KOH absorbs CO2. Volume of CO2 produced = 224−64=160 mL
Unreacted O2=64 mL, so O2 consumed = 264−64=200 mL
For hydrocarbon CxHy: HydrocarbonCO2=80160=2⇒x=2
O2 consumed per mole = x+4y=80200=2.5
2+4y=2.5⇒y=2
Hydrocarbon is C2H2
JEE Main 2026 — Chemistry Physical Chemistry
80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to 273 K occupy 224 mL. When the system is treated with KOH solution, the volume decreases to 64 mL. The formula of the hydrocarbon is :
Held on 21 Jan 2026 · Verified 6 Jul 2026.
C2H2
C2H4
C2H6
C4H10
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