Question

Calculate the percentage of pure iron in 10kg of Ferric oxide of 80% purity.
[ Fe = 56, O = 16]

plz answer ASAP​

Answer 1

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

Answer 2

Answer:

56%

Explanation:

160 g of pure Fe2O3 contains 112 g of pure iron.

Therefore, 8000g. of pure Fe2O3 contains: (112×8000)/100 = 5600g or 5.6 kg

Therefore, % of pure Fe in 10kg of Fe2O3 =

(5.6× 100) / 10 = 56% of pure Iron.