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

The reaction is as follow;

$$\begin{gathered} \underset{\frac{27}{12}}{\mathrm{C}}+\underset{\frac{88}{32}}{{\mathrm{O}}_2}\to {\mathrm{CO}}_2 \\ \mathrm{n}=2.25,\mathrm{n}=2.75 \end{gathered}$$

We can observe that the limiting reagent in this case is C.

Moles of ${\mathrm{CO}}_2$ in produced= Moles of C $=\frac{27}{12}=2.25$

Volume of ${\mathrm{CO}}_2$ at STP$=2.25\times 22.4=50.4\mathrm{L}$

Ratio of C and O in ${\mathrm{CO}}_2$$=12:32=3:8$

Moles of unreacted ${\mathrm{O}}_2$​=2.75−2.25=0.5 mol.
Volume of unreacted ${\mathrm{O}}_2$​ at STP =0.5×22.4=11.2 L.