of gaseous hydrocarbon was burnt completely in of at STP. The volume remaining was . On treatment wi — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question
Question
$10\text{ mL}$ of gaseous hydrocarbon was burnt completely in $80\text{ mL}$ of $O_2$ at STP. The volume remaining was $70\text{ mL}$. On treatment with $KOH$, the volume decreased to $50\text{ mL}$. The formula of the hydrocarbon is:
Answer: C
💡 Solution & Explanation
$\text{Volume of } CO_2 \text{ produced} = 70 - 50 = 20\text{ mL}$. $\text{Volume of } O_2 \text{ used} = 80 - 50 = 30\text{ mL}$. <br>Reaction: $C_xH_y + (x + y/4)O_2 \rightarrow xCO_2 + (y/2)H_2O$. <br>$x = V_{CO_2} / V_{HC} = 20/10 = 2$. <br>$x + y/4 = V_{O_2} / V_{HC} = 30/10 = 3 \Rightarrow 2 + y/4 = 3 \Rightarrow y = 4$. <br>Wait, let's re-calculate: $20/10 = 2$. $O_2$ used is $30\text{ mL}$. $2 + y/4 = 3 \Rightarrow y=4$. Formula is $C_2H_4$. Option A.
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