How many grams of aluminum would be needed to react with 56 kg of ammonium perchlorate?

Reaction of Ammonium Perchlorate and Aluminium is as follow,

3 Al + 3 NH₄ClO₄ → Al₂O₃ + AlCl₃ + 3 NO + 6 H₂O

According to Equation,

0.352 Kg (3 moles) of NH₄ClO₄ required = 0.081 Kg of Al for reaction

So,

56 Kg of NH₄ClO₄ will require = X Kg of Al

Solving for X,

X = (56 Kg * 0.081 Kg) : 0.352 Kg

X = 12.88 Kg of Aluminium

Or,

X = 12800 g of Aluminium

Result:

12800 g of Aluminium is required to react with 56 Kg of Ammonium Perchlorate.