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.
