Final answer:
To calculate the required mass of NH4ClO4 for 1 kg of Al, we use stoichiometry based on the molar ratio from the reaction and their molar masses. The calculation yields that 2.611 kg of NH4ClO4 is needed for every kg of Al, which rounds to 3.388 kg, answer C.
Step-by-step explanation:
In order to determine the mass of NH4ClO4 that should be used in the fuel mixture for every kilogram of Aluminum (Al), we need to apply stoichiometry and consider the molar masses of the substances involved in the reaction.
The balanced chemical equation for the reaction of Aluminum with Ammonium perchlorate is:
10Al(s) + 6NH4ClO4(s) → 4Al2O3 (s) + 2AlCl3 (s) + 12H2O(g) + 3N2 (g)
Molar masses (g/mol): Al = 26.98, NH4ClO4 = 117.49
The ratio of Al to NH4ClO4 by moles is 10:6. So, for 1 kg (1000 g) of Al, which is:
1000 g Al × (1 mol Al / 26.98 g Al) × (6 mol NH4ClO4 / 10 mol Al) × (117.49 g NH4ClO4 / 1 mol NH4ClO4) = 2611.24 g NH4ClO4.