Question

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

Answer 1

`"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)