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.

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