# How many grams of Al_2O_3 would form if 12.5 moles of Al burned?

Mar 4, 2016

Over $600$ $g$ of aluminum oxide would result.

#### Explanation:

The stoichiometric equation for the combustion of aluminum is:

$2 A l \left(s\right) + \frac{3}{2} {O}_{2} \left(g\right) \rightarrow A {l}_{2} {O}_{3} \left(s\right)$

If $12.5$ $m o l$ of aluminum were consumed, clearly $6.25$ $m o l$ aluminum oxide would result.

$\text{Mass of aluminum oxide}$ $=$ $6.25 \cdot m o l \times 101.96 \cdot g \cdot m o {l}^{-} 1$ $=$ ?? $g$.