We must consider the relation between Al and \(O_2\):
\(\frac{3\ mol\ O_2}{4\ mol\ Al} = \frac{4,5\ mol\ O_2}{x}\ \to\ x = \frac{4,5\ mol\ O_2\cdot 4\ mol\ Al}{3\ mol\ O_2} = \bf 6\ mol\ Al\)
As you can see, the number of moles of Al is double to the number of moles of \(Al_2O_3\), keeping constant the proportion of both substances.