$\underset{\text{(Propene)} (3 \times 12 +6 =42)}{C _{3} H _{6}}+\frac{9}{2} O _{2} \longrightarrow 3 CO _{2}+\underset{(3 \times 18=54)}{3 H _{2} O }$
Given, $H _{2} O ($ formed $)=27\, g$
To find the mass of propene,
The mass ratio between $C _{3} H _{6}$ and $H _{2} O$
$=42: 54$
$\because 54\, g$ of $H _{2} O$ require $=42\, g C _{3} H _{6} $
$\therefore 27\, g$ of $H _{2} O$ require
$= \frac{42 \times 27}{54}=21\, g \cdot C _{3} H _{6}$