20 ml of a mixture of `C_(2)H_(2)` and CO was exploded with 30 ml of oxygen. The gases after the reaction had a volume of 34 ml. On treatment with KOH, 8 ml of oxygen gas remain unreacted. Which of the following options show a correct composition of the mixture? 

Options

A. `V_("ethyne")= 6ml` and `V_("co")= 14 ml`
B. `V_("ethyne")= 4ml` and `V_("co")= 16 ml`
C. `V_("ethyne")= 2ml` and `V_("co")= 18 ml`
D. `V_("ethyne")= 5ml` and `V_("co")= 15 ml`

Explanation

To solve the problem step by step, we will analyze the given information and perform the necessary calculations.

### Step 1: Define Variables
Let:
- \( x \) = volume of \( C_2H_2 \) (ethyne) in ml
- \( 20 - x \) = volume of CO in ml

### Step 2: Write the Reaction Equations
1. The combustion of ethyne:
\[ C_2H_2 + \frac{5}{2} O_2 \rightarrow 2 CO_2 + H_2O \]
This means 1 ml of \( C_2H_2 \) reacts with 2.5 ml of \( O_2 \).

2. The combustion of carbon monoxide:
\[ 2 CO + O_2 \rightarrow 2 CO_2 \]
This means 1 ml of CO reacts with 0.5 ml of \( O_2 \).

### Step 3: Calculate the Total Volume After Reaction
The total volume after the reaction is given as 34 ml. The volume of gases after the reaction can be expressed as:
\[ \text{Volume of products} = \text{Volume of } CO_2 + \text{Volume of unreacted } O_2 \]
From the reactions:
- Volume of \( CO_2 \) produced from \( C_2H_2 \) = \( 2x \) (since 1 ml of \( C_2H_2 \) produces 2 ml of \( CO_2 \)).
- Volume of \( CO_2 \) produced from \( CO \) = \( 20 - x \) (since 1 ml of CO produces 1 ml of \( CO_2 \)).
- Unreacted \( O_2 \) = 8 ml.

Thus, the equation becomes:
\[ 2x + (20 - x) + 8 = 34 \]

### Step 4: Simplify the Equation
Now, simplify the equation:
\[ 2x + 20 - x + 8 = 34 \]
\[ x + 28 = 34 \]
\[ x = 34 - 28 \]
\[ x = 6 \]

### Step 5: Calculate Volumes of Each Component
Now that we have \( x \):
- Volume of \( C_2H_2 \) = \( x = 6 \) ml
- Volume of CO = \( 20 - x = 20 - 6 = 14 \) ml

### Step 6: Conclusion
The composition of the mixture is:
- \( C_2H_2 \) = 6 ml
- CO = 14 ml

### Final Answer
The correct option showing the composition of the mixture is:
- 6 ml of \( C_2H_2 \) and 14 ml of CO.