The formula of a gaseous hydrocarbon which requires 6 times of its...

Combustion reaction.

${C}_{x}{H}_{y ( g )} + \left(x + \dfrac{y}{4}\right){O}_{2}\left(g\right) \to {xCO}_{2}\left(g\right) + \dfrac{y}{2}{H}_{2}O\left(ℓ\right)$

$\begin{cases}V & 6V & - \\ - & - & Vx = 4V\end{cases}$

$⇒ x = 4$

Since: \left(I\right){Vo}_{2} = 6 \times {V}_{{C}_{x}{H}_{y}}$</p><p>$⇒ V\left(x + \dfrac{y}{4}\right) = 6V$</p><p>$⇒ \left(x + \dfrac{y}{4}\right) = 6 ⇒ 4 + \dfrac{y}{4} = 6$</p><p>$⇒ y = 8