Question

Calculate  the  amount  of  aluminum  required  to  get  1120Kg  of  iron  by  the  reaction.
2Al + Fe2O3 --> Al2O3 + 2Fe?

Answer 1

According to the balanced equation.
2 moles of Al(54kg) gives 2 moles of Fe(112kg).
So for 1120 kg of Fe we need  540 kg of Al.

Answer 2

As per the equation
2 mol of Aluminium + 1 mole of Iron Oxide -----> 1 mole of Aluminium Oxide + 2 moles of Iron

1 mole of aluminium = 27 U
1 mole of Iron Oxide =  56*2 +16*3 = 160U

1 mole of Aluminium Oxide = 27*2 + 16*3 = 102U

54U  +  160U  →  102U+112U 54  g  +  160g  →  102g+112g As  per  the  balanced  equation  54g For  112g  of  iron,  we  required  54g  Aluminum  for  1120  kg  of  iron  required  aluminum  is-

(1120*1000) *54 =540000 gram 
so it is 540 kg