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

Volume of \(O_{2}\) used

\(=375 \times \frac{20}{100}=75 m l\)

From the reaction of combination

\(m l C_{x} H_{y}\) requires \(=\frac{4 x+y}{4} m l O_{2}\)

\(15\left(\frac{4 x+y}{4}\right)=75\)

So, \(4 x+y=20\)

\(x=3\)

\(y=8\)

\(\therefore C_{3} H_{8}\)