Given the balanced equation: 3fe3 + (aq) + al (s) → 3fe2 + (aq) + al3 + (aq) what is the total number of moles of electrons lost by 2 moles of al (s) ?

Answer is: the total number of moles of electrons lost by two moles of aluminium are six moles (6 mol).

Oxidation half reaction: Al° → Al³⁺ + 3e⁻, lost of electrons.

Reduction half reaction: Fe³⁺ + e⁻→ Fe²⁺ / ·3; 3Fe³⁺ + 3e⁻→ 3Fe²⁺; gain of electrons.

One mole of aluminium lost 3 moles of electrons, so 2 moles of alumiunim:

2 · 3 mol = 6 mol.