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%

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%

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.

查看解析

登录即可查看完整答案

类似问题

Calcium oxide reacts with excess carbon dioxide to produce calcium carbonate. Junior Chemist isolates 5.9 g of calcium carbonate at the end of her experiment. Earlier she had calculated (from stoichiometry) a yield of 12.8 g. What is her % yield?

2.10 Determine percentage yields for reactions

If the reaction in question Q16 produces 24.6 g of sulfur trioxide, what is the % yield of the reaction

Consider the following equation 2 Mg (s) + O2 (g) → 2 MgO (s) When 40.0 g of O2 completely react with Mg, the actual yield of MgO is 83.6 g. What is the percent yield? Given molar mass of MgO is 40.31 g/mol.

The formula to calculate the yield percentage is

The formula to calculate the yield percentage is

Sn3(PO4)4 + 6 Na2CO3 → 3 Sn(CO3)2 + 4 Na3PO4 36 grams of tin (IV) phosphate Sn3(PO4)4 is mixed with an excess of sodium carbonate Na2CO3. In the lab 29.8 grams of tin (IV) carbonate Sn(CO3)2 are actually formed. What is the percent yield?

The percentage yield of a reaction is 24.3%. If an experimental value of 114.86 g of product is obtained, what would be the theoretical value?

更多留学生实用工具

考试浏览器助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单