For complete neutralization of $HCl$ with $NaOH,$ the heat of neutralization is ...... $kJ \ mol^{-1}$

If $C$ (diamond) $\rightarrow C$ (graphite) $+ X \ kJ \ mol^{-1}$,$C$ (diamond) $+ O_{2(g)} \rightarrow CO_{2(g)} + Y \ kJ \ mol^{-1}$,and $C$ (graphite) $+ O_{2(g)} \rightarrow CO_{2(g)} + Z \ kJ \ mol^{-1}$,at constant temperature,then which of the following relations is correct?

The enthalpy of combustion at $25\,^{\circ}C$ of $H_2$,cyclohexene $(C_6H_{10})$ and cyclohexane $(C_6H_{12})$ are $-241$,$-3800$ and $-3920 \ kJ/mol$ respectively. The heat of hydrogenation of cyclohexene is.....$kJ/mol$.

Calculate the standard enthalpy change of the following reaction: $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_{2}O_{(\ell)}$ if $\Delta_{f} H^{\circ}(CH_{4}) = -75 \ kJ \ mol^{-1}$,$\Delta_{f} H^{\circ}(CO_{2}) = -390 \ kJ \ mol^{-1}$,and $\Delta_{f} H^{\circ}(H_{2}O) = -286 \ kJ \ mol^{-1}$.

$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(g)}$
$B.E. (H-H) = x_1$; $B.E. (O=O) = x_2$;
$B.E. (O-H) = x_3$
Heat of vaporisation of water $= x_4$,then $\Delta H_f$ [heat of formation of liquid water] is:

Based on the following thermochemical reactions:
$H_2O_{(g)} + C_{(s)} \rightarrow CO_{(g)} + H_{2(g)} ; \Delta H = 131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)} ; \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(g)} ; \Delta H = -242 \ kJ$
$C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} ; \Delta H = x \ kJ$
The value of $x$ will be: