Question

calculate the amount of Al required to get 1120kg of Fe Al+Fe2O2-Al2o3+Fe​

Answer 1

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