`A` mixture of `100ml` of `CO,CO_(2)` and `O_(2)` was sparked. When the resulting gaseous mixture was passed through `KOH` solution, contraction in volume was found to be `80ml`, the composition of initial mixture may be (in the same order)

Options

A. `30ml, 60ml, 10ml`
B. `30ml, 50ml, 20ml`
C. `30ml, 60ml, 20ml`
D. `30ml, 40ml, 30ml`

Explanation

To solve the problem, we need to determine the composition of the initial mixture of gases (CO, CO₂, and O₂) based on the contraction in volume observed after passing the resulting gaseous mixture through KOH solution. The contraction in volume is given as 80 ml.

### Step-by-Step Solution:

1. **Define Variables:**
Let:
- \( X \) = volume of CO (in ml)
- \( Y \) = volume of CO₂ (in ml)
- \( Z \) = volume of O₂ (in ml)

From the problem, we know:
\[ X + Y + Z = 100 \, \text{ml} \quad \text{(1)} \]

2. **Understanding the Reaction:**
The reaction that occurs when CO and O₂ are sparked is:
\[ 2 \, \text{CO} + \text{O}_2 \rightarrow 2 \, \text{CO}_2 \]
This means that 1 volume of O₂ reacts with 2 volumes of CO to produce 2 volumes of CO₂.

3. **Case 1: CO as the Limiting Reagent**
- If CO is the limiting reagent, then all of it will react.
- The volume of CO₂ produced will be equal to the volume of CO used, which is \( X \).
- The total volume of CO₂ after the reaction will be:
\[ \text{Total CO₂} = X + Y \quad \text{(2)} \]
- Since KOH absorbs CO₂, the contraction in volume will equal the volume of CO₂ produced:
\[ X + Y = 80 \, \text{ml} \quad \text{(3)} \]

4. **Substituting into Equation (1):**
From equation (1):
\[ X + Y + Z = 100 \]
Substituting equation (3) into this gives:
\[ 80 + Z = 100 \]
Therefore:
\[ Z = 20 \, \text{ml} \quad \text{(4)} \]

5. **Finding Values of X and Y:**
From equation (3), we have:
\[ X + Y = 80 \, \text{ml} \]
We can check the options provided to see which combinations of \( X \) and \( Y \) satisfy this equation along with \( Z = 20 \, \text{ml} \).

6. **Case 2: O₂ as the Limiting Reagent**
- If O₂ is the limiting reagent, then all of it will react.
- The volume of CO₂ produced will be:
\[ \text{Total CO₂} = 2Z + Y \quad \text{(5)} \]
- Since KOH absorbs CO₂, we have:
\[ 2Z + Y = 80 \, \text{ml} \quad \text{(6)} \]

7. **Substituting into Equation (1):**
From equation (1):
\[ X + Y + Z = 100 \]
We can express \( X \) in terms of \( Z \) and \( Y \):
\[ X = 100 - Y - Z \quad \text{(7)} \]
Substituting equation (6) into this gives:
\[ X = 100 - (80 - 2Z) - Z \]
Simplifying:
\[ X = 100 - 80 + 2Z - Z = 20 + Z \]
Thus:
\[ X - Z = 20 \quad \text{(8)} \]

8. **Finding Values of X and Z:**
We can check the options provided to see which combinations of \( X \) and \( Z \) satisfy equation (8).

9. **Final Check:**
After checking the options against both cases, we find that the valid combinations for the initial mixture are:
- \( X = 30 \, \text{ml}, Y = 50 \, \text{ml}, Z = 20 \, \text{ml} \) (from Case 1)
- \( X = 30 \, \text{ml}, Y = 20 \, \text{ml}, Z = 30 \, \text{ml} \) (from Case 2)

### Conclusion:
The composition of the initial mixture may be:
- **Option 1:** \( 30 \, \text{ml} \, \text{CO}, 50 \, \text{ml} \, \text{CO}_2, 20 \, \text{ml} \, \text{O}_2 \)
- **Option 2:** \( 30 \, \text{ml} \, \text{CO}, 20 \, \text{ml} \, \text{CO}_2, 30 \, \text{ml} \, \text{O}_2 \)