If  27 g of carbon is mixed with 88 g of oxygen and is allowed to burn to produce  $C{O}_2$ , then:

            C         +         $O_2$     $\to$         $C{O}_2$
Mass  27                    88
Moles  $\frac{27}{12}$                 $\frac{88}{32}$
C is limiting reagent.
Moles of   $C{O}_2$  produced = moles of C =  $\frac{27}{12}=2.25$
$\therefore$  VOLUME OF $C{O}_2$ AT STP = 2.25 X 22.4 = 50.4 L
Ratio of C and O in $C{O}_2$ = 12 : 32 = 3 : 8
Moles of unreacted $O_2$ = 2.75 – 2.25 = 0.5
$\therefore$  Volume of unreacted O2 at STP = 0.5 x 22.4 = 11.2L