Question

When \(1\ dm^3\) of \(CO_2\) gas is passed over hot coke, the volume of gaseous mixture after complete reaction at STP becomes \(1.4\ dm^3\). The composition of the gaseous mixture at STP is:

• A. \(0.6\ dm^3\) of \(CO\), \(0.8\ dm^3\) of \(CO_2\)
• B. \(0.6\ dm^3\) of \(CO\), \(0.9\ dm^3\) of \(CO_2\)
• C. \(0.8\ dm^3\) of \(CO\), \(0.6\ dm^3\) of \(CO_2\)
• D. \(0.8\ dm^3\) of \(CO\), \(0.7\ dm^3\) of \(CO_2\)

Hint:

For gas volume problems at same temperature and pressure, use mole ratio directly as volume ratio.

Answer 1

The Correct Option is C

Step 1: Write the reaction.
When carbon dioxide is passed over hot coke, the reaction is: \[ \ce{CO2 + C -> 2CO}. \]

Step 2: Let volume of \(CO_2\) reacted be \(x\ dm^3\).
Initial volume of \(CO_2\) is: \[ 1\ dm^3. \] If \(x\ dm^3\) of \(CO_2\) reacts, then unreacted \(CO_2\) is: \[ 1-x. \]

Step 3: Calculate volume of CO formed.
From the reaction: \[ 1\text{ volume of }CO_2 \rightarrow 2\text{ volumes of }CO. \] So \(x\ dm^3\) of \(CO_2\) gives: \[ 2x\ dm^3 \] of \(CO\).

Step 4: Use final volume.
Final gaseous mixture contains: \[ (1-x)\ dm^3 \text{ of } CO_2 \] and: \[ 2x\ dm^3 \text{ of } CO. \] Total volume: \[ (1-x)+2x. \] Given total volume: \[ 1.4\ dm^3. \] So: \[ 1-x+2x=1.4. \] \[ 1+x=1.4. \] \[ x=0.4. \]

Step 5: Find final volumes.
Volume of \(CO\): \[ 2x=2(0.4)=0.8\ dm^3. \] Volume of unreacted \(CO_2\): \[ 1-x=1-0.4=0.6\ dm^3. \] Therefore, the gaseous mixture contains: \[ 0.8\ dm^3 \text{ of } CO \] and: \[ 0.6\ dm^3 \text{ of } CO_2. \]