How much heat in $kJ$ is produced by the combustion of $15.5 \, g$ of propane? ${C_3H_8} + 5{O_2} \to 3{CO_2} + 4{H_2O}; \Delta{H^o} = -2219 \, kJ/mol$

The compounds with negative heat of formation are known as:

Calculate the enthalpy of formation of ethylene in $kJ\,mol^{-1}$ from the following data:
$C_{(graphite)} + O_{2(g)} \to CO_{2(g)} ; \Delta H = -393.5\,kJ$
$H_{2(g)} + 1/2 O_{2(g)} \to H_2O_{(l)} ; \Delta H = -286.2\,kJ$
$C_2H_{4(g)} + 3 O_{2(g)} \to 2 CO_{2(g)} + 2 H_2O_{(l)} ; \Delta H = -1410.8\,kJ$

The bond dissociation energies of $XY$,$X_2$,and $Y_2$ (all diatomic molecules) are in the ratio $1 : 1 : 0.5$. If the enthalpy of formation of $XY$ is $\Delta_fH = -200 \ kJ \ mol^{-1}$,find the bond dissociation energy of $X_2$ in $kJ \ mol^{-1}$.

If bond enthalpies of $N \equiv N$,$H - H$ and $N - H$ bonds are $x_1, x_2$ and $x_3$ respectively,$\Delta H_f^o$ for $NH_3$ will be

State Hess's Law of Constant Heat Summation.