`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. 30 mL,60mL,10mL
B. 30 mL,50mL, 20mL
C. 50 mL,30mL,20mL
D. 20 mL,70 mL, 10 mL

Answer

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 the reaction and subsequent passage through KOH solution. Here’s a step-by-step solution:

### Step 1: Understand the Reaction
When the mixture of gases is sparked, carbon monoxide (CO) reacts with oxygen (O₂) to form carbon dioxide (CO₂):
\[ 2 \text{CO} + \text{O}_2 \rightarrow 2 \text{CO}_2 \]

### Step 2: Analyze the Volume Contraction
The problem states that there is a contraction in volume of 80 ml after passing the gas mixture through KOH solution. KOH absorbs CO₂, which means that the volume of CO₂ produced must equal the volume contraction observed:
\[ \text{Volume of } CO_2 = 80 \text{ ml} \]

### Step 3: Set Up the Initial Conditions
Let’s denote the initial volumes of CO, CO₂, and O₂ in the mixture as:
- Volume of CO = x ml
- Volume of CO₂ = y ml
- Volume of O₂ = z ml

According to the problem, the total volume is:
\[ x + y + z = 100 \text{ ml} \]

### Step 4: Determine the Limiting Reagent
From the reaction stoichiometry, we know that:
- For every 2 ml of CO, 1 ml of O₂ is required to produce 2 ml of CO₂.

Thus, if we denote the amount of CO consumed as \( a \) ml, the amount of O₂ consumed will be \( \frac{a}{2} \) ml, and the amount of CO₂ produced will be \( a \) ml.

### Step 5: Relate the Volumes
Since we know that the volume of CO₂ produced must equal the volume contraction (80 ml), we can set:
\[ a = 80 \text{ ml} \]

### Step 6: Substitute into the Volume Equation
Now we can express the volumes in terms of \( x \), \( y \), and \( z \):
- CO consumed: \( 80 \text{ ml} \)
- O₂ consumed: \( \frac{80}{2} = 40 \text{ ml} \)
- CO₂ produced: \( 80 \text{ ml} \)

Now, we can substitute these values into the total volume equation:
\[ (x - 80) + (y + 80) + (z - 40) = 100 \]

### Step 7: Solve for the Initial Volumes
We can rearrange this equation:
\[ x + y + z - 40 = 100 \]
\[ x + y + z = 140 \]

This contradicts our initial total volume of 100 ml. Thus, we need to check the initial assumptions and try different combinations of x, y, and z to find a valid set of volumes.

### Step 8: Testing Combinations
We can test different combinations of CO, CO₂, and O₂ to see which one satisfies the conditions:
1. **Combination 1:** CO = 30 ml, CO₂ = 60 ml, O₂ = 10 ml
2. **Combination 2:** CO = 30 ml, CO₂ = 30 ml, O₂ = 40 ml
3. **Combination 3:** CO = 50 ml, CO₂ = 20 ml, O₂ = 30 ml
4. **Combination 4:** CO = 20 ml, CO₂ = 70 ml, O₂ = 10 ml

After testing these combinations, we find that:
- **Combination 1** leads to a valid reaction and contraction.
- **Combination 2** also leads to a valid reaction and contraction.

### Conclusion
The possible compositions of the initial mixture are:
1. CO = 30 ml, CO₂ = 60 ml, O₂ = 10 ml
2. CO = 30 ml, CO₂ = 30 ml, O₂ = 40 ml