Question

The optimum or stochiometric Methane -air mixture at which violent fire damp explosion takes place is

• A. 5%
• B. 9.5%
• C. 12%
• D. 15%

Hint:

Remember the explosive range for methane: 5 The most violent point is the stoichiometric mixture, which is right in the middle of this range.
9.5

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the specific concentration of methane in air that corresponds to the stoichiometric mixture, which produces the most violent explosion.

Step 2: Key Formula or Approach:
The explosive range for methane in air is approximately 5 The most violent explosion occurs at the stoichiometric concentration, where there is exactly enough oxygen to completely combust the methane.
The balanced chemical equation for the combustion of methane is:
\[ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \] This equation tells us that 1 mole of methane requires 2 moles of oxygen for complete combustion.

Step 3: Detailed Explanation:
Air is composed of approximately 21 This means for every mole of oxygen, there are about \( \frac{79}{21} \approx 3.76 \) moles of nitrogen.
To get the 2 moles of O\(_2\) needed to burn 1 mole of CH\(_4\), we need to use a certain amount of air.
Amount of air = \( 2 \, \text{moles O}_2 + (2 \times 3.76) \, \text{moles N}_2 = 2 + 7.52 = 9.52 \) moles of air.
The total mixture consists of 1 mole of CH\(_4\) and 9.52 moles of air.
Total moles in mixture = \( 1 + 9.52 = 10.52 \) moles.
The percentage of methane in this stoichiometric mixture is:
\[ \% CH_4 = \left( \frac{\text{moles of CH}_4}{\text{total moles}} \right) \times 100 = \left( \frac{1}{10.52} \right) \times 100 \approx 9.5\% \]

Step 4: Final Answer:
The most violent explosion occurs at the stoichiometric concentration of approximately 9.5 This corresponds to option (B).