Question

15 $mL$ of a gaseous hydrocarbon required for complete combustion 357 $mL$ air (containing $21 \% O _{2}$ by volume) and the gaseous products occupied $327 mL$. What is the formula of the hydrocarbon?

Answer 1

SOLUTION : Volume of $O_{2}$ reacted $=\frac{357 \times 21}{100} mL =75 mL$

Volume of $N_{2}$ left $=(357-75) mL =282 mL$

Volume of $CO _{2}$ formed $=(327-282) mL =45 mL$

We know that

$15 mL \quad 15\left(x+\frac{y}{4}\right) mL O _{2} 15 x mL CO _{2} \quad$ oml $H _{2} O$

Vol. of $CO _{2}$ formed: $15 x=45 ; \therefore x=3$

Vol. of $O_{2}$ required: $15\left(x+\frac{y}{4}\right)=75 ; x+\frac{y}{4}=5 ; \frac{y}{4}=2 ; y=8$

Hence formula of hydrocarbon $= C _{ 3 } H _{ 8 }$.