Discussion Forum

At 300 K and I atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O₂ by volume for complete combustion. After combustion, the gases occupy 330 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is

Answer:

Explanation:

$C_{x}H_{y}(g) + \left(x+\frac{y}{4}\right) O_{2}(g)\rightarrow xCO_{2}(g)+\frac{y}{2}H_{2}O(l)$

O₂ used = 20% of 375 = 75 ml

Inert part of the air = 80% of 375 = 300 ml

Total volume of gases = CO₂ + inert pair of air =30+300=330 ml

$\frac{x}{1}=\frac{30}{15}$

x=2

$\frac{x+\frac{y}{4}}{1}=\frac{75}{15}$

$x+\frac{y}{4}=5$

x=2,y=12

ie C₂H₁₂