How many moles of nitrogen gas would be produced if 8.77 moles of copper(11) 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 8.77 moles of copper(11) 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)
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Chapter6: Thermochemisty
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![**Problem Statement:**
How many moles of nitrogen gas would be produced if 8.77 moles of copper(II) oxide were reacted with excess ammonia in the following chemical reaction?
**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:**
In this balanced chemical equation, we can see the relationship between the reactants and products:
- 2 moles of ammonia (\(\text{NH}_3\)) react with 3 moles of copper(II) oxide (\(\text{CuO}\)).
- This reaction produces 3 moles of copper (\(\text{Cu}\)), 1 mole of nitrogen gas (\(\text{N}_2\)), and 3 moles of water (\(\text{H}_2\text{O}\)).
Given that we have 8.77 moles of \(\text{CuO}\) and an excess of \(\text{NH}_3\), we need to determine how many moles of \(\text{N}_2\) are produced.
**Stoichiometric Calculation:**
1. According to the balanced equation, 3 moles of \(\text{CuO}\) produce 1 mole of \(\text{N}_2\).
2. Therefore, the moles of \(\text{N}_2\) produced can be calculated as:
\[
\frac{1 \, \text{mol} \, \text{N}_2}{3 \, \text{mol} \, \text{CuO}} \times 8.77 \, \text{mol} \, \text{CuO} = \frac{8.77}{3} \, \text{mol} \, \text{N}_2
\]
3. Simplifying the calculation:
\[
\frac{8.77}{3} \approx 2.92 \, \text{mol} \, \text{N}_2
\]
**Conclusion:**
Therefore, 2.92 moles of nitrogen gas (\(\text{N}_2\))](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbc3b72c7-eb99-406f-83fe-82fc41b2a8b5%2Fb1b6face-abc4-4454-bf45-a506a857d022%2Fns3jn7p_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
How many moles of nitrogen gas would be produced if 8.77 moles of copper(II) oxide were reacted with excess ammonia in the following chemical reaction?
**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:**
In this balanced chemical equation, we can see the relationship between the reactants and products:
- 2 moles of ammonia (\(\text{NH}_3\)) react with 3 moles of copper(II) oxide (\(\text{CuO}\)).
- This reaction produces 3 moles of copper (\(\text{Cu}\)), 1 mole of nitrogen gas (\(\text{N}_2\)), and 3 moles of water (\(\text{H}_2\text{O}\)).
Given that we have 8.77 moles of \(\text{CuO}\) and an excess of \(\text{NH}_3\), we need to determine how many moles of \(\text{N}_2\) are produced.
**Stoichiometric Calculation:**
1. According to the balanced equation, 3 moles of \(\text{CuO}\) produce 1 mole of \(\text{N}_2\).
2. Therefore, the moles of \(\text{N}_2\) produced can be calculated as:
\[
\frac{1 \, \text{mol} \, \text{N}_2}{3 \, \text{mol} \, \text{CuO}} \times 8.77 \, \text{mol} \, \text{CuO} = \frac{8.77}{3} \, \text{mol} \, \text{N}_2
\]
3. Simplifying the calculation:
\[
\frac{8.77}{3} \approx 2.92 \, \text{mol} \, \text{N}_2
\]
**Conclusion:**
Therefore, 2.92 moles of nitrogen gas (\(\text{N}_2\))
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