How many moles of NO, would be required to produce 5.15 moles of HNO in the presence of excess water in the following chemical reaction? 3 NO₂(g) + H₂O (1) → 2 HNO₂(g) + NO(g)
How many moles of NO, would be required to produce 5.15 moles of HNO in the presence of excess water in the following chemical reaction? 3 NO₂(g) + H₂O (1) → 2 HNO₂(g) + NO(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
![**Chemical Reaction Problem**
**Objective:**
Determine the number of moles of nitrogen dioxide (NO₂) required to produce 5.15 moles of nitric acid (HNO₃) in the presence of excess water.
**Chemical Reaction:**
\[ 3 \text{NO}_2(g) + \text{H}_2\text{O}(l) \rightarrow 2 \text{HNO}_3(g) + \text{NO}(g) \]
**Explanation:**
In the given balanced chemical equation, 3 moles of NO₂ react with water to produce 2 moles of HNO₃ and 1 mole of NO. This information can be used to calculate the required moles of NO₂ using stoichiometry.
**Stoichiometry Calculation:**
To calculate the required moles of NO₂:
- From the balanced equation, 2 moles of HNO₃ are produced from 3 moles of NO₂.
- Therefore, to produce 5.15 moles of HNO₃, the moles of NO₂ needed can be calculated using the ratio from the equation:
\[ \text{Moles of NO}_2 = \left(\frac{3}{2}\right) \times 5.15 \]
**Calculation:**
- Moles of NO₂ = 7.725
Hence, 7.725 moles of NO₂ are required to produce 5.15 moles of HNO₃.
This problem helps demonstrate the application of stoichiometry in predicting the quantities of reactants required for a specific amount of product formation in a chemical reaction.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F70274774-6f46-4808-8916-795560a9e0a1%2Fdafe494a-c3d4-427b-99c3-84e67f5bfafc%2F2zb6a74_processed.png&w=3840&q=75)
Transcribed Image Text:**Chemical Reaction Problem**
**Objective:**
Determine the number of moles of nitrogen dioxide (NO₂) required to produce 5.15 moles of nitric acid (HNO₃) in the presence of excess water.
**Chemical Reaction:**
\[ 3 \text{NO}_2(g) + \text{H}_2\text{O}(l) \rightarrow 2 \text{HNO}_3(g) + \text{NO}(g) \]
**Explanation:**
In the given balanced chemical equation, 3 moles of NO₂ react with water to produce 2 moles of HNO₃ and 1 mole of NO. This information can be used to calculate the required moles of NO₂ using stoichiometry.
**Stoichiometry Calculation:**
To calculate the required moles of NO₂:
- From the balanced equation, 2 moles of HNO₃ are produced from 3 moles of NO₂.
- Therefore, to produce 5.15 moles of HNO₃, the moles of NO₂ needed can be calculated using the ratio from the equation:
\[ \text{Moles of NO}_2 = \left(\frac{3}{2}\right) \times 5.15 \]
**Calculation:**
- Moles of NO₂ = 7.725
Hence, 7.725 moles of NO₂ are required to produce 5.15 moles of HNO₃.
This problem helps demonstrate the application of stoichiometry in predicting the quantities of reactants required for a specific amount of product formation in a chemical reaction.
![**Problem: Stoichiometry of a Reaction**
**Question:**
How many moles of ammonia would be required to react exactly with 0.276 moles of copper(II) oxide in the following chemical reaction?
\[ 2 \text{NH}_3(g) + 3 \text{CuO}(s) \rightarrow 3 \text{Cu}(s) + \text{N}_2(g) + 3 \text{H}_2\text{O}(g) \]
**Explanation:**
To solve this problem, we need to use stoichiometry, which is the calculation of reactants and products in chemical reactions. The balanced chemical equation provides the molar ratio between the reactants and products.
1. **Identify the molar ratio:**
From the balanced equation, 2 moles of \( \text{NH}_3 \) react with 3 moles of \( \text{CuO} \).
2. **Calculate the moles of ammonia needed:**
- Given moles of \( \text{CuO} \): 0.276 moles
- Molar ratio of \( \text{NH}_3 \) to \( \text{CuO} \) is 2:3.
\[
\text{Moles of } \text{NH}_3 = \left(\frac{2}{3}\right) \times 0.276 = 0.184 \text{ moles}
\]
**Conclusion:**
0.184 moles of ammonia are required to react with 0.276 moles of copper(II) oxide according to the given chemical equation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F70274774-6f46-4808-8916-795560a9e0a1%2Fdafe494a-c3d4-427b-99c3-84e67f5bfafc%2F66uq04h_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem: Stoichiometry of a Reaction**
**Question:**
How many moles of ammonia would be required to react exactly with 0.276 moles of copper(II) oxide in the following chemical reaction?
\[ 2 \text{NH}_3(g) + 3 \text{CuO}(s) \rightarrow 3 \text{Cu}(s) + \text{N}_2(g) + 3 \text{H}_2\text{O}(g) \]
**Explanation:**
To solve this problem, we need to use stoichiometry, which is the calculation of reactants and products in chemical reactions. The balanced chemical equation provides the molar ratio between the reactants and products.
1. **Identify the molar ratio:**
From the balanced equation, 2 moles of \( \text{NH}_3 \) react with 3 moles of \( \text{CuO} \).
2. **Calculate the moles of ammonia needed:**
- Given moles of \( \text{CuO} \): 0.276 moles
- Molar ratio of \( \text{NH}_3 \) to \( \text{CuO} \) is 2:3.
\[
\text{Moles of } \text{NH}_3 = \left(\frac{2}{3}\right) \times 0.276 = 0.184 \text{ moles}
\]
**Conclusion:**
0.184 moles of ammonia are required to react with 0.276 moles of copper(II) oxide according to the given chemical equation.
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