9 " mL of " a mixture of methane and ethylene was exploded with 30 mL (excess) of oxygen. After cooling, the volume was 21.0 mL. Further treatment with caustic potash solution reduced the volume to 7.0 mL. Determine the composition of the mixture.

Explanation

To determine the composition of the mixture of methane (CH₄) and ethylene (C₂H₄) in the given problem, we can follow these steps:

### Step 1: Define Variables
Let the volume of methane (CH₄) in the mixture be \( x \) mL. Therefore, the volume of ethylene (C₂H₄) will be \( 9 - x \) mL since the total volume of the mixture is 9 mL.

### Step 2: Write the Combustion Reactions
1. The combustion of methane (CH₄) with oxygen (O₂) can be represented as:
\[ CH₄ + 2O₂ \rightarrow CO₂ + 2H₂O \]
From this reaction, 1 volume of CH₄ reacts with 2 volumes of O₂ to produce 1 volume of CO₂.

2. The combustion of ethylene (C₂H₄) with oxygen (O₂) is represented as:
\[ C₂H₄ + 3O₂ \rightarrow 2CO₂ + 2H₂O \]
From this reaction, 1 volume of C₂H₄ reacts with 3 volumes of O₂ to produce 2 volumes of CO₂.

### Step 3: Calculate the Total Volume of CO₂ Produced
After the combustion, the total volume of gas is 21 mL. This volume includes unreacted oxygen and the produced CO₂ and H₂O.

Let’s denote:
- The volume of CO₂ produced from methane as \( \frac{x}{1} \) (1 volume of CO₂ for each volume of CH₄).
- The volume of CO₂ produced from ethylene as \( 2(9 - x) \) (2 volumes of CO₂ for each volume of C₂H₄).

Thus, the total volume of CO₂ produced is:
\[ \text{Total CO₂} = x + 2(9 - x) = x + 18 - 2x = 18 - x \]

### Step 4: Relate CO₂ Volume to KOH Absorption
After treatment with caustic potash (KOH), the volume is reduced to 7 mL. The volume of CO₂ absorbed by KOH is:
\[ \text{Volume of CO₂ absorbed} = 21 - 7 = 14 \text{ mL} \]

### Step 5: Set Up the Equation
From the above, we can set up the equation:
\[ 18 - x = 14 \]

### Step 6: Solve for \( x \)
Now, solve for \( x \):
\[ 18 - x = 14 \\ x = 18 - 14 \\ x = 4 \text{ mL} \]

### Step 7: Determine the Volume of Ethylene
Now, substitute \( x \) back to find the volume of ethylene:
\[ \text{Volume of C₂H₄} = 9 - x = 9 - 4 = 5 \text{ mL} \]

### Conclusion
The composition of the mixture is:
- Volume of methane (CH₄) = 4 mL
- Volume of ethylene (C₂H₄) = 5 mL