Air contains 20% by volume of oxygen. The volume of air required for the compoete combustion of 1L of methane under the same conditions is

Options

A. 2L
B. 4L
C. 10L
D. 0.4L

Answer

To find the volume of air required for the complete combustion of 1 liter of methane (CH₄), we can follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of methane.
The balanced equation for the combustion of methane is:
\[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \]

### Step 2: Determine the volume of oxygen required for the combustion of 1 liter of methane.
From the balanced equation, we see that 1 mole of methane reacts with 2 moles of oxygen. Therefore, for 1 liter of methane, we need:
\[ 2 \text{ liters of } O_2 \]

### Step 3: Calculate the volume of air needed to provide the required oxygen.
We know that air contains 20% oxygen by volume. Let the volume of air required be \( V \) liters. Since 20% of air is oxygen, we can express the volume of oxygen in terms of the volume of air:
\[ \text{Volume of } O_2 = 0.20 \times V \]

### Step 4: Set up the equation to find the volume of air.
We need 2 liters of oxygen for the combustion of 1 liter of methane, so we can set up the equation:
\[ 0.20 \times V = 2 \text{ liters} \]

### Step 5: Solve for \( V \).
To find \( V \), we rearrange the equation:
\[ V = \frac{2 \text{ liters}}{0.20} \]
\[ V = 10 \text{ liters} \]

### Conclusion:
The volume of air required for the complete combustion of 1 liter of methane is **10 liters**.

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