A mixture of 15 ml of CO & `CO_2` is mixed with V ml (excess) of oxygen and electrically sparked. The volume after explosion was found to be (V+12) ml. What would be the residual volume if 25 ml of the original mixture is exposed to an alkali?

Options

A. 7ml
B. 12ml
C. 10ml
D. 9ml

Answer

To solve the problem step by step, we will analyze the reaction of carbon monoxide (CO) and carbon dioxide (CO₂) with oxygen (O₂) and determine the residual volume after the reaction with alkali.

### Step-by-Step Solution:

1. **Understanding the Mixture**:
We have a mixture of 15 ml of CO and CO₂. Let’s denote the volume of CO as \( x \) ml and the volume of CO₂ as \( 15 - x \) ml.

2. **Reaction with Oxygen**:
The reaction that occurs when CO is mixed with O₂ is:
\[ 2 \text{CO} + \text{O}_2 \rightarrow 2 \text{CO}_2 \]
This means that 2 volumes of CO react with 1 volume of O₂ to produce 2 volumes of CO₂.

3. **Volume After Explosion**:
After the explosion with \( V \) ml of excess oxygen, the total volume is found to be \( V + 12 \) ml. This indicates that some of the CO has reacted with O₂.

4. **Calculating CO Consumption**:
Since only CO reacts with O₂, we can set up the equation based on the volume of CO consumed:
- Let \( y \) ml of CO react with \( \frac{y}{2} \) ml of O₂.
- The volume of CO₂ produced from this reaction is \( y \) ml.

5. **Setting Up the Equation**:
The volume after the explosion is given as:
\[ V + 12 = (15 - y) + (y + \frac{y}{2}) + (V - \frac{y}{2}) \]
Simplifying this gives:
\[ V + 12 = 15 + V - \frac{y}{2} \]
Rearranging leads to:
\[ 12 = 15 - \frac{y}{2} \]
Thus, we find:
\[ \frac{y}{2} = 3 \quad \Rightarrow \quad y = 6 \text{ ml} \]

6. **Finding the Volume of CO**:
Since \( y = 6 \) ml of CO reacted, the remaining volume of CO is:
\[ x = 15 - 6 = 9 \text{ ml} \]

7. **Calculating Residual Volume**:
Now, if we expose 25 ml of the original mixture to an alkali, we need to determine how much CO is present in that volume:
- The proportion of CO in the original mixture is:
\[ \frac{9 \text{ ml}}{15 \text{ ml}} = \frac{3}{5} \]
- Therefore, in 25 ml of the original mixture, the volume of CO is:
\[ \text{Volume of CO} = \frac{3}{5} \times 25 = 15 \text{ ml} \]

8. **Final Residual Volume Calculation**:
When the 25 ml mixture is exposed to alkali, the CO will react completely, leaving:
\[ \text{Residual Volume} = 25 \text{ ml} - 15 \text{ ml} = 10 \text{ ml} \]

### Final Answer:
The residual volume after 25 ml of the original mixture is exposed to an alkali is **10 ml**.