Heat of combustion of $C_2{H}_4$ is  -337 kcal. If 11.2lit $O_2$ is used at STP, in the combustion, calculate heat liberated in K cal

${\mathrm{C}}_2{\mathrm{H}}_4+3{\mathrm{O}}_2\to 2{\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}+337\mathrm{kcal}$

$\begin{array}{l}1\mathrm{mole}\mathrm{of}{\mathrm{C}}_2{\mathrm{H}}_4\mathrm{required}3\times 22.4\mathrm{lt}\mathrm{of}{\mathrm{O}}_2\mathrm{at}\mathrm{STP}\\ \mathrm{Relases}337\mathrm{Kcal}\end{array}$

$\begin{array}{l}\mathrm{If}11.2\mathrm{lt}{\mathrm{O}}_2\mathrm{is}\mathrm{used},\mathrm{heat}\mathrm{liberated}=\frac{337}{3\times 22.4}\times 11.2\mathrm{Kcal}\\ =\frac{337}{6}\\ =56\mathrm{Kcal}\end{array}$