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: Understanding the Question:
This is a stoichiometry problem involving gaseous reactants and products. $CO_2$ gas reacts with solid carbon (coke) to form Carbon Monoxide ($CO$). We are given the initial volume of $CO_2$ and the total volume of the gas mixture after the reaction. We must find the individual volumes of $CO_2$ and $CO$ in that final mixture.
Step 2: Key Formula or Approach:
The chemical equation is: $CO_2(g) + C(s) \rightarrow 2CO(g)$. According to Avogadro's Law, at constant temperature and pressure, the volume of a gas is proportional to its moles. Therefore, we can treat volumes as stoichiometric ratios. Crucially, the solid Carbon ($C(s)$) has negligible volume and is not part of the gas mixture calculation. Step 3: Detailed Explanation:

Initial Volume: $V_{CO_2} = 1 dm^3$.
Let $x$ be the volume of $CO_2$ that reacts.
Volumes remaining/formed: - Volume of $CO_2$ left $= (1 - x) dm^3$. - Volume of $CO$ produced $= 2x dm^3$ (Stoichiometry is $1:2$).
Total volume of mixture: \[ V_{total} = (1 - x) + 2x = 1.4 \] \[ 1 + x = 1.4 \] \[ x = 0.4 dm^3 \]
Final Composition: - Volume of $CO_2$ remaining $= 1 - 0.4 = 0.6 dm^3$. - Volume of $CO$ formed $= 2(0.4) = 0.8 dm^3$.
This matches option (A).
Step 4: Final Answer:
The composition of the mixture is $0.8 dm^3$ of $CO$ and $0.6 dm^3$ of $CO_2$.