Answer:
Option D
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