At 300K and 1atm,10 mL of a hydrocarbon requires 55 mL of O₂ for complete combustion and 40 mL of CO₂ is formed. The hydrocarbon is:
(All measurements are made at same P and Z)

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C₂H₈

C₄H₆

C₄H₁₀

C₅H₁₀

answer is B.

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Detailed Solution

Let the hydrocarbon be C_aH_b
$\begin{array}{l} C_{a} H_{b} + (a + \frac{b}{4}) O_{2} ⟶ {aCO}_{2} + \frac{b}{2} H_{2} O \\ 10 55 \\ - - 10 a \\ 10 \times [a + \frac{b}{4}] = 55 \\ and 10 a = 40 \\ ∴ a = 4 and b = 6 \end{array}$

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