A mixture of 26 ml of H₂ and 24ml of O₂ is exploded . After cooling the tube contains unreacted 

$\underset{ \underset{\frac{26}{2}=13\mathrm{Limiting} \mathrm{reragent} }{26\mathrm{ml}}}{2{\mathrm{H}}_2} + \underset{\underset{\frac{24}{1}=24}{24\mathrm{ml}}}{\hspace{0.17em}{\mathrm{O}}_2 } \stackrel{ }{\to } 2{\mathrm{H}}_2\mathrm{O}\hspace{0.17em}$

Volume of H₂ required double than O₂.

26 vol of H₂ reat with 13 vol of O₂ .

Vol of O₂ left = 24-13=11 vol = 11 ml .