On subjecting 10ml mixture of `N_(2)` and CO to repeated electric spark to form `CO_(2)` and NO, 7 ml of `O_(2)` was required for combustion. What was the mole precent of CO in the mixture ? (All volumes were measured under identical conditions)

Options

A. 4
B. 6
C. 40
D. 60

Answer

To solve the problem step by step, we will follow the information provided in the question and apply stoichiometry principles.

### Step 1: Understanding the Reaction
We have a mixture of nitrogen (N₂) and carbon monoxide (CO). When subjected to electric sparks, CO reacts with oxygen (O₂) to form carbon dioxide (CO₂), and nitrogen (N₂) reacts to form nitrogen oxides (NO). The relevant reactions are:

1. **For CO combustion:**
\[ \text{CO} + \frac{1}{2} \text{O}_2 \rightarrow \text{CO}_2 \]

2. **For N₂ reaction:**
\[ \text{N}_2 + \text{O}_2 \rightarrow 2 \text{NO} \]

### Step 2: Setting Up the Variables
Let’s denote the volume of CO in the mixture as \( x \) ml. Since the total volume of the gas mixture is 10 ml, the volume of N₂ will be:
\[ \text{Volume of } N_2 = 10 - x \text{ ml} \]

### Step 3: Calculating Oxygen Required
From the reactions, we can determine how much O₂ is required for combustion:

- For \( x \) ml of CO, the volume of O₂ required is:
\[ \text{O}_2 \text{ required for CO} = \frac{x}{2} \text{ ml} \]

- For \( 10 - x \) ml of N₂, the volume of O₂ required is:
\[ \text{O}_2 \text{ required for } N_2 = 10 - x \text{ ml} \]

### Step 4: Total Oxygen Required
According to the problem, the total volume of O₂ required for combustion is 7 ml. Therefore, we can set up the equation:
\[ \frac{x}{2} + (10 - x) = 7 \]

### Step 5: Solving the Equation
Now, we solve the equation:
\[ \frac{x}{2} + 10 - x = 7 \]
\[ \frac{x}{2} - x = 7 - 10 \]
\[ \frac{x}{2} - \frac{2x}{2} = -3 \]
\[ -\frac{x}{2} = -3 \]
\[ x = 6 \text{ ml} \]

### Step 6: Finding Volume of N₂
Now, we can find the volume of N₂:
\[ \text{Volume of } N_2 = 10 - x = 10 - 6 = 4 \text{ ml} \]

### Step 7: Calculating Mole Percent of CO
The mole percent of CO in the mixture can be calculated using the formula:
\[ \text{Mole percent of CO} = \left( \frac{\text{Volume of CO}}{\text{Total volume}} \right) \times 100 \]
Substituting the values:
\[ \text{Mole percent of CO} = \left( \frac{6}{10} \right) \times 100 = 60\% \]

### Final Answer
The mole percent of CO in the mixture is **60%**.

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