JEE Main 2019 — Some Basic Concepts Of Chemistry Question with Solution

C_xH_y + $$\left(x+\frac{y}{4}\right)$$O₂ $$\to$$ xCO₂ + $$\frac{y}{2}$$H₂O

According to the above equation,

1 mL of hydrocarbon = x mL of CO₂ is produced

$$\therefore$$ 10 mL of hydrocarbon = 10x mL of CO₂ is produced.

According to question,

From 10 mL of hydrocarbon 40 mL of CO₂ is produced.

$$\therefore$$ 10x = 40

$$\Rightarrow$$ x = 4

Also from the above equation,

1 mL of C_xH_y react with $$\left(x+\frac{y}{4}\right)$$ mL of O₂.

$$\therefore$$ 10 mL of C_xH_y react with 10$$\left(x+\frac{y}{4}\right)$$ mL of O₂.

According to question,

10$$\left(x+\frac{y}{4}\right)$$ = 55

$$\Rightarrow$$ $$40+\frac{10y}{4}=55$$

$$\Rightarrow$$ y = 6

$$\therefore$$ Hydrocarbon is C₄H₆