Question

The space shuttle carries about 7.35×10⁴ kg of solid aluminum fuel, which is oxidized with ammonium perchlorate according to the following reaction:
10Al(s)+6NH4ClO4(s)→4Al2O3(s)+2AlCl3(s)+12H2O(g)+3N2(g)
The space shuttle also carries about 6.02×10⁵ kg of oxygen (which reacts with hydrogen to form gaseous water).

Part A

Assuming that aluminum and oxygen are the limiting reactants, determine the total energy produced by these fuels. (ΔH∘ffor solid ammonium perchlorate is -295 kJ/mol.)

Part B

Suppose that a future space shuttle was powered by matter-antimatter annihilation. The matter could be normal hydrogen (containing a proton and an electron) and the antimatter could be antihydrogen (containing an antiproton and a positron). What mass of antimatter would be required to produce the energy equivalent of the aluminum and oxygen fuel currently carried on the space shuttle?

Answer 1

Part-A: mass of Al(s) = 7.35*10⁴ Kg * (1000 g/1 Kg) = 7.35*10⁷ g

H^o = (H_f^o of all products) - (H_f^o of all reactants)

=> H^o = [4 mol*H_f^o(Al2O3,s) + 2 mol*H_f^o(AlCl3,s) + 12 mol*H_f^o(H2O,g) + 3 mol*H_f^o(N2,g)] -

[6 mol*H_f^o(NH4ClO4,s) + 10 mol*H_f^o(Al,s)]

=> H^o = [4 mol*(‐1675.27 kJ/mol) + 2 mol*(‐705.63 kJ/mol) + 12 mol*(‐241.82 kJ/mol) + 3*0] - [6 mol*(-295 kJ/mol) + 10*0]

=> H^o = - 9244.18 kJ

9244.18 kJ is produced by 10 mol Al(s).

Total moles of Al(s) = mass/atomic mass of Al(s) =   7.35*10⁷​​​​​​​ g / 26.982 g/mol = 2724038.25 mol

=> Total energy produced = 2724038.25 mol * (- 9244.18 kJ / 10 mol) = - 2.52*10⁹ kJ (Answer)

Part-B:

E = 2.52*10⁹ kJ * (1000 J/1kJ) = m*C²

=> 2.52*10¹² J = m * (3*10⁸ m/s)²

=> m = 2.52*10¹² J / 9*10¹⁶ m²s^-2)

=> m = 2.8*10^-5​​​​​​​ Kg (Answer)