Question

If 11.9 g of iron(iii) oxide (rust) is produced from a certain amount of iron, how many grams of oxygen are needed for this reaction?

Answer 1

Answer is: 3.58 grams of oxygen are needed.

Balanced chemical reaction: 4Fe + 3O₂ → 2Fe₂O₃.

m(Fe₂O₃) = 11.9 g; mass of rust.

n(Fe₂O₃) = m(Fe₂O₃) ÷ M(Fe₂O₃).

n(Fe₂O₃) = 11.9 g ÷ 159.7 g/mol.

n(Fe₂O₃) = 0.0745 mol; amount of rust.

From chemical reaction: n(Fe₂O₃) : n(O₂) = 2 : 3.

n(O₂) = 3 · 0.0745 mol ÷ 2.

n(O₂) = 0.111 mol; amount of oxygen.

m(O₂) = n(O₂) · M(O₂).

m(O₂) = 0.111 mol · 32 g/mol.

m(O₂) = 3.577 g; mass of oxygen.