When `CO_(2)(g)` is passed over red hot coke it partially gets reduced to `CO(g)`. Upon passing `0.5` litre of `CO_(2)(g)` over red hot coke, the total volume of the gases increased to `700mL`. The composition of the gaseous mixture at `STP` is :-

Options

A. `CO_(2)=200mL:CO=500mL`
B. `CO_(2)=350mL:CO=350mL`
C. `CO_(2)=0.0mL:CO=700mL`
D. `CO_(2)=300mL:CO=400mL`

Answer

To solve the problem, we need to analyze the reaction that occurs when carbon dioxide (CO₂) is passed over red hot coke (which is primarily carbon, C). The reaction can be represented as follows:

\[ \text{CO}_2(g) + \text{C}(s) \rightleftharpoons 2\text{CO}(g) \]

### Step-by-Step Solution:

1. **Identify the Initial Volume of CO₂**:
- We start with 0.5 liters (or 500 mL) of CO₂ gas.

2. **Determine the Final Volume of Gases**:
- After passing the CO₂ over red hot coke, the total volume of gases is 700 mL.

3. **Calculate the Volume Increase**:
- The increase in volume is given by:
\[ \text{Volume Increase} = \text{Final Volume} - \text{Initial Volume} = 700 \, \text{mL} - 500 \, \text{mL} = 200 \, \text{mL} \]

4. **Determine the Volume of CO Produced**:
- According to the reaction, for every 1 volume of CO₂ that reacts, 2 volumes of CO are produced. Let \( x \) be the volume of CO₂ that reacts.
- The volume of CO produced will be \( 2x \).
- The total volume of gases after the reaction can be expressed as:
\[ \text{Total Volume} = \text{Unreacted CO}_2 + \text{CO produced} = (0.5 - x) + 2x \]
- Setting this equal to the final volume:
\[ 0.5 - x + 2x = 0.7 \]
- Simplifying this equation:
\[ 0.5 + x = 0.7 \]
\[ x = 0.7 - 0.5 = 0.2 \, \text{L} \]

5. **Calculate the Volume of CO₂ that Reacted**:
- From our calculation, \( x = 0.2 \, \text{L} \) of CO₂ reacted.

6. **Calculate the Volume of CO Produced**:
- Since 1 volume of CO₂ produces 2 volumes of CO:
\[ \text{Volume of CO produced} = 2 \times 0.2 = 0.4 \, \text{L} \, (or \, 400 \, \text{mL}) \]

7. **Calculate the Volume of Unreacted CO₂**:
- The volume of unreacted CO₂ is:
\[ \text{Unreacted CO}_2 = 0.5 - 0.2 = 0.3 \, \text{L} \, (or \, 300 \, \text{mL}) \]

8. **Final Composition of Gaseous Mixture**:
- The final composition of the gaseous mixture is:
- Volume of CO₂ = 300 mL
- Volume of CO = 400 mL

### Summary of the Composition:
- Total volume of gases = 700 mL
- Composition:
- CO₂: 300 mL
- CO: 400 mL