Heat of combustion of C₂H₄ is -337 K.Cal. If 5.6 lit O₂ is used at STP in the combustion, heat liberated is ___ K.Cal

C₂H₄ + 3O₂ $\to$2CO₂ + 3H₂O; $∆\mathrm{H}$ = - 337 Kcal

3 $\times$22.4 lit of O₂ liberates 337 KCal

5.6 lit of O₂ = ?

heat liberated = $\frac{337\times 5.6}{3\times 22.4}$=$\frac{337}{12}$=28.08 Kcal