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