For the complete combustion of 4 litres of CO at NTP,the required volume of `O_(2)` at NTP is :-

Options

A. 4 litres
B. 8 litres
C. 2litres
D. 1litre

Explanation

To solve the problem of determining the required volume of \( O_2 \) for the complete combustion of 4 liters of \( CO \) at NTP, we will follow these steps:

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation**:
The combustion of carbon monoxide (\( CO \)) can be represented by the following balanced chemical equation:
\[ 2 \, CO + O_2 \rightarrow 2 \, CO_2 \]
This equation shows that 2 moles (or volumes) of \( CO \) react with 1 mole (or volume) of \( O_2 \) to produce 2 moles (or volumes) of \( CO_2 \).

2. **Determine the Volume Ratio**:
From the balanced equation, we observe that:
- 2 volumes of \( CO \) require 1 volume of \( O_2 \).
- This means that for every 2 liters of \( CO \), we need 1 liter of \( O_2 \).

3. **Calculate the Required Volume of \( O_2 \)**:
Given that we have 4 liters of \( CO \), we can set up a proportion based on the volume ratio:
\[ \text{Volume of } O_2 = \left( \frac{1 \, \text{liter of } O_2}{2 \, \text{liters of } CO} \right) \times 4 \, \text{liters of } CO \]
Simplifying this gives:
\[ \text{Volume of } O_2 = \frac{4}{2} = 2 \, \text{liters} \]

4. **Conclusion**:
Therefore, the required volume of \( O_2 \) for the complete combustion of 4 liters of \( CO \) at NTP is:
\[ \boxed{2 \, \text{liters}} \]