Question

In a 108.9 g sample of iron(III) oxide, how many grams of iron are present?

Answer 1

Molar mass of Fe2O3,
MM = 2*MM(Fe) + 3*MM(O)
= 2*55.85 + 3*16.0
= 159.7 g/mol


mass(Fe2O3)= 108.9 g

number of mol of Fe2O3,
n = mass of Fe2O3/molar mass of Fe2O3
=(108.9 g)/(159.7 g/mol)
= 0.6819 mol
This is number of moles of Fe2O3

one mole of Fe2O3 contains 2 mole of Fe
So, number of moles of Fe = 2 * number of moles of Fe2O3
= 2 * 0.6819
= 1.364

Molar mass of Fe = 55.85 g/mol

mass of Fe,
m = number of mol * molar mass
= 1.364 mol * 55.85 g/mol
= 76.2 g

Answer: 76.2 g