How many atoms of oxygen would there be in a 2.500 kg sample of iron(II) nitrate (Fe(NO3)2)? NA = \frac{6.022x10^{23}objects}{1mol} 1mol 6.022x10 23 objects

5.02×10²⁵ atoms of O

Explanation:

Our compound is: Fe(NO₃)₂

We know that 1 mol of Fe(NO₃)₂ has 1 mol of Iron(II), and 2 moles of nitrates (where you have 2 moles of N and 6 moles of O)

Let's convert the mass to moles and firslty, let's convert the mass from kg to g → 2.500 kg . 1000 g / 1kg = 2500 g

We convert the moles to mass → 2500 g / 179.85 g/mol = 13.90 moles

Now we can propose this rule of three:

1 mol of Fe(NO₃)₂ has 6 moles of O

Then, 13.90 moles of Fe(NO₃)₂ will  have (13.90 . 6) /1 = 83.4 moles of O

Let's count the atoms: 83.4 mol . 6.02×10²³atoms  /1mol = 5.02×10²⁵ atoms of O

i think it's c. i'm not sure but i think so.