer:
0.25