d
Since the compound undergoing combustion is an alkane. Hence the combustion reaction can be written as

\(\underset{5L}{\mathop{{{C}_{n}}{{H}_{2n+2}}}}\,+\underset{25L}{\mathop{\left( \frac{3n+1}{2} \right){{O}_{2}}}}\,\to \) \(nC{{O}_{2}}+(n+1){{H}_{2}}O\)

since volumes are measured at constant \(T\) & \(P\). Hence according to Avogadro's law 

Volume \(\propto \) mole 

\(\therefore \,{{n}_{alkane}}=\left( \frac{2}{3n+1} \right)\times {{n}_{{{O}_{2}}}}\)

\(5 = \frac{2}{{3n + 1}} \times 25\)

\(\therefore \,n = 3\)

Hence alkane is propane \((C_3H_8)\)