Question 27 of 32 If 10.0 moles of O₂ are reacted with excess NO in the reaction below, and only 8.4 mol of NO₂ were collected, then what is the percent yield for the reaction? 2 NO (g) + O₂ (g) → 2 NO₂ (g)
Question 27 of 32 If 10.0 moles of O₂ are reacted with excess NO in the reaction below, and only 8.4 mol of NO₂ were collected, then 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
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![### Percent Yield Calculation in Chemical Reactions
**Question:**
If 10.0 moles of \(O_2\) are reacted with excess \(NO\) in the reaction below, and only 8.4 mol of \(NO_2\) were collected, then what is the percent yield for the reaction?
**Chemical Equation:**
\[ 2NO(g) + O_2(g) \rightarrow 2NO_2(g) \]
**Details:**
Given:
- 10.0 moles of \(O_2\)
- Collected: 8.4 moles of \(NO_2\)
**Calculation:**
1. **Determine the theoretical yield:**
According to the balanced equation, 1 mole of \(O_2\) produces 2 moles of \(NO_2\). Therefore, 10 moles of \(O_2\) would theoretically produce:
\[ 10 \text{ moles } O_2 \times \frac{2 \text{ moles } NO_2}{1 \text{ mole } O_2} = 20 \text{ moles } NO_2 \]
2. **Percent Yield Formula:**
\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \]
In this case, the actual yield is 8.4 moles \(NO_2\).
\[ \text{Percent Yield} = \left( \frac{8.4 \text{ moles } NO_2}{20 \text{ moles } NO_2} \right) \times 100 \]
3. **Calculate the Percent Yield:**
\[ \text{Percent Yield} = 42\% \]
**Conclusion:**
The percent yield for the reaction is 42%.
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**Interactive Components:**
- **Calculator Interface:** Basic calculator interface showing numerical buttons (1-9, 0) and operations like \(/, \times, +, -, C\), and \(\times 10\) for ease of use in percent yield calculations.
---
This problem helps students understand how to calculate the percent yield from a chemical reaction by applying stoichiometry and interpreting balanced chemical equations. It's a key concept in assessing the efficiency of reactions in laboratory and industrial settings.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6895f9cd-8521-4c9a-bc62-df11c4ac3cae%2F8c422097-27ea-4d67-ad41-d377f42749da%2Fkwk3lc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Percent Yield Calculation in Chemical Reactions
**Question:**
If 10.0 moles of \(O_2\) are reacted with excess \(NO\) in the reaction below, and only 8.4 mol of \(NO_2\) were collected, then what is the percent yield for the reaction?
**Chemical Equation:**
\[ 2NO(g) + O_2(g) \rightarrow 2NO_2(g) \]
**Details:**
Given:
- 10.0 moles of \(O_2\)
- Collected: 8.4 moles of \(NO_2\)
**Calculation:**
1. **Determine the theoretical yield:**
According to the balanced equation, 1 mole of \(O_2\) produces 2 moles of \(NO_2\). Therefore, 10 moles of \(O_2\) would theoretically produce:
\[ 10 \text{ moles } O_2 \times \frac{2 \text{ moles } NO_2}{1 \text{ mole } O_2} = 20 \text{ moles } NO_2 \]
2. **Percent Yield Formula:**
\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \]
In this case, the actual yield is 8.4 moles \(NO_2\).
\[ \text{Percent Yield} = \left( \frac{8.4 \text{ moles } NO_2}{20 \text{ moles } NO_2} \right) \times 100 \]
3. **Calculate the Percent Yield:**
\[ \text{Percent Yield} = 42\% \]
**Conclusion:**
The percent yield for the reaction is 42%.
---
**Interactive Components:**
- **Calculator Interface:** Basic calculator interface showing numerical buttons (1-9, 0) and operations like \(/, \times, +, -, C\), and \(\times 10\) for ease of use in percent yield calculations.
---
This problem helps students understand how to calculate the percent yield from a chemical reaction by applying stoichiometry and interpreting balanced chemical equations. It's a key concept in assessing the efficiency of reactions in laboratory and industrial settings.
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