Enthalpy of formation of methane is $-75 \ kJ / mol$. What is the enthalpy change for formation of $24 \ g$ of methane (in $kJ$)?

From the following data at $25^{\circ} C$,calculate the $\Delta_{r} H^0$ for the reaction $H_2O_{(g)} \rightarrow 2 H_{(g)} + O_{(g)}$:

$1/2 H_{2(g)} + 1/2 O_{2(g)} \rightarrow OH_{(g)}$	$\Delta H = 42.09 \ kJ \ mol^{-1}$
$H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(g)}$	$\Delta H = -242 \ kJ \ mol^{-1}$
$H_{2(g)} \rightarrow 2 H_{(g)}$	$\Delta H = 436 \ kJ \ mol^{-1}$
$O_{2(g)} \rightarrow 2 O_{(g)}$	$\Delta H = 496 \ kJ \ mol^{-1}$

The enthalpies of formation of $Al_{2}O_{3}$ and $Cr_{2}O_{3}$ are $-1596 \ kJ$ and $-1134 \ kJ$ respectively. $\Delta H$ for the reaction $2Al + Cr_{2}O_{3} \to 2Cr + Al_{2}O_{3}$ is.......$kJ$

The bond enthalpies of heavy hydrogen $(D-D)$, oxygen $(O=O)$, and heavy water $(D-O)$ are $+400$, $+498$, and $+490 \ kJ \ mol^{-1}$, respectively. The $\Delta_{r} H^{\circ}$ of the reaction to produce $D_2O$ is:

$XeF_{2(g)} + H_{2(g)} \to 2HF_{(g)} + Xe_{(g)}$,$\Delta H^o = -430 \ kJ$
Bond energy:
$H-H = 435 \ kJ/mol$
$H-F = 565 \ kJ/mol$
Calculate the average bond energy of the $Xe-F$ bond in $kJ/mol$.

At $298 \ K$ and constant pressure, the heat of formation of $C_2H_{2(g)}$ and $C_6H_{6(g)}$ is $230 \ kJ \ mol^{-1}$ and $85 \ kJ \ mol^{-1}$ respectively. What will be the change in heat for the reaction?
$3 \ C_2H_{2(g)} \rightarrow C_6H_{6(g)}$