NEET 2026 (Phase 1) · ChemistryPrevious Year Question
When 1 dm$^3$ of \ce{CO2} gas is passed over hot coke, the volume of the 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$^3$ \ce{CO}, 0.8 dm$^3$ \ce{CO2}
- B.0.8 dm$^3$ \ce{CO}, 0.6 dm$^3$ \ce{CO2}✓
- C.0.6 dm$^3$ \ce{CO}, 0.8 dm$^3$ \ce{CO2}
- D.0.6 dm$^3$ \ce{CO}, 0.4 dm$^3$ \ce{CO2}
Correct Answer
(B) 0.8 dm$^3$ \ce{CO}, 0.6 dm$^3$ \ce{CO2}
Solution & Explanation
\textbf{Answer:} (B) 0.8 dm$^3$ \ce{CO}, 0.6 dm$^3$ \ce{CO2} \textbf{Solution:} \ce{CO2 + C(s) -> 2CO}. By Gay-Lussac's law, volumes behave like moles. Let $x$ dm$^3$ of \ce{CO2} react. Remaining \ce{CO2} $= 1-x$; \ce{CO} formed $= 2x$. Total volume $= (1-x) + 2x = 1 + x = 1.4 \Rightarrow x = 0.4$. \ce{CO} $= 2(0.4) = 0.8$ dm$^3$; \ce{CO2} left $= 1 - 0.4 = 0.6$ dm$^3$.
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