Question:
One mole of $O_2$ (g) was passed over hot coke. At the end of the reaction, 40 % of $O_2$ (g) was unreacted. What is the volume (in L) of reaction mixture at STP (273.15 K and 1 bar)? (Assume only CO (g) is formed in the reaction)
One mole of $O_2$ (g) was passed over hot coke. At the end of the reaction, 40 % of $O_2$ (g) was unreacted. What is the volume (in L) of reaction mixture at STP (273.15 K and 1 bar)? (Assume only CO (g) is formed in the reaction)
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Calculate remaining reactants and product moles, then use molar volume at STP.
Updated On: Jun 10, 2026
- 22.7
- 72.64
- 36.32
- 45.4
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The Correct Option is C
Solution and Explanation
Step 1: Concept
Stoichiometry: $O_2(g) + 2C(s) \rightarrow 2CO(g)$.
Step 2: Analysis
1 mole $O_2$ initially. 40 Reacted $O_2 = 0.6$ moles. CO produced = $2 \times 0.6 = 1.2$ moles. Total moles = $0.4 + 1.2 = 1.6$ moles. Volume at STP (1 bar, 273.15 K) $\approx 22.7$ L/mol. Total Volume = $1.6 \times 22.7 = 36.32$ L.
Step 3: Conclusion
The volume is 36.32 L.
Final Answer: (C)
Stoichiometry: $O_2(g) + 2C(s) \rightarrow 2CO(g)$.
Step 2: Analysis
1 mole $O_2$ initially. 40 Reacted $O_2 = 0.6$ moles. CO produced = $2 \times 0.6 = 1.2$ moles. Total moles = $0.4 + 1.2 = 1.6$ moles. Volume at STP (1 bar, 273.15 K) $\approx 22.7$ L/mol. Total Volume = $1.6 \times 22.7 = 36.32$ L.
Step 3: Conclusion
The volume is 36.32 L.
Final Answer: (C)
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