# Given the reaction 2Al + 3Cl_2 -> 2AlCl_3, how many grams of Al are needed to produce 90.50 grams of AlCl_3, if we have an unlimited supply of Cl_2?

Mar 4, 2016

18.31 grams of Al are necessary to produce 90.50 grams of $\text{AlCl"_3}$.

#### Explanation:

${\text{2Al + 3Cl}}_{2}$$\rightarrow$$\text{2AlCl"_3}$

This is a stoichiometry problem converting grams of $\text{AlCl"_3}$ to grams of $\text{Al}$.

${\text{grams AlCl}}_{3}$$\rightarrow$$\text{moles AlCl"_3}$$\rightarrow$$\text{moles Al}$$\rightarrow$$\text{grams Al}$

"grams of given" xx (1"mol given")/("molar mass of given")xx("mol of want")/("mol of given")xx("molar mass of want")/(1"mol of want")

$\text{Molar mass Al=26.9815385 g/mol}$ (periodic table)
$\text{Molar mass AlCl"_3"=133.340539 g/mol}$
https://pubchem.ncbi.nlm.nih.gov/compound/24012#section=Top

$\text{90.50 g AlCl"_3xx(1"mol AlCl"_3)/(133.340539"g AlCl"_3)xx(2"mol Al")/(2"mol AlCl"_3)xx(26.9815385"g Al")/(1"mol Al")="18.31 grams of Al}$ rounded to four significant figures