# 2Al + 3Cl_2 -> 2AlCl_3 How many grams of chlorine are required to produce 4 moles of aluminum chloride?

Approx. $425 \cdot g$
$2 A l \left(s\right) + 3 C {l}_{2} \left(g\right) \rightarrow 2 A l C {l}_{3} \left(s\right)$
The equation explicitly tells us that $54 \cdot g$ of aluminum metal reacts with $6 \times 35.45 \cdot g$ $C {l}_{2}$ gas to give $266.7 \cdot g$ of $\text{aluminum trichloride}$. From where am I getting these numbers?