The volume of oxygen at STP required to burn 2.4g of carbon completely is-  

$\begin{array}{l}\mathrm{C}\left(\mathrm{S}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{CO}}_2\left(\mathrm{g}\right)\\ \text{ moles }=1\text{ mole }1\text{ mole }1\text{ mole }\\ \text{ weight }=12\mathrm{gm}32\mathrm{gm}44\mathrm{gm}\\ 12\mathrm{gm}\text{ of }\mathrm{C}\text{ requires }\to 1\text{ mole of }{\mathrm{O}}_2\\ \therefore 2.4\mathrm{gm}\text{ of }\mathrm{C}\text{ will require }\\ \to \frac{1}{12}\times 2.4\text{ mole of }{\mathrm{O}}_2\\ \text{ volume of }2.4/12\text{ mole }{\mathrm{O}}_2\text{ at }\mathrm{STP}\\ =\frac{22.4\times 2.4}{12}\text{ litre }\\ =4.48\text{ litre }\end{array}$