1. Find moles of iron

Given mass of Fe = $558.5\,\text{g}$ and atomic mass of Fe = $55.85\,\text{g mol}^{-1}$.

$n(\text{Fe}) = \frac{558.5}{55.85} = 10\,\text{mol}$

2. Find number of atoms in 10 mol Fe

Using Avogadro number $N_A = 6.023 \times 10^{23}$,

$\text{Number of atoms} = 10 \times 6.023 \times 10^{23} = 6.023 \times 10^{24}$

3. Check each option

• Option A: twice that in $60\,\text{g}$ carbon

For carbon:

$n(\text{C}) = \frac{60}{12} = 5\,\text{mol}$

So number of atoms in $60\,\text{g}$ carbon is:

$5N_A$

Twice this is:

$2 \times 5N_A = 10N_A$

This matches iron atoms:

$10N_A = 6.023 \times 10^{24}$

So A is correct.

• Option B: $6.023 \times 10^{22}$

Actual value is $6.023 \times 10^{24}$, so B is incorrect.

• Option C: half that in $8\,\text{g}$ He

For helium:

$n(\text{He}) = \frac{8}{4} = 2\,\text{mol}$

Atoms in $8\,\text{g}$ He = $2N_A$, and half of this is $N_A$.

But Fe atoms = $10N_A$.

So C is incorrect.

• Option D: $558.5 \times 6.023 \times 10^{23}$

This is far too large. Correct expression is:

$\frac{558.5}{55.85} \times 6.023 \times 10^{23}$

So D is incorrect.

4. Final answer

The number of atoms in $558.5\,\text{g}$ Fe is $10N_A = 6.023 \times 10^{24}$, which corresponds to Option A.