Question

The amount of aluminium required to get 112g of iron in the following equation is ................... g. 2Al (s) + Fe2 O3(s)  Al 2 O3(s) + 2Fe(s)

Answer 1

Answer:

48 grams of aluminium

Explanation:

2 mole of Al + 1 mole of Iron oxide -> mole of Aluminium oxide + 2 mole of Iron

1 mole of Al =27u

1mole of Fe3O4= 56×2+16×3 = 160u

1mole of Al2O3 = 27×2+16×3 = 102u

54u + 160u - - 102 u + 112 u

54 g + 160g = 160 g

Therefore,

we need 160 - 112 = 48 grams of Aluminium