Reaction-enthalpy for synthesis of 40 g hydrazine


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> Calculate the reaction-enthalpy for the synthesis of 40 g hydrazine (rocketfuel): $$\ce{4NH3(g) + Cl2 (g) -> N2H4(l) + 2 NH4Cl(s)}$$
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**Attempt at solution:** 


First we have to account for the decomposition of four moles of ammonia into its elements. 


Since $\Delta H\_\mathrm f^o(\ce{NH3})$ = $-46.1 \mathrm{~kJ~mol^{-1}}$, we do: 


$$\Delta H\_1 = 4(+46.1) = 184.4 \mathrm{~kJ}$$ 


$\ce{Cl2}$ is in its standard state so we can ignore that. 


We know that $\Delta H\_f^o(\ce{N2H4})$ = $50.6\mathrm{~kJ~mol^{-1}}$. Therefore, $\Delta H\_2 = \pu{50.6 kJ}$. 


Then, $\Delta H\_\mathrm f^o(\ce{NH4Cl})$ = $-314.2\mathrm{~kJ~mol^{-1}}$ to give


$$\Delta H\_3 = 2(-314.4) = -628.8~\mathrm{kJ}$$


Finally, 
\begin{align}
\Delta H\_{\mathrm{rxn}} &= \Delta H\_1 + \Delta H\_2 + \Delta H\_3\\
&= 184.4 + 50.6 + (-628.8) = -393.8~\mathrm{kJ}
\end{align}


However, my textbook says the correct answer is $-492.3~\mathrm{kJ}$.


Can someone tell me what I did wrong, please? Also, I didn't account for the $40\mathrm{~g}$. How does that affect the problem?




You are on the right track. Basically, you are just missing the last step. Your answer for the enthalpy of reaction should be $ΔH\_{rxn}=−393.8\frac{\mathrm{kJ}}{\mathrm{mol}}$. So all you have to do is find the moles of hydrazine  

$n=\frac{40}{32.0452}=1.248\space\mathrm{moles}$ and multiply it by your $ΔH\_{rxn}$.


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