In the reaction below, 7.0 mol of NO and 5.0 mol of O2 are reacted together. The reaction generates 3.0 mol of NO2. What is the percent yield for the reaction? 2 NO (g) + O2 (g) → 2 NO2 (g)
In the reaction below, 7.0 mol of NO and 5.0 mol of O2 are reacted together. The reaction generates 3.0 mol of NO2. What is the percent yield for the reaction? 2 NO (g) + O2 (g) → 2 NO2 (g)
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter6: Thermochemistry
Section: Chapter Questions
Problem 45E: One of the components of polluted air is NO. It is formed in the high-temperature environment of...
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![### Reaction Yield Calculation
**Question 1 of 4**
In the reaction below, 7.0 mol of NO and 5.0 mol of O₂ are reacted together. The reaction generates 3.0 mol of NO₂. What is the percent yield for the reaction?
\[ 2 \, \text{NO (g)} + \text{O}_2 \, \text{(g)} \rightarrow 2 \, \text{NO}_2 \, \text{(g)} \]
**Calculation Method:**
To determine the percent yield of the reaction, follow these steps:
1. **Determine the Limiting Reactant:**
- From the balanced equation, 2 moles of NO react with 1 mole of O₂ to produce 2 moles of NO₂.
- Start with 7.0 mol of NO and 5.0 mol of O₂.
- Calculate the moles of NO₂ that can be produced from each reactant.
2. **Moles of NO₂ from NO:**
- \( \frac{7.0 \text{ mol NO}}{2 \text{ mol NO}} \times 2 \text{ mol NO}_2 = 7.0 \, \text{mol NO}_2 \)
3. **Moles of NO₂ from O₂:**
- \( \frac{5.0 \text{ mol O}_2}{1 \text{ mol O}_2} \times 2 \text{ mol NO}_2 = 10.0 \, \text{mol NO}_2 \)
4. **Identify Limiting Reactant:**
- NO is the limiting reactant because it produces fewer moles of NO₂ (7.0 mol).
5. **Calculate Percent Yield:**
- Actual yield of NO₂ from reaction = 3.0 mol
- Theoretical yield of NO₂ based on limiting reactant = 7.0 mol
- Percent yield = \( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \)
- Percent yield = \( \frac{3.0 \, \text{mol}}{7.0 \, \text{mol}} \times 100\% = 42.9\% \)
The percent yield of the reaction is **42.9%**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F675d014f-fd61-4129-8894-e08fca4ddce6%2Fa3315327-8c70-422f-8f4e-057a3cfaf7ee%2F7cr9qyeq_processed.png&w=3840&q=75)
Transcribed Image Text:### Reaction Yield Calculation
**Question 1 of 4**
In the reaction below, 7.0 mol of NO and 5.0 mol of O₂ are reacted together. The reaction generates 3.0 mol of NO₂. What is the percent yield for the reaction?
\[ 2 \, \text{NO (g)} + \text{O}_2 \, \text{(g)} \rightarrow 2 \, \text{NO}_2 \, \text{(g)} \]
**Calculation Method:**
To determine the percent yield of the reaction, follow these steps:
1. **Determine the Limiting Reactant:**
- From the balanced equation, 2 moles of NO react with 1 mole of O₂ to produce 2 moles of NO₂.
- Start with 7.0 mol of NO and 5.0 mol of O₂.
- Calculate the moles of NO₂ that can be produced from each reactant.
2. **Moles of NO₂ from NO:**
- \( \frac{7.0 \text{ mol NO}}{2 \text{ mol NO}} \times 2 \text{ mol NO}_2 = 7.0 \, \text{mol NO}_2 \)
3. **Moles of NO₂ from O₂:**
- \( \frac{5.0 \text{ mol O}_2}{1 \text{ mol O}_2} \times 2 \text{ mol NO}_2 = 10.0 \, \text{mol NO}_2 \)
4. **Identify Limiting Reactant:**
- NO is the limiting reactant because it produces fewer moles of NO₂ (7.0 mol).
5. **Calculate Percent Yield:**
- Actual yield of NO₂ from reaction = 3.0 mol
- Theoretical yield of NO₂ based on limiting reactant = 7.0 mol
- Percent yield = \( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \)
- Percent yield = \( \frac{3.0 \, \text{mol}}{7.0 \, \text{mol}} \times 100\% = 42.9\% \)
The percent yield of the reaction is **42.9%**
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