Question

10mL $C_xH_y$ reacts with 80ml of excess $O_2$, after rxn volume of gases left is 70ml. These gases passes thorough KOH solution, 50ml of gas finally comes out. Find formula of compound.

• A. CH4
• B. C2H4
• C. C2H6
• D. C3H8

Answer 1

Answer: (B)

C2H4

Answer 2

The balanced combustion equation for a hydrocarbon $C_xH_y$ is: $C_xH_y(g) + (x + y/4)O_2(g) \rightarrow xCO_2(g) + (y/2)H_2O(l)$

1. Volume of $O_2$ reacted: Initial volume of $O_2$ = 80 mL Volume of gas after reaction and cooling = 70 mL Volume of gas after passing through KOH = 50 mL (This is unreacted $O_2$) Volume of $O_2$ reacted = Initial $O_2$ - Unreacted $O_2$ = 80 mL - 50 mL = 30 mL.

2. Volume of $CO_2$ produced: KOH absorbs $CO_2$. The decrease in volume after passing through KOH is the volume of $CO_2$ produced. Volume of $CO_2$ produced = Volume of gas before KOH - Volume of gas after KOH Volume of $CO_2$ produced = 70 mL - 50 mL = 20 mL.

3. Determining x and y using Gay-Lussac's Law: From the balanced equation, 10 mL of $C_xH_y$ produces 20 mL of $CO_2$. By Gay-Lussac's Law, the ratio of volumes is equal to the ratio of stoichiometric coefficients. For $CO_2$: $\frac{10 \text{ mL } C_xH_y}{1} = \frac{20 \text{ mL } CO_2}{x} \Rightarrow x = \frac{20}{10} = 2$.

   Also from the balanced equation, 10 mL of $C_xH_y$ reacts with 30 mL of $O_2$. For $O_2$: $\frac{10 \text{ mL } C_xH_y}{1} = \frac{30 \text{ mL } O_2 \text{ reacted}}{x + y/4} \Rightarrow x + y/4 = \frac{30}{10} = 3$. Substitute $x=2$: $2 + y/4 = 3 \Rightarrow y/4 = 1 \Rightarrow y = 4$.

4. Molecular Formula: The molecular formula is $C_xH_y = C_2H_4$.