Question

Calculate the percentage of pure iron in 10 Kg of iron (III) oxide [ Fe2O3] of 80 % purity. [3]

[ Fe = 56, O = 16]​

Answer 1

Answer:

ferric oxide e.g Fe3O4

in 10kg iron ore Fe3O4 contain 10x 80/100 = 8kg

now,

1mole of Fe3O4 contain 3 mole of Fe

now,

no of mole of Fe3O4=8000g/(56 x 3+16 x 4) = 8000/232 =34.5

so, no of mole of Fe =3 x 34.5

=103.5 mole

we know ,

mole =weight/molecular weight

103.5 = weight/56

weight =103.5 x 56 g =5796 g =5.796 kg (approx )

Hope this helps u!<3