Answer
$3.38$ moles of $O_2$ are needed to react with 4.50 moles of Al.
Work Step by Step
$4 Al(s) + 3O_2(g) --\gt 2Al_2O_3(s)$
1. Find the mole-mole conversion factors:
According to the coefficients: 4 moles $Al$ = 3 moles $O_2$
$\frac{ 4 \space moles \space Al}{ 3 \space moles \space O_2}$ and $\frac{ 3 \space moles \space O_2}{ 4 \space moles \space Al}$
2. Use the mole-mole conversion factors to calculate the amount of equivalent $O_2$ in moles:
$ 4.50$ moles $Al \times \frac{ 3 \space moles \space O_2}{ 4 \space moles \space Al} = 3.38$ moles $O_2$
