The combustion of ammonia in the presence of oxygen yields NO2 and H2O:


(Ch. 3) The combustion of 43.9 g of ammonia with 258 g of oxygen produces __________ g of NO2.

To find the mass of \( \text{NO}_2 \) produced in the combustion of ammonia, let's follow these steps:

### **Step 1: Write the Balanced Chemical Equation**
The combustion of ammonia (\( \text{NH}_3 \)) with oxygen (\( \text{O}_2 \)) can be written as:

\[
4 \text{NH}_3 + 7 \text{O}_2 \rightarrow 4 \text{NO}_2 + 6 \text{H}_2\text{O}
\]

This equation tells us that 4 moles of ammonia react with 7 moles of oxygen to produce 4 moles of nitrogen dioxide and 6 moles of water.

### **Step 2: Calculate the Molar Masses**
- **Ammonia (\( \text{NH}_3 \)):**  
  \( \text{N} \) = 14.01 g/mol, \( \text{H} \) = 1.008 g/mol  
  Molar mass of \( \text{NH}_3 \) = \( 14.01 + (3 \times 1.008) = 17.034 \) g/mol

- **Oxygen (\( \text{O}_2 \)):**  
  \( \text{O}_2 \) = \( 2 \times 16.00 = 32.00 \) g/mol

- **Nitrogen Dioxide (\( \text{NO}_2 \)):**  
  \( \text{N} \) = 14.01 g/mol, \( \text{O} \) = \( 2 \times 16.00 = 32.00 \) g/mol  
  Molar mass of \( \text{NO}_2 \) = \( 14.01 + 32.00 = 46.01 \) g/mol

### **Step 3: Determine the Limiting Reagent**
- **Moles of Ammonia:**  
  \[
  \text{Moles of } \text{NH}_3 = \frac{43.9 \, \text{g}}{17.034 \, \text{g/mol}} \approx 2.576 \, \text{moles}
  \]

- **Moles of Oxygen:**  
  \[
  \text{Moles of } \text{O}_2 = \frac{258 \, \text{g}}{32.00 \, \text{g/mol}} = 8.0625 \, \text{moles}
  \]

### **Step 4: Use Stoichiometry to Find the Limiting Reagent**
From the balanced equation:
- 4 moles of \( \text{NH}_3 \) require 7 moles of \( \text{O}_2 \).
- To react completely with 2.576 moles of \( \text{NH}_3 \), we need:
  \[
  \text{Required moles of } \text{O}_2 = \frac{7}{4} \times 2.576 = 4.507 \, \text{moles}
  \]

Since we have 8.0625 moles of \( \text{O}_2 \), which is more than the 4.507 moles required, ammonia (\( \text{NH}_3 \)) is the limiting reagent.

### **Step 5: Calculate the Moles of \( \text{NO}_2 \) Produced**
From the balanced equation:
- 4 moles of \( \text{NH}_3 \) produce 4 moles of \( \text{NO}_2 \).
- Therefore, 2.576 moles of \( \text{NH}_3 \) will produce 2.576 moles of \( \text{NO}_2 \).

### **Step 6: Convert Moles of \( \text{NO}_2 \) to Grams**
\[
\text{Mass of } \text{NO}_2 = 2.576 \, \text{moles} \times 46.01 \, \text{g/mol} \approx 118.48 \, \text{g}
\]

### **Final Answer**
The combustion of 43.9 g of ammonia with 258 g of oxygen produces approximately **118.48 g of \( \text{NO}_2 \)**.

Would you like to go through any specific part of this process in more detail?

### Relative Questions to Expand Knowledge
1. What would happen if oxygen was the limiting reagent instead of ammonia?
2. How would the amount of water produced be calculated in this reaction?
3. How does the combustion of ammonia differ from other combustion reactions, such as methane?
4. What is the significance of identifying the limiting reagent in a chemical reaction?
5. How would the mass of \( \text{NO}_2 \) change if the amount of ammonia were doubled?

### Tip
Always identify the limiting reagent in a chemical reaction to accurately determine the amount of products formed.

The combustion of 43.9 g of ammonia with 258 g of oxygen produces __________ g of NO2.

Understand the combustion of ammonia with oxygen and how to calculate the mass of NO2 produced. This problem covers key concepts like stoichiometry, limiting reagents, and the law of conservation of mass. Step-by-step solution provided for students in Grades 10-12.

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