# Using the equation, "4Fe + 3O"_2rarr"2Fe"_2"O"_3, if 96.0 g of oxygen reacts, what mass of iron was oxidized?

Jun 17, 2016

The mass of iron oxidized is 223 g.

#### Explanation:

"4Fe(s) + 3O"_2("g")$\rightarrow$$\text{2Fe"_2"O"_3}$

Determine the molar masses of oxygen gas and iron using their atomic masses from the periodic table in g/mol.

${\text{O}}_{2} :$(2xx15.998 "g/mol")="31.998 g/mol"

$\text{Fe} :$$\text{55.845 g/mol}$

Now you need to determine the number of moles of oxygen gas that are in 96.0 g of oxygen gas by dividing the given mass of ${\text{O}}_{2}$ by its molar mass.

$96.0 \cancel{\text{g O"_2xx(1"mol O"_2)/(31.998cancel"g O"_2)="3.00 mol O"_2}}$

Now you need to determine the moles of Fe by multiplying the mole ratio for iron and oxygen from the balanced equation, so that iron is in the numerator and oxygen gas is in the denominator. This gives the moles of iron.

$3.00 \cancel{\text{mol O"_2xx(4"mol Fe")/(3cancel"mol O"_2)="4.00 mol Fe}}$

Then you can determine the mass of iron needed to react with 96.0 g of oxygen gas by multiplying the moles of iron by its molar mass.

$4.00 \cancel{\text{mol Fe"xx(55.845"g Fe")/(1cancel"mol Fe")="223 g Fe}}$ rounded to three significant figures

You can combine all of these steps as follows:

$96.0 \cancel{\text{g O"_2xx(1cancel"mol O"_2)/(31.998cancel"g O"_2)xx(4cancel"mol Fe")/(3cancel"mol O"_2)xx(55.845"g Fe")/(1cancel"mol Fe")="223 g Fe}}$