Question

Does the reaction in Exercise 76 or that in Exercise 77 produce more energy per kilogram of reactant mixture (stoichiometric amounts)?

Short answer

The reaction in Exercise 77 produces more energy per kilogram of reactant mixture enthalpy.

Step-by-step solution

Step 1: Calculation of energy produced per kilogram of the reaction mixture in exercise 76

In exercise 77, 3 moles of aluminium react with 3 moles of ammonium perchlorate to produce energy:

$3\mathrm{Al}\left(\mathrm{s}\right)+3{\mathrm{NH}}_4{\mathrm{ClO}}_4\left(\mathrm{s}\right)\to {\mathrm{Al}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)+{\mathrm{AlCl}}_3\left(\mathrm{s}\right)+3\mathrm{NO}\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

$$\begin{gathered} 1\mathrm{mole}\mathrm{of}\mathrm{Aluminium}=26.8\mathrm{g} \\ 3\mathrm{moles}\mathrm{of}\mathrm{Aluminium}=3\times 26.8\mathrm{g}=80.94\mathrm{g} \\ 1\mathrm{mole}\mathrm{of}{\mathrm{NH}}_4{\mathrm{ClO}}_4=117.49\mathrm{g} \\ 3\mathrm{moles}\mathrm{of}{\mathrm{NH}}_4{\mathrm{ClO}}_4=3\times 117.49\mathrm{g}=352.47\mathrm{g} \\ \mathrm{Total}\mathrm{mass}\mathrm{of}\mathrm{reaction}\mathrm{mixture}=433.41\mathrm{g} \\ 433.41\mathrm{g}\mathrm{of}\mathrm{fuel}\mathrm{produces}\mathrm{energy}=-2677\mathrm{kJ}/\mathrm{mol} \\ 1\mathrm{Kg}\mathrm{of}\mathrm{fuel}\mathrm{will}\mathrm{produce}\mathrm{energy}=1\times {10}^{-3}\mathrm{g}\times \frac{-2677\mathrm{kJ}/\mathrm{mol}}{433.41\mathrm{g}}=-6177\mathrm{kJ}/\mathrm{mol} \end{gathered}$$

Step 2: Calculation of energy produced per kilogram of the reaction mixture in exercise 77.

$4{\mathrm{N}}_2{\mathrm{H}}_3{\mathrm{CH}}_3\left(\mathrm{l}\right)+5{\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{l}\right)\to 12{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)+9{\mathrm{N}}_2\left(\mathrm{g}\right)+4{\mathrm{CO}}_2\left(\mathrm{g}\right)$

In exercise 77, 4moles of N₂H₃CH₃react with 5 moles ofN₂O₄to produce energy:

$$\begin{gathered} 1\mathrm{mole}\mathrm{of}{\mathrm{N}}_2{\mathrm{H}}_3{\mathrm{CH}}_3=46.08\mathrm{g} \\ 4\mathrm{moles}\mathrm{of}{\mathrm{N}}_2{\mathrm{H}}_3{\mathrm{CH}}_3=4\times 46.08\mathrm{g}=184.32\mathrm{g} \\ 1\mathrm{mole}\mathrm{of}{\mathrm{N}}_2{\mathrm{O}}_4=92.02\mathrm{g} \\ 5\mathrm{moles}\mathrm{of}{\mathrm{N}}_2{\mathrm{O}}_4=5\times 92.02\mathrm{g}=460.1\mathrm{g} \\ \mathrm{Total}\mathrm{mass}\mathrm{of}\mathrm{reaction}\mathrm{mixture}=644.42\mathrm{g} \\ 644.42\mathrm{g}\mathrm{of}\mathrm{fuel}\mathrm{produces}\mathrm{energy}=-4594\mathrm{kJ}/\mathrm{mol} \\ 1\mathrm{Kg}\mathrm{of}\mathrm{fuel}\mathrm{will}\mathrm{produce}\mathrm{energy}=1\times {10}^{-3}\mathrm{g}\times \frac{-4594\mathrm{kJ}/\mathrm{mol}}{644.42\mathrm{g}}=-7128.89\mathrm{kJ}/\mathrm{mol} \end{gathered}$$

Result:

Hence the enthalpy of the reaction per kilogram of ${\mathrm{N}}_2{\mathrm{H}}_3{\mathrm{CH}}_3\left(\mathrm{l}\right)+{\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{l}\right)$is $-7128.89\mathrm{kJ}/\mathrm{mol}$which is more than the mixture of $\mathrm{Al}\left(\mathrm{s}\right)+{\mathrm{NH}}_4{\mathrm{ClO}}_4$ .