`O_2` required for complete oxidation of 1 litre of ethane at NTP is

Options

A. 3.5 lit.
B. 0.156 mole
C. 5.00 g
D. 7.0 litre

Answer

To determine the amount of \( O_2 \) required for the complete oxidation of 1 liter of ethane (\( C_2H_6 \)) at Normal Temperature and Pressure (NTP), we can follow these steps:

### Step-by-Step Solution:

1. **Write the balanced chemical equation for the combustion of ethane**:
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 equation**:
To balance the equation, we need to ensure that the number of atoms for each element is the same on both sides. 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 \( \frac{7}{2} \) moles of oxygen.

3. **Determine the volume of \( O_2 \) required**:
At NTP, 1 mole of any gas occupies 22.4 liters. Since we are dealing with 1 liter of ethane, we can find the volume of oxygen required using the stoichiometry from the balanced equation:
- 1 mole of \( C_2H_6 \) requires \( \frac{7}{2} \) moles of \( O_2 \).
- Therefore, for 1 liter of \( C_2H_6 \), the volume of \( O_2 \) required is:
\[ \text{Volume of } O_2 = \frac{7}{2} \text{ liters} = 3.5 \text{ liters} \]

4. **Conclusion**:
Thus, the amount of \( O_2 \) required for the complete oxidation of 1 liter of ethane at NTP is \( 3.5 \) liters.

### Final Answer:
The answer is \( 3.5 \) liters of \( O_2 \).

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