$\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 $(C_3H_8)$