The heat of neutralization of a strong acid by a strong base is nearly equal to

Using the data provided,calculate the bond energy $(kJ \ mol^{-1})$ of a $C \equiv C$ bond in $C_{2}H_{2}$. (Take the bond energy of a $C-H$ bond as $350 \ kJ \ mol^{-1}$)
$2C_{(s)} + H_{2(g)} \longrightarrow C_{2}H_{2(g)} \quad \Delta H = 225 \ kJ \ mol^{-1}$
$2C_{(s)} \longrightarrow 2C_{(g)} \quad \Delta H = 1410 \ kJ \ mol^{-1}$
$H_{2(g)} \longrightarrow 2H_{(g)} \quad \Delta H = 330 \ kJ \ mol^{-1}$

Which of the following reactions represents the standard heat of formation of the product?

The standard heat of formation,in $kcal / mol$ of $Ba^{2+}$ is
[Given : standard heat of formation of $SO_4^{2-}$ ion $(aq) = -216 \ kcal / mol$,
Standard heat of crystallisation of $BaSO_{4(s)} = -4.5 \ kcal / mol$,standard heat of formation of $BaSO_{4(s)} = -349 \ kcal / mol$]

The equation $\frac{1}{2} H_2 + \frac{1}{2} Cl_2 \to HCl$ $(\Delta H_{298} = -22.060 \ kcal)$ means:

Calculate the heat of formation for propene $(C_3H_6)$ using the following thermochemical equations:
$(i) C_{(s)} + O_{2(g)} \to CO_{2(g)}; \Delta H_1 = -94.05 \ k.cal/mole$
$(ii) H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(l)}; \Delta H_2 = -68.32 \ k.cal/mole$
$(iii) C_3H_{6(g)} + \frac{9}{2} O_{2(g)} \to 3 CO_{2(g)} + 3 H_2O_{(l)}; \Delta H_3 = -499.7 \ k.cal/mole$
(Note: The original question provided propane combustion data; assuming propene combustion data for consistency).