QJEEM21Ch9d9f: JEE Main 2021 Chemistry - Class 11 JEEM21Ch9d9f Solution: The formula of a gaseous hydrocarbon which requires 6 times of…

Combustion reaction :

$$C_x{H}_y(g)+\left(x+\frac{y}{4}\right)O_2(g)\to xC{O}_2(g)+\frac{y}{2}{H}_{2}O(l)$$

Suppose, volume of C_xH_y is V and volume of O₂ is 6 times greater than C_xH_y = 6V

then volume of xCO₂ $$\Rightarrow$$ V_x = 4 V

x = 4

Since, $$V_{O_2}=6\times V_{C_x{H}_y}$$

$$V\left(x+\frac{y}{4}\right)$$ = 6V

$$\left(x+\frac{y}{4}\right)=6$$ ..... (i)

Put value of x = 4 in Eq. (i) we get,

$$4+\frac{y}{4}=6\Rightarrow y=8$$