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12 mL of a hydrocarbon is exploded with 50 mL of excess oxygen. After cooling, the volume is 32 mL. After treatment with KOH, the volume reduces to 8 mL. Determine the molecular formula of the hydrocarbon.

1. C3H8
2. C3H6
3. C2H4
4. C2H6

Correct answer: C2H6

Solution

Let hydrocarbon be CxHy. Combustion: CxHy + (x + y/4)O2 -> x CO2 + (y/2) H2O. Volumes at constant T,P (H2O condensed on cooling): CO2 = 24 mL (from 12 mL CxHy, so x=2). O2 used = initial O2 - remaining O2. Remaining after KOH = 8 mL (O2). Total gas at start = 50+12 = 62 mL. After explosion (water condensed): 32 mL (CO2 + O2). CO2=24, O2 remaining = 8. O2 used = 50 - 8 = 42 mL. From 12 mL: O2/CxHy = 42/12 = 3.5. So (x+y/4) = 3.5. With x=2: y/4 = 1.5 => y = 6. Formula: C2H6.

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