How much volume of air will be needed for complete combustion of 1 litre of ethane?

Options

A. 13.5 L
B. 3.5 L
C. 17.5 L
D. 20.5 L

Answer

To determine the volume of air needed for the complete combustion of 1 liter of ethane (C2H6), we can follow these steps:

### Step-by-Step Solution:

1. **Write the Balanced Combustion Reaction**:
The complete combustion of ethane can be represented by the following equation:
\[ C_2H_6 + O_2 \rightarrow CO_2 + H_2O \]

2. **Balance the Reaction**:
To balance the equation, we need to determine the stoichiometric coefficients. The balanced equation for the combustion of ethane is:
\[ C_2H_6 + \frac{7}{2} O_2 \rightarrow 2 CO_2 + 3 H_2O \]
This indicates that 1 mole of ethane reacts with 3.5 moles of oxygen.

3. **Calculate the Volume of Oxygen Required**:
Since the volume of a gas at constant temperature and pressure is directly proportional to the number of moles (from the ideal gas law), we can say:
\[ \text{Volume of } O_2 = 3.5 \text{ liters (for 1 liter of } C_2H_6) \]

4. **Determine the Volume of Air Needed**:
Air is composed of approximately 21% oxygen. To find the total volume of air needed to provide the required volume of oxygen, we can use the formula:
\[ \text{Volume of air} = \frac{\text{Volume of } O_2}{\text{Fraction of } O_2 \text{ in air}} = \frac{3.5 \text{ liters}}{0.21} \]
Calculating this gives:
\[ \text{Volume of air} \approx 16.67 \text{ liters} \]

### Final Answer:
Thus, the volume of air needed for the complete combustion of 1 liter of ethane is approximately **16.67 liters**.

---