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#?

1 Answer
Mar 4, 2016

Answer:

18.31 grams of Al are necessary to produce 90.50 grams of #"AlCl"_3"#.

Explanation:

#"2Al + 3Cl"_2##rarr##"2AlCl"_3"#

This is a stoichiometry problem converting grams of #"AlCl"_3"# to grams of #"Al"#.

#"grams AlCl"_3##rarr##"moles AlCl"_3"##rarr##"moles Al"##rarr##"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")#

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

#"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