# What mass of alumina will result if 14*mol of aluminum metal are oxidized?

$2 A l \left(s\right) + \frac{3}{2} {O}_{2} \left(g\right) \rightarrow A {l}_{2} {O}_{3} \left(s\right)$
Given the stoichiometric equation, if $14 \cdot m o l$ of metal are oxidized. then $7 \cdot m o l$ of $\text{alumina}$ will be generated............
$\text{Mass} = 7 \cdot m o l \times 101.96 \cdot g \cdot m o {l}^{-} 1 \cong 714 \cdot g$..........