The formula of a gaseous hydrocarbon which requires 6 times of its own volume of $$O_2$$ for complete oxidation and produces 4 times its own volume of $$CO_2$$ is $$C_xH_y$$. The value of y is ______.

Let the gaseous hydrocarbon be $$C_xH_y$$. The balanced combustion reaction is:

$$C_xH_y + \left(x + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O$$

Since all volumes are measured under the same conditions of temperature and pressure, by Avogadro's law, the ratio of volumes equals the ratio of moles for gaseous species.

From the given information, 1 volume of $$C_xH_y$$ requires 6 volumes of $$O_2$$. So:

$$x + \frac{y}{4} = 6 \quad \cdots (i)$$

Also, 1 volume of $$C_xH_y$$ produces 4 volumes of $$CO_2$$. So:

$$x = 4 \quad \cdots (ii)$$

Substituting $$x = 4$$ in equation (i):

$$4 + \frac{y}{4} = 6$$

$$\frac{y}{4} = 2$$

$$y = 8$$

The hydrocarbon is $$C_4H_8$$ and the value of $$y$$ is $$\textbf{8}$$.