Question

An iron nail rusts when exposed to oxygen. According to the following reaction, how many grams of iron(III) oxide will be formed upon the complete reaction of 32.0 grams of oxygen gas with excess iron?
iron (s) + oxygen (g) $\rightarrow$ iron(III) oxide (s)
_______ grams iron(III) oxide

Answer 1

Answer:

53.2

Explanation:

The balanced reaction is:

2Fe(s) + 3O₂(g) → Fe₂O₃

It means that 3 moles of oxygen form 1 mol of iron(III) oxide. The molar masses are: Fe = 55.8 g/mol and O = 16 g/mol. So

O₂ = 2x 16 = 32 g/mol

Fe₂O₃ = 2x55.8 + 3x16 = 159.6 g/mol

So, 32 g of O₂ corresponds to 1 mol of O₂. The stoichiometry calculus must be (always in moles):

3 mol of O₂ ------------------------ 1 mol of Fe₂O₃

1 mol of O₂  ------------------------ x

By a direct simple three rule:

3x = 1

x = 1/3 mol of Fe₂O₃

The mass is the molar mass multiplied by the number of moles, so:

m = 159.6x (1/3)

m = 53.2 g iron (III) oxide.