What will be the enthalpy of formation of benzene in $kJ$ for the reaction: $6C_{(s)} + 3H_{2(g)} \rightarrow C_6H_{6(l)}$? Given that the enthalpy of combustion of benzene is $-3268 \ kJ$, the enthalpy of formation of $CO_{2(g)}$ is $-393.5 \ kJ$, and the enthalpy of formation of $H_2O_{(l)}$ is $-285.8 \ kJ$.

Consider the following data:
$(i)$ $2Al(s) + 6HCl(aq) \to Al_2Cl_6(aq) + 3H_2(g) + 1200 \text{ kJ/mol}$
(ii) $H_2(g) + Cl_2(g) \to 2HCl(g) + 164 \text{ kJ/mol}$
(iii) $HCl(g) + aq \to HCl(aq) + 83 \text{ kJ/mol}$
(iv) $Al_2Cl_6(s) + aq \to Al_2Cl_6(aq) + 663 \text{ kJ/mol}$
The enthalpy of formation of anhydrous solid $Al_2Cl_6$ is:

The value of the enthalpy of neutralization of an acid and a base is significant only when $......$

Calculate $\Delta H$ in $kJ$ for the following reaction:
$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$
Given that:
$H_2O_{(g)} + C_{(s)} \longrightarrow CO_{(g)} + H_{2(g)} ; \Delta H = +131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(g)} ; \Delta H = -242 \ kJ$

Which of the following statements is correct about the heat of combustion?

On the basis of the following reactions,which one is correct?
$C_{(gr)} + O_{2_{(g)}} \to CO_{2_{(g)}}, \Delta H = x \ kJ/mol$
$C_{(gr)} + \frac{1}{2} O_{2_{(g)}} \to CO_{(g)}, \Delta H = y \ kJ/mol$
$CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \to CO_{2_{(g)}}, \Delta H = z \ kJ/mol$