Question

How many Fe atoms are contained in 787 g of Fe?

Answer 1

Answer:

8.49 * 10^24 atoms

Explanation:

First we calculate the number of moles of Fe in 787 grams...which is

n = given mass/ molecular mass

n = 787/55.8

n = 14.10 moles

We know that 1 mol contains 6.022 * 10 ^ 23 atoms.

therefore 14.10 will contain 14.10 * 6.022 * 10^23 atoms = 84.9 * 10^23 = 8.49 * 10^24 atoms.