The chemical formula for rust is #Fe_2O_3#. How many moles of #Fe# are present in 1.0 kg of the compound?

1 Answer
Jan 15, 2016

Answer:

#"1.0 kg Fe"_2"O"_3"# contains #"13 mol Fe ions"#.

Explanation:

#"1 mol Fe"_2"O"_3"# contains #"2 mol Fe ions and 3 mol O ions"#

#"1.0 kg=1000 g"#

#"Molar mass Fe"_2"O"_3="159.6882 g/mol"#
http://pubchem.ncbi.nlm.nih.gov/compound/14833#section=Top

Determine the moles of #"Fe"_2"O"_3"# by dividing the given mass by its molar mass, then multiply times the mol ratio between #"Fe"# ions and #"Fe"_2"O"_3"# to get the moles of #"Fe"# ions in #"1000 g Fe"_2"O"_3"#.

#1000cancel("g Fe"_2"O"_3)xx(1cancel("mol Fe"_2"O"_3))/(159.6882cancel("g Fe"_2"O"_3))xx(2"mol Fe")/(1cancel("mol Fe"_2"O"_3")##="13 mol Fe ions"#
(rounded to two significant figures)