Q101 / 180 · NEET 2026 (Cancelled) · Chemistry
When $1$ dm$^3$ of $\text{CO}_2$ 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$0.8$ dm$^3$ of CO, $0.8$ dm$^3$ of $\text{CO}_2$
- Correct answerB$0.8$ dm$^3$ of CO, $0.6$ dm$^3$ of $\text{CO}_2$
- C$0.6$ dm$^3$ of CO, $0.8$ dm$^3$ of $\text{CO}_2$
- D$0.6$ dm$^3$ of CO, $0.4$ dm$^3$ of $\text{CO}_2$
Solution
The reaction of carbon dioxide with hot coke is given by:
$\text{CO}_2(g) + \text{C}(s) \rightarrow 2\text{CO}(g)$
Initial volume of $\text{CO}_2 = 1$ dm$^3$
Let $x$ dm$^3$ of $\text{CO}_2$ react with hot coke.
Volume of $\text{CO}_2$ remaining $= 1 - x$ dm$^3$
Volume of $\text{CO}$ formed $= 2x$ dm$^3$
Total volume of the gaseous mixture $= (1 - x) + 2x = 1 + x$ dm$^3$
Given that the total volume after the reaction is $1.4$ dm$^3$:
$1 + x = 1.4$
$x = 0.4$ dm$^3$
Volume of $\text{CO}_2$ remaining $= 1 - 0.4 = 0.6$ dm$^3$
Volume of $\text{CO}$ formed $= 2 \times 0.4 = 0.8$ dm$^3$
The composition of the gaseous mixture is $0.8$ dm$^3$ of CO and $0.6$ dm$^3$ of $\text{CO}_2$.
Answer: $0.8$ dm$^3$ of CO, $0.6$ dm$^3$ of $\text{CO}_2$
$\text{CO}_2(g) + \text{C}(s) \rightarrow 2\text{CO}(g)$
Initial volume of $\text{CO}_2 = 1$ dm$^3$
Let $x$ dm$^3$ of $\text{CO}_2$ react with hot coke.
Volume of $\text{CO}_2$ remaining $= 1 - x$ dm$^3$
Volume of $\text{CO}$ formed $= 2x$ dm$^3$
Total volume of the gaseous mixture $= (1 - x) + 2x = 1 + x$ dm$^3$
Given that the total volume after the reaction is $1.4$ dm$^3$:
$1 + x = 1.4$
$x = 0.4$ dm$^3$
Volume of $\text{CO}_2$ remaining $= 1 - 0.4 = 0.6$ dm$^3$
Volume of $\text{CO}$ formed $= 2 \times 0.4 = 0.8$ dm$^3$
The composition of the gaseous mixture is $0.8$ dm$^3$ of CO and $0.6$ dm$^3$ of $\text{CO}_2$.
Answer: $0.8$ dm$^3$ of CO, $0.6$ dm$^3$ of $\text{CO}_2$
Hence, the correct answer is option: 'B'.