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

Jan 15, 2016

$\text{1.0 kg Fe"_2"O"_3}$ contains $\text{13 mol Fe ions}$.

#### Explanation:

$\text{1 mol Fe"_2"O"_3}$ contains $\text{2 mol Fe ions and 3 mol O ions}$

$\text{1.0 kg=1000 g}$

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

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

$1000 \cancel{\text{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}}$$= \text{13 mol Fe ions}$
(rounded to two significant figures)