Answer to Question #285682 in General Chemistry for rii

Question

How many grams of iron (III) oxide are produced from 45.0 L of oxygen gas?

Answer 1

Iron(III) oxide is a product of the oxidation of iron.

4 Fe + 3 O2 + 2 H2O → 4 FeO(OH)

The resulting hydrated iron(III) oxide, written here as FeO(OH), dehydrates around 200 °C.

2 FeO(OH) → Fe2O3 + H2O

Overall the equation reads

4Fe + 3O2 → 2Fe2O3

3 mol O2 produce 2 mol Fe2O3

Mol O2 in 45.0 L at STP :

At STP 1 mol O2 has volume = 22.4 L

Mol O2 = 45.0 L/ 22.4 L/mol = 2.0089 mol O2

This will produce 2.0089 * 2/3 = 1.3393 mol Fe2O3

Molar mass Fe2O3 = 159.6882 g/mol

Mass Fe produced = 1.3393 mol * 159.6882 g/mol = 213.870 Fe2O3 produced