Question

Calculate the number of oxygen atoms in a 70.0 g sample of iron(iii) oxide fe2o3. Be sure your answer has a unit symbol if necessary, and round it to 3 significant digits.

Answer 1

To find the number of oxygen atoms in a 70.0 g sample of Fe2O3, calculate the moles of Fe2O3, use the mole ratio to find the moles of oxygen, and then convert to atoms using Avogadro's number. There are 7.91 x 10^23 oxygen atoms in the sample to three significant digits.

Step-by-step explanation:

To calculate the number of oxygen atoms in a 70.0 g sample of iron(III) oxide (Fe2O3), first, we need the molar mass of Fe2O3. Iron has a molar mass of 55.85 g/mol, and oxygen has a molar mass of 16.00 g/mol. The molar mass of Fe2O3 is thus 2(55.85) + 3(16.00) = 159.7 g/mol.

Next, we convert the mass of Fe2O3 to moles using the formula moles = mass / molar mass:

Moles of Fe2O3 = 70.0 g / 159.7 g/mol = 0.438 mol.

Since there are 3 moles of oxygen for every mole of Fe2O3, we multiply the moles of Fe2O3 by 3 to get the moles of oxygen:

Moles of O = 0.438 mol * 3 = 1.314 mol.

Finally, we convert moles of oxygen to atoms using Avogadro's number (6.022 x 1023 atoms/mol):

Atoms of O = 1.314 mol * 6.022 x 1023 atoms/mol = 7.91 x 1023 atoms.

Thus, to three significant digits, there are 7.91 x 1023 oxygen atoms in a 70.0 g sample of Fe2O3.