How much alumina is formed if 4.2*mol of aluminum is oxidized?

Why, $2.1 \cdot m o l$....................
$2 A l \left(s\right) + \frac{3}{2} {O}_{2} \left(g\right) \rightarrow A {l}_{2} {O}_{3} \left(s\right)$
Look at the stoichiometry of the oxide: ONE mole of alumina contains 2 moles of aluminum, and 3 moles of oxygen ATOMS. We have $4.2 \cdot m o l$ of aluminum, burnt in excess oxygen, and thus $2.1 \cdot m o l$ alumina result.