How many iron atoms are present in 55.845grams of Iron?

Amount of Fe = 55.845/55.845 = 1mol

There is 1 mol of Fe in a 55.845g sample.

1 mol of Fe contains 6.02 x 1023 atoms (avogadro constant).

Therefore there are 6.02 x 1023 atoms in 55.845g of iron.