Stoichiometry of Iron Rusting: Fe to Fe2O3

Medium

Problem

Iron reacts with oxygen to form iron(III) oxide: 4Fe + 3O2 -> 2Fe2O3. If 11.2 g of iron reacts completely with excess oxygen, what mass of Fe2O3 is produced?

Given

Balanced equation: 4Fe + 3O2 -> 2Fe2O3
Mass of iron: m(Fe) = 11.2 g
Oxygen is in excess, so iron is the limiting reactant
M(Fe) = 55.845 g/mol

Find

The mass of Fe2O3 formed in grams.

Principle

A balanced equation gives mole ratios, not mass ratios. Grams of one substance cannot jump straight to grams of another: the path is grams of reactant -> moles of reactant -> moles of product (via the balanced mole ratio) -> grams of product. Each conversion factor equals one mole-equivalent, so the chemistry is preserved at every hop.

Formula

n_{\mathrm{Fe}} = \frac{m}{M},\quad n_{\mathrm{Fe_2O_3}} = n_{\mathrm{Fe}} \times \frac{2}{4},\quad m = nM

n(Fe) = m / M ; n(Fe2O3) = n(Fe) x (2/4) ; m(Fe2O3) = n x M

Solution

Moles of iron: n(Fe) = 11.2 g / 55.845 g/mol = 0.2006 mol Fe.
Mole ratio from the equation: 4 mol Fe produce 2 mol Fe2O3, so multiply by 2/4. n(Fe2O3) = 0.2006 mol x (2/4) = 0.1003 mol Fe2O3.
Molar mass of the product: M(Fe2O3) = 2(55.845) + 3(15.999) = 111.69 + 47.997 = 159.687 g/mol, rounded to 159.69 g/mol.
Mass of product: m = 0.1003 mol x 159.69 g/mol = 16.017 g, rounded to 16.0 g Fe2O3.

Approximately 16.0 g of Fe2O3.

Unit check

g/(g/mol) gives mol; mol times the unitless ratio 2/4 gives mol of product; mol times g/mol gives grams of product. Each unit cancels exactly as the three-step path requires.

What this means physically

Just 11.2 g of iron can gain about 4.8 g of oxygen mass to make 16.0 g of rust. Chemically this is why rusted structures gain weight — the oxide incorporates oxygen from the air, and stoichiometry predicts exactly how much.

Common mistake

Students often apply the 4:2 ratio to grams directly (multiplying 11.2 g by 2/4), which is wrong because the ratio comes from moles. Grams must always be converted to moles before any equation ratio is used.