# What mass of aluminum chloride could theoretically be made from 8.1 g of aluminum and excess chlorine?

Approx. $40 \cdot g$
$\text{Moles of aluminum } = \frac{8.1 \cdot g}{26.98 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.300 \cdot m o l$
$A l \left(s\right) + \frac{3}{2} C {l}_{2} \left(g\right) \rightarrow A l C {l}_{3} \left(s\right)$
Given the 1:1 stoichiometry between Al and the product, if all the aluminum were oxidized, $0.300 \cdot m o l \times 133.34 \cdot g \cdot m o {l}^{-} 1 \text{ aluminum trichloride}$ would result, approx. 40 g.