On repeated sparking, `10 mL` of a mixture of carbon monoxide and nitrogen required `7 mL` of oxygen for combustion. What was the volume of nitrogen? (All volumes are measured under identical conditions).

Options

A. `7//2 mL`
B. `4 mL`
C. `7 mL`
D. `17//2 mL`

Explanation

To solve the problem, we need to determine the volume of nitrogen in a mixture of carbon monoxide (CO) and nitrogen (N2) that requires 7 mL of oxygen (O2) for combustion.

### Step-by-Step Solution:

1. **Identify the Total Volume of the Mixture**:
The total volume of the gas mixture is given as 10 mL. This mixture consists of carbon monoxide and nitrogen.

2. **Define Variables**:
Let the volume of carbon monoxide be \( x \) mL. Therefore, the volume of nitrogen will be \( 10 - x \) mL, since the total volume is 10 mL.

3. **Write the Combustion Reactions**:
The combustion reactions are as follows:
- For carbon monoxide:
\[ \text{CO} + \frac{1}{2} \text{O}_2 \rightarrow \text{CO}_2 \]
This means that 1 volume of CO requires 0.5 volumes of O2.

- For nitrogen:
\[ \text{N}_2 + \text{O}_2 \rightarrow 2 \text{NO} \]
This means that 1 volume of N2 requires 1 volume of O2.

4. **Calculate the Volume of Oxygen Required**:
- The volume of oxygen required for \( x \) mL of CO is:
\[ \frac{x}{2} \text{ mL of O}_2 \]
- The volume of oxygen required for \( 10 - x \) mL of N2 is:
\[ 10 - x \text{ mL of O}_2 \]

5. **Set Up the Equation**:
The total volume of oxygen required for combustion is given as 7 mL. Therefore, we can set up the equation:
\[ \frac{x}{2} + (10 - x) = 7 \]

6. **Solve the Equation**:
- Combine the terms:
\[ \frac{x}{2} + 10 - x = 7 \]
- Multiply through by 2 to eliminate the fraction:
\[ x + 20 - 2x = 14 \]
- Rearranging gives:
\[ -x + 20 = 14 \]
- Solving for \( x \):
\[ -x = 14 - 20 \]
\[ -x = -6 \implies x = 6 \text{ mL} \]

7. **Calculate the Volume of Nitrogen**:
Since \( x \) represents the volume of carbon monoxide, the volume of nitrogen is:
\[ 10 - x = 10 - 6 = 4 \text{ mL} \]

### Final Answer:
The volume of nitrogen in the mixture is **4 mL**.

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