`20mL` of a mixture of `CO` and `H_(2)` were mixed excess of `O_(2)` and exploded & cooled. There was a volume contraction of 23mL. All volume measurements corresponds to room temperature `(27^(@)C)` and one atmospheric pressure. Determine the volume ratio `(V_(1):V_(2) "of" Co` anf `H_(2)` in the original mixture .

Options

A. `6.5:13.5`
B. `5:15`
C. `2:3`
D. `7:13`

Answer

To solve the problem, we need to determine the volume ratio of carbon monoxide (CO) and hydrogen (H₂) in a mixture that undergoes a reaction with excess oxygen (O₂) resulting in a volume contraction. Here’s a step-by-step solution:

### Step 1: Define Variables
Let:
- \( V_1 \) = volume of CO in the mixture
- \( V_2 \) = volume of H₂ in the mixture

From the problem, we know:
- The total volume of the mixture is \( V_1 + V_2 = 20 \, \text{mL} \) (Equation 1)

### Step 2: Write the Chemical Reactions
1. The reaction of carbon monoxide with oxygen:
\[ 2 \, \text{CO} + \text{O}_2 \rightarrow 2 \, \text{CO}_2 \]
- Volume change: \( V_1 + \frac{V_1}{2} - V_1 = \frac{V_1}{2} \)

2. The reaction of hydrogen with oxygen:
\[ 2 \, \text{H}_2 + \text{O}_2 \rightarrow 2 \, \text{H}_2\text{O} \]
- Volume change: \( V_2 + \frac{V_2}{2} - V_2 = \frac{V_2}{2} \)

### Step 3: Calculate Total Volume Contraction
The total volume contraction after the reactions is given as 23 mL. Therefore, we can express this as:
\[ \frac{V_1}{2} + \frac{V_2}{2} = 23 \, \text{mL} \quad \text{(Equation 2)} \]

### Step 4: Simplify Equation 2
Multiply Equation 2 by 2 to eliminate the fractions:
\[ V_1 + V_2 = 46 \, \text{mL} \quad \text{(Equation 3)} \]

### Step 5: Solve the System of Equations
Now we have two equations:
1. \( V_1 + V_2 = 20 \, \text{mL} \) (Equation 1)
2. \( V_1 + V_2 = 46 \, \text{mL} \) (Equation 3)

Since Equation 3 cannot be correct (it contradicts Equation 1), we need to consider the volume of water formed. The water formed will condense and reduce the volume further. The correct contraction should account for the fact that water is in liquid form and does not contribute to the gas volume.

### Step 6: Adjust for Water Volume
The volume contraction due to water is effectively \( V_2 \) (as it condenses). Thus, we need to adjust our contraction equation:
\[ \frac{V_1}{2} + \frac{V_2}{2} + V_2 = 23 \, \text{mL} \]
This simplifies to:
\[ \frac{V_1}{2} + \frac{3V_2}{2} = 23 \quad \text{(Equation 4)} \]

### Step 7: Solve Equations 1 and 4
Now we have:
1. \( V_1 + V_2 = 20 \) (Equation 1)
2. \( V_1 + 3V_2 = 46 \) (Equation 4)

Subtract Equation 1 from Equation 4:
\[ (V_1 + 3V_2) - (V_1 + V_2) = 46 - 20 \]
This simplifies to:
\[ 2V_2 = 26 \implies V_2 = 13 \, \text{mL} \]

### Step 8: Find \( V_1 \)
Substituting \( V_2 \) back into Equation 1:
\[ V_1 + 13 = 20 \implies V_1 = 7 \, \text{mL} \]

### Step 9: Determine the Volume Ratio
The volume ratio \( V_1 : V_2 \) is:
\[ V_1 : V_2 = 7 : 13 \]

### Final Answer
The volume ratio of carbon monoxide to hydrogen in the original mixture is \( 7 : 13 \).

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