# How many grams of aluminum are required to produce 8.70 moles of aluminum chloride in the reaction 2Al + 3Cl_2 -> 2AlCl_3?

Mar 29, 2016

$234.9 g$

#### Explanation:

The balanced chemical equation represents the mole ratio in which the chemicals combine, and in this case illustrates that $2 m o l A l$ produces $2 m o l A l C {l}_{3}$, hence these 2 chemicals are in a $1 : 1$ ratio.

Thus, to produce $8.70 m o l$ aluminium chloride, it will require $8.70 m o l$ aluminium.

But this quantity of $A l$ has a mass in grams of

$m = n \times {M}_{r}$

$= 8.70 m o l \times 27 g / m o l$

$= 234.9 g$