If 10.0 moles of O2 are reacted with excess NO in the reaction below, and only 7.0 mol of NO2 were collected, then what is the percent yield for the reaction? 2 NO (g) + O2 (g) → 2 NO2 (g) ->
If 10.0 moles of O2 are reacted with excess NO in the reaction below, and only 7.0 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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![**Question 6 of 37**
If 10.0 moles of O₂ are reacted with excess NO in the reaction below, and only 7.0 mol of NO₂ were collected, then what is the percent yield for the reaction?
\[2 \text{NO} (g) + \text{O}_2 (g) \rightarrow 2 \text{NO}_2 (g)\]
**Explanation:**
This question involves a chemical reaction where nitrogen monoxide (NO) reacts with oxygen (O₂) to form nitrogen dioxide (NO₂). The stoichiometry of the reaction is such that 2 moles of NO react with 1 mole of O₂ to produce 2 moles of NO₂.
To find the percent yield:
1. **Determine the theoretical yield:**
- According to the balanced equation, 1 mole of O₂ produces 2 moles of NO₂. Therefore, 10.0 moles of O₂ should theoretically produce 20.0 moles of NO₂.
2. **Calculate the percent yield:**
- Percent yield is given by \(\frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%\).
- The actual yield is 7.0 moles of NO₂.
- Therefore, percent yield = \(\frac{7.0 \, \text{mol}}{20.0 \, \text{mol}} \times 100\% = 35\%\).
This information helps to understand reaction efficiency in practical scenarios compared to theoretical calculations.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F81a44892-e725-436e-9c8e-21f599a0ba84%2Fd4b6426c-909d-4bff-a858-61bedfd6192a%2Ft7nerza.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 6 of 37**
If 10.0 moles of O₂ are reacted with excess NO in the reaction below, and only 7.0 mol of NO₂ were collected, then what is the percent yield for the reaction?
\[2 \text{NO} (g) + \text{O}_2 (g) \rightarrow 2 \text{NO}_2 (g)\]
**Explanation:**
This question involves a chemical reaction where nitrogen monoxide (NO) reacts with oxygen (O₂) to form nitrogen dioxide (NO₂). The stoichiometry of the reaction is such that 2 moles of NO react with 1 mole of O₂ to produce 2 moles of NO₂.
To find the percent yield:
1. **Determine the theoretical yield:**
- According to the balanced equation, 1 mole of O₂ produces 2 moles of NO₂. Therefore, 10.0 moles of O₂ should theoretically produce 20.0 moles of NO₂.
2. **Calculate the percent yield:**
- Percent yield is given by \(\frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%\).
- The actual yield is 7.0 moles of NO₂.
- Therefore, percent yield = \(\frac{7.0 \, \text{mol}}{20.0 \, \text{mol}} \times 100\% = 35\%\).
This information helps to understand reaction efficiency in practical scenarios compared to theoretical calculations.
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