For the complete combustion of 4 litre ethane, how much oxygen is required ?

Options

A. 14 litre
B. 4 litre
C. 8 litre
D. 12 litre

Answer

To find out how much oxygen is required for the complete combustion of 4 liters of ethane (C₂H₆), we can follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of ethane.
The complete combustion of ethane can be represented by the following equation:
\[ \text{C}_2\text{H}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \]
Balancing the equation:
- There are 2 carbon atoms in ethane, so we need 2 CO₂.
- There are 6 hydrogen atoms in ethane, so we need 3 H₂O.
- The balanced equation becomes:
\[ \text{C}_2\text{H}_6 + \frac{7}{2} \text{O}_2 \rightarrow 2 \text{CO}_2 + 3 \text{H}_2\text{O} \]

### Step 2: Determine the stoichiometric coefficients.
From the balanced equation, we see that:
- 1 mole of C₂H₆ reacts with \( \frac{7}{2} \) moles of O₂.

### Step 3: Calculate the volume of oxygen required for 4 liters of ethane.
Using the ideal gas law, we know that at standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters.

Given that we have 4 liters of ethane, we can calculate the required volume of oxygen:
1. First, find the number of moles of ethane:
\[ \text{Moles of C}_2\text{H}_6 = \frac{4 \text{ L}}{22.4 \text{ L/mol}} = \frac{4}{22.4} \text{ moles} \]

2. Now, using the stoichiometric ratio from the balanced equation:
\[ \text{Moles of O}_2 = \frac{7}{2} \times \text{Moles of C}_2\text{H}_6 \]
Substituting the moles of ethane:
\[ \text{Moles of O}_2 = \frac{7}{2} \times \frac{4}{22.4} = \frac{14}{22.4} \text{ moles} \]

3. Now, convert moles of O₂ back to liters:
\[ \text{Volume of O}_2 = \text{Moles of O}_2 \times 22.4 \text{ L/mol} = \frac{14}{22.4} \times 22.4 = 14 \text{ L} \]

### Final Answer:
The volume of oxygen required for the complete combustion of 4 liters of ethane is **14 liters**.

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