Question at position 9 P4O6 (molar mass = 219.9 g/mol) is formed by the reaction of P4 with O2.P4(s) + 3O2(g) → P4O6(s) If a mixture of 75.3 g of P4 (molar mass = 124 g/mol) and excess O2 produce 43.3 g of P4O6, (molar mass = 219.88. g/mol) what is the percent yield for the reaction? 38.0% 57.5% 48.8% 16.3 %32.4%单项选择题
A
38.0%
B
57.5%
C
48.8%
D
16.3 %
E
32.4%

Question

Question at position 9 P4O6 (molar mass = 219.9 g/mol) is formed by the reaction of P4 with O2.P4(s) + 3O2(g) → P4O6(s) If a mixture of 75.3 g of P4 (molar mass = 124 g/mol) and excess O2 produce 43.3 g of P4O6, (molar mass = 219.88. g/mol) what is the percent yield for the reaction? 38.0% 57.5% 48.8% 16.3 %32.4%单项选择题

A

38.0%

B

57.5%

C

48.8%

D

16.3 %

E

32.4%

BlackTom AI · Accepted Answer

First, restate the given data and the reaction: P4(s) + 3 O2(g) → P4O6(s). You’re starting with 75.3 g of P4 (molar mass 124.0 g/mol) and excess O2, and the actual isolated product is 43.3 g of P4O6 (molar mass 219.88 g/mol).

Option A: 38.0%.
- The theoretical yield is determined from the limiting reactant, which here is P4 since O2 is in excess. Convert the given P4 mass to moles: 75.3 g ÷ 124.0 g/mol ≈ 0.607 mol P4. The reaction has a 1:1 mole ratio between P4 and P4O6, so the maximum moles of P4O6 produced would be ≈ 0.607 mol. Convert to grams: 0.607 mol × 219.88 g/mol ≈ 133.4 g theoretical yield. Percent yield would be (43.3 g actual / 133.4 g theoretical) × 100 ≈ 32.5%. The option 38.0% is higher than this rough calculation, which would require either a higher actual yield or a smaller theoretical yield, both unsupported by the data.

Option B: 57.5%.
- Using the same theoretical yield of ≈133.4 g, the implied actual yield would be 0.575 × 133.4 g ≈ 76.7 g. Since the observed actual yield is only 43.3 g, this option would overestimate the result given the stoichiometry and masses provided.

Option C: 48.8%.
- This would correspond to an actual yield near 0.488 × 133.4 g ≈ 65.1 g. Again, the experimental yield (43.3 g) is notably less than this value, indicating this percentage is not consistent with the given masses.

Option D: 16.3%.
- A percent yield as low as 16.3% would imply an actual product mass around 0.163 × 133.4 g ≈ 21.7 g, which is far smaller than the reported 43.3 g. This would require a substantial loss or side reaction not described in the data.

Option E: 32.4%.
- Theoretical yield is still about 133.4 g as established from the starting P4 and the 1:1 stoichiometry. Calculating percent yield as (43.3 g / 133.4 g) × 100 ≈ 32.5% gives a result that closely matches 32.4% when rounded, aligning well with the given masses and reaction stoichiometry.

Across all options, the core approach is to determine the theoretical yield from the limiting reactant, then compare the actual yield to that theoretical value. The key potential pitfall would be miscomputing the molar masses or misapplying the 1:1 ratio between P4 and P4O6, which would skew the percent yield significantly above or below the calculated value. The remaining distractors fail to align with the mass-based stoichiometric calculation derived from the provided P4 mass and product mass.

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