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.