How many moles of nitrogen gas would be produced if 6.03 moles of copper(II) oxide were reacted with excess ammonia in the following chemical reaction? 2 NH3(g) + 3 CuO (s) → 3 Cu(s) + N2(g) + 3 H2O(g)
How many moles of nitrogen gas would be produced if 6.03 moles of copper(II) oxide were reacted with excess ammonia in the following chemical reaction? 2 NH3(g) + 3 CuO (s) → 3 Cu(s) + N2(g) + 3 H2O(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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![**Stoichiometry Problem: Determining Moles of Nitrogen Gas Produced**
In this problem, we explore a chemical reaction to determine how many moles of nitrogen gas (\(N_2\)) will be produced when 6.03 moles of copper(II) oxide (\(CuO\)) react with an excess of ammonia (\(NH_3\)).
**Chemical Reaction:**
\[ 2 \, NH_3(g) + 3 \, CuO(s) \rightarrow 3 \, Cu(s) + N_2(g) + 3 \, H_2O(g) \]
**Given Information:**
- Moles of \(CuO\): 6.03 moles
- \(NH_3\) is in excess, ensuring complete reaction of \(CuO\).
**Objective:**
Calculate the moles of \(N_2\) produced.
**Calculation Steps:**
1. **Stoichiometric Ratio:**
- From the balanced equation, 3 moles of \(CuO\) produce 1 mole of \(N_2\).
2. **Determine Moles of \(N_2\):**
- Use the stoichiometric ratio to calculate moles of \(N_2\) from \(CuO\).
\[
\text{Moles of } N_2 = \left( \frac{1 \, \text{mole } N_2}{3 \, \text{moles } CuO} \right) \times 6.03 \, \text{moles } CuO
\]
\[
\text{Moles of } N_2 = 2.01 \, \text{moles}
\]
**Interactive Calculator Functionality:**
The diagram on the right side illustrates a simple calculator interface that can be used to input numbers and compute the result. This calculator aids in inputting numerical data to verify calculations and ensure accuracy in stoichiometric computations.
**Conclusion:**
After performing the calculation, 6.03 moles of copper(II) oxide will yield 2.01 moles of nitrogen gas under the given conditions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F439e481c-fa44-496c-b8e8-64fcf7f72bb8%2F5971e614-2007-45de-b6a9-7296171d5689%2F4fa5nq_processed.png&w=3840&q=75)
Transcribed Image Text:**Stoichiometry Problem: Determining Moles of Nitrogen Gas Produced**
In this problem, we explore a chemical reaction to determine how many moles of nitrogen gas (\(N_2\)) will be produced when 6.03 moles of copper(II) oxide (\(CuO\)) react with an excess of ammonia (\(NH_3\)).
**Chemical Reaction:**
\[ 2 \, NH_3(g) + 3 \, CuO(s) \rightarrow 3 \, Cu(s) + N_2(g) + 3 \, H_2O(g) \]
**Given Information:**
- Moles of \(CuO\): 6.03 moles
- \(NH_3\) is in excess, ensuring complete reaction of \(CuO\).
**Objective:**
Calculate the moles of \(N_2\) produced.
**Calculation Steps:**
1. **Stoichiometric Ratio:**
- From the balanced equation, 3 moles of \(CuO\) produce 1 mole of \(N_2\).
2. **Determine Moles of \(N_2\):**
- Use the stoichiometric ratio to calculate moles of \(N_2\) from \(CuO\).
\[
\text{Moles of } N_2 = \left( \frac{1 \, \text{mole } N_2}{3 \, \text{moles } CuO} \right) \times 6.03 \, \text{moles } CuO
\]
\[
\text{Moles of } N_2 = 2.01 \, \text{moles}
\]
**Interactive Calculator Functionality:**
The diagram on the right side illustrates a simple calculator interface that can be used to input numbers and compute the result. This calculator aids in inputting numerical data to verify calculations and ensure accuracy in stoichiometric computations.
**Conclusion:**
After performing the calculation, 6.03 moles of copper(II) oxide will yield 2.01 moles of nitrogen gas under the given conditions.
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