If 27 gram of Carbon is mixed with 88 gram of Oxygen and is allowed to burn to produce $C{O}_2$then:

 

$C+O_2\to C{O}_2$

12 gram of carbon combines with 32 gram of oxygen to form 44 gram of Carbon dioxide.

Then 27 gram of carbon combine with $\frac{27\times 32}{12}$gram of oxygen = 72 grams 

Oxygen is in excess, so oxygen is excess reagent and carbon is limiting reagent.

Combining mass ratio of  and O= 12: 32 = 3: 8

         At STP 32 gram of $O_2$ occupy = 22.4 L

Then, 72 gram of $O_2$ occupy = $\frac{72\times 22.4}{32}$$=50.4litre$

But given mass of  $O_2=88grams$

Volume of 88 grams of $O_2$at STP $=\frac{88\times 22.4}{32}=61.6litreof{O}_2$
 

$\therefore$Unreacted volume of oxygen, $O_2=61.6-50.4=11.2litre$