How many atoms of oxygen are in 30.5 grams of Fe2O3​?

 

Answer: 7.65 x 10^22 atoms of oxygen in 30.5 grams of Fe2O3

To determine the number of atoms of oxygen in 30.5 grams of Fe2O3, we first need to calculate the number of moles of Fe2O3, and then use the mole ratio between Fe2O3 and O atoms to calculate the number of O atoms.

The molecular weight of Fe2O3 can be calculated by adding the atomic weights of two Fe atoms and three O atoms:

Molecular weight of Fe2O3 = (2 x atomic weight of Fe) + (3 x atomic weight of O)

= (2 x 55.85 g/mol) + (3 x 16.00 g/mol)

= 159.69 g/mol

So, 30.5 g of Fe2O3 is equal to:

Number of moles = Mass / Molecular weight

= 30.5 g / 159.69 g/mol

= 0.191 moles

The mole ratio between Fe2O3 and O atoms is 3:2. This means that for every 3 moles of Fe2O3, there are 2 moles of O atoms. Therefore, the number of moles of O atoms in 0.191 moles of Fe2O3 is:

Number of moles of O atoms = 2/3 x 0.191 moles

= 0.127 moles

Finally, we can use Avogadro's number to convert the number of moles of O atoms into the actual number of O atoms:

Number of O atoms = Number of moles x Avogadro's number

= 0.127 moles x 6.022 x 10^23 atoms/mol

= 7.65 x 10^22 atoms

Therefore, there are 7.65 x 10^22 atoms of oxygen in 30.5 grams of Fe2O3.