Heat of combustion of ${\mathrm{C}}_2{\mathrm{H}}_4$ is $-337\mathrm{ }\mathrm{K}$. Cal. If $5.6$ lit ${\mathrm{O}}_2$ is used at STP in the combustion, heat liberated is ..... K.Cal

 The given reaction is ${\mathrm{C}}_2{\mathrm{H}}_4+{\mathrm{O}}_2⟶2{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}$

 Balancing the equation

 ${\mathrm{C}}_2{\mathrm{H}}_4+3{\mathrm{O}}_2⟶2{\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}$

 At STP 1 mol of gas occupies 22.L,

 ${\mathrm{C}}_2{\mathrm{H}}_4=1\times 22.4=22.4 \mathrm{L}$

 ${\mathrm{O}}_2=3\times 22.4=67.2 \mathrm{L}$

 ${\mathrm{CO}}_2=2\times 22.4=44.8 \mathrm{L}$

 ${\mathrm{H}}_2\mathrm{O}=2\times 22.4=44.8 \mathrm{L}$

 For $67.2\mathrm{ }\mathrm{L}$ of oxygen heat of combustion is $-337\mathrm{Kcal}$, then for $5.6\mathrm{ }\mathrm{L}$ is

 $\frac{-337\times 5.66}{67.2}=-28.08\mathrm{kCal}$

Hence the correct option is (A).