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To solve this problem, let's go through the required steps.

### Step 1: Write the Balanced Chemical Reactions

1. **Methane (CH₄)**:
   \[
   \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}
   \]

2. **Ethane (C₂H₆)**:
   \[
   \text{C}_2\text{H}_6 + \frac{7}{2}\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}
   \]

3. **Propane (C₃H₈)**:
   \[
   \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}
   \]

### Step 2: Determine the Theoretical Oxygen Requirement (O₂)

Given:
- Basis of 100 moles of fuel mixture.

The mole fractions of each component in the natural gas are:
- \( \text{CH}_4 = 92\% \), so \( n_{\text{CH}_4} = 0.92 \times 100 = 92 \, \text{mol} \)
- \( \text{C}_2\text{H}_6 = 6.5\% \), so \( n_{\text{C}_2\text{H}_6} = 0.065 \times 100 = 6.5 \, \text{mol} \)
- \( \text{C}_3\text{H}_8 = 1.5\% \), so \( n_{\text{C}_3\text{H}_8} = 0.015 \times 100 = 1.5 \, \text{mol} \)

The theoretical oxygen for each component is calculated as:
1. For CH₄: \( 92 \, \text{mol CH}_4 \times 2 \, \text{mol O}_2/\text{mol CH}_4 = 184 \, \text{mol O}_2 \)
2. For C₂H₆: \( 6.5 \, \text{mol C}_2\text{H}_6 \times 3.5 \, \text{mol O}_2/\text{mol C}_2\text{H}_6 = 22.75 \, \text{mol O}_2 \)
3. For C₃H₈: \( 1.5 \, \text{mol C}_3\text{H}_8 \times 5 \, \text{mol O}_2/\text{mol C}_3\text{H}_8 = 7.5 \, \text{mol O}_2 \)

Adding these together:
\[
\text{Theoretical O}_2 = 184 + 22.75 + 7.5 = 214.25 \, \text{mol O}_2
\]

### Step 3: Account for Excess Air

Since there is a 40% excess air supply, we multiply the theoretical oxygen by 1.4 to find the total amount of oxygen provided.

\[
\text{Total O}_2 = 214.25 \times 1.4 = 299.95 \, \text{mol O}_2
\]

Since air is composed of approximately 21% oxygen and 79% nitrogen by volume, we can calculate the total moles of air (A) required:

\[
A = \frac{\text{Total O}_2}{0.21} = \frac{299.95}{0.21} \approx 1428.33 \, \text{mol air}
\]

Thus, **1428.33 moles of air** are fed into the combustion chamber per 100 moles of natural gas.

Would you like more detailed steps on any part of this solution, or do you have specific questions? Here are some related questions that might help deepen your understanding:

1. How do you derive the mole fractions of each gas component from their percentages?
2. Why is oxygen demand calculated per component separately before summing?
3. What would be the impact of varying the excess air percentage?
4. How would the calculation change if there were another hydrocarbon in the fuel mix?
5. What if nitrogen in air was not considered inert?

**Tip:** Always check the excess air percentage, as it directly affects the oxygen demand and total air requirement in combustion calculations.

Natural gas, containing 92% CH₄, 6.5% C₂H₆, and 1.5% C₃H₈, is combusted with 40% excess air. Determine the moles of air fed per 100 mol of natural gas fed into the combustion chamber.

Calculate the moles of air required for the combustion of natural gas containing CH₄, C₂H₆, and C₃H₈ with 40% excess air. This solution covers stoichiometry and combustion reaction concepts, including the theoretical oxygen requirement and total air calculation.

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