Question

The formula of a gaseous hydrocarbon which requires 6 times of its own volume of O2 for complete oxidation and produces 4 times its own volume of CO2 is CxHy. The value of y is _________

Hint:

For any hydrocarbon $C_xH_y$, the volume of $O_2$ required is always $(x + y/4)$ times the volume of the hydrocarbon.

Answer 1

Correct Answer: 8

Step 1: The general combustion equation for a hydrocarbon is: $$C_xH_y + (x + \frac{y}{4})O_2 \to xCO_2 + \frac{y}{2}H_2O$$
Step 2: According to Gay-Lussac's law of combining volumes, the volume ratios are equal to the stoichiometric coefficients.
Step 3: Given $x = 4$ (volume of $CO_2$ produced).
Step 4: Given $x + \frac{y}{4} = 6$ (volume of $O_2$ required).
Step 5: Substitute $x=4$ into the oxygen equation: $4 + \frac{y}{4} = 6 \Rightarrow \frac{y}{4} = 2 \Rightarrow y = 8$. The hydrocarbon is $C_4H_8$.