`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 find the amount of oxygen (`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. **Understanding the Reaction**:
The complete combustion of ethane can be represented by the following balanced equation:
\[ C_2H_6 + O_2 \rightarrow CO_2 + H_2O \]
Balancing the equation:
\[ C_2H_6 + \frac{7}{2} O_2 \rightarrow 2 CO_2 + 3 H_2O \]
This shows that 1 mole of ethane reacts with \( \frac{7}{2} \) moles of oxygen.

2. **Calculating Moles of Ethane**:
At NTP (Normal Temperature and Pressure), 1 mole of gas occupies 22.4 liters. Therefore, the number of moles of ethane in 1 liter is:
\[ \text{Number of moles of } C_2H_6 = \frac{1 \text{ L}}{22.4 \text{ L/mol}} = 0.0446 \text{ moles} \]

3. **Calculating Moles of Oxygen Required**:
From the balanced equation, we know that 1 mole of ethane requires \( \frac{7}{2} \) moles of oxygen. Thus, the moles of oxygen required for 0.0446 moles of ethane is:
\[ \text{Moles of } O_2 = 0.0446 \text{ moles } C_2H_6 \times \frac{7}{2} = 0.1561 \text{ moles} \]

4. **Calculating Volume of Oxygen**:
To find the volume of oxygen required at NTP:
\[ \text{Volume of } O_2 = \text{Moles of } O_2 \times 22.4 \text{ L/mol} = 0.1561 \text{ moles} \times 22.4 \text{ L/mol} \approx 3.5 \text{ L} \]

5. **Calculating Mass of Oxygen**:
The molar mass of oxygen (`O_2`) is 32 g/mol. Therefore, the mass of oxygen required is:
\[ \text{Mass of } O_2 = \text{Moles of } O_2 \times \text{Molar mass of } O_2 = 0.1561 \text{ moles} \times 32 \text{ g/mol} \approx 4.99 \text{ g} \]

### Final Results:
- Volume of `O_2` required: **3.5 liters**
- Mass of `O_2` required: **5 grams**