Combustion volume ratio identifies the alkane Steps:

• General combustion: CnH2n+2 + ((3n+1)/2)O2 → nCO2 + (n+1)H2O (gaseous products at RTP).
• Volume ratio O2:alkane = 130 cm³ : 20 cm³ = 13:2.
• Set ((3n+1)/2) = 13/2, so 3n+1=13, 3n=12, n=4.
• n=4 corresponds to butane (C4H10).

Why A is correct:

• Butane's equation: C4H10 + (13/2)O2 → 4CO2 + 5H2O requires exactly 6.5 volumes O2 per volume butane, matching the data.

Why the others are wrong:

• B. Ethane (C2H6, n=2) requires 3.5 volumes O2 (too low).
• C. Methane (CH4, n=1) requires 2 volumes O2 (too low).
• D. Propane (C3H8, n=3) requires 5 volumes O2 (too low).

Final answer: A