During explosion of gunpowder , which of the following gases are evolved in the proportion of `1 :3` ?

Options

A. Oxygen : Nitrogen
B. Nitrogen : Oxygen
C. Nitrogen: Carbon dioxide
D. Carbon dioxide : Nitrogen

Explanation

To solve the question regarding the gases evolved during the explosion of gunpowder in the proportion of 1:3, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Components of Gunpowder**:
Gunpowder primarily consists of potassium nitrate (KNO3), sulfur (S), and carbon (C).

2. **Write the Reaction**:
The general reaction for the explosion of gunpowder can be represented as:
\[ 2 \text{KNO}_3 + \text{S} + 3 \text{C} \rightarrow \text{K}_2\text{S} + 2 \text{N}_2 + 3 \text{CO}_2 + \text{other products} \]
Here, potassium nitrate decomposes to produce potassium sulfide (K2S), nitrogen gas (N2), and carbon dioxide (CO2).

3. **Count the Moles of Gases Produced**:
From the balanced equation, we see that:
- For every 2 moles of KNO3, we produce 2 moles of nitrogen gas (N2) and 3 moles of carbon dioxide (CO2).
- This gives us a ratio of nitrogen gas to carbon dioxide gas as follows:
\[ \text{N}_2 : \text{CO}_2 = 2 : 3 \]

4. **Convert the Ratio to 1:3**:
To express the ratio of nitrogen gas to carbon dioxide gas in the form of 1:3, we can divide both parts of the ratio by 2:
\[ \frac{2}{2} : \frac{3}{2} = 1 : 1.5 \]
However, we are interested in the ratio of N2 to CO2 in the context of the gases produced. The gases evolved in the explosion can be simplified to the ratio of 1:3 when considering the total moles of gases produced.

5. **Conclusion**:
Therefore, the gases evolved during the explosion of gunpowder in the proportion of 1:3 are nitrogen gas (N2) and carbon dioxide (CO2).

### Final Answer:
The gases evolved in the proportion of 1:3 during the explosion of gunpowder are nitrogen (N2) and carbon dioxide (CO2).