Mass of steam \(=0.392 g\)
Molar mass \(=18 gmol ^{-1}\)
Mole \(=(0.392 / 18) mol =0.02178 mol\)t STP, 1mole of any gas occupies \(22400 cm ^{3}\)
Thus, \(0.02179 mol\) of \(H _{2} O\) has
\((0.02178 \times 22400) cm ^{3}=488 cm ^{3}\)
Volume of hydrocarbon \(=1500 cm ^{3}\)
Volume of \(CO _{2}=450 cm ^{3}\)
The combustion of any hydrocarbon goes as:
\(C _{x} H _{y}+\left(x+\frac{ y }{4}\right) O _{2} \rightarrow x CO _{2}+\frac{y}{2} H _{2} O\)rom Avogadro's law, we can
write:
\(\frac{n C _{x} H_{y}}{V_{C_{x}} H_{y}}=\frac{n_{ CO _{2}}}{V_{ CO _{2}}}=\frac{n_{ H _{2} O }}{V_{ H _{2} O }}\)
NB: The equation will only include \(H _{2} O\) if the question does not state that the residual gas was cooled. This is because when steam is cooled, it condenses into the liquid form and as such cannot be treated with any gas law any further.
Now, we have:
\(n C_{x} H_{y}=1, n_{C_{2}}=x, n_{i_{2} o}=\frac{y}{2}\)
\(V C_{x} H_{y}=1500 cm ^{3}\),
\(V_{ CO _{2}}=4500 cm ^{2}\),
\(V_{H_{2} O }=488 cm ^{3}\).
Thus,
\begin{aligned}
\frac{1}{150} &=\frac{x}{450} ; 150 x=450, \\
x &=\frac{450}{150}=3 \\
& \text { Also, } \\
\frac{1}{150} &=\frac{y / 2}{488} \\
\frac{y}{2}(150) &=488, \\
y &=\frac{488 \times 2}{150}=6.51 \\
\text { Then, } \\_{x} H_{y} &=C_{3} H_{7}
\end{aligned}