# Question #8bddf

May 20, 2017

$2.6 {\text{g Fe"_2"O}}_{3}$

#### Explanation:

The reaction for this problem is:

$4 {\text{Fe"_3"O"_4 + "O"_2 -> 6"Fe"_2"O}}_{3}$

The molar mass of ${\text{Fe"_3"O}}_{4}$ is 231.533 g/mol

So, you have $2.5 {\text{g" times (1" mol")/(231.533"g") = 0.0108" mol " "Fe"_3"O}}_{4}$

Now multiply this by the $6 : 4$ ratio of ${\text{Fe"_2"O}}_{3}$ to ${\text{Fe"_3"O}}_{4}$ to get $0.0162 {\text{ mol ""Fe"_2"O}}_{3}$.

Finally multiply this by the molar mass of $\text{Fe"_2"O"_3 - 159.69 " g/mol}$.

$0.0162 {\text{ mol" times 159.69 " g/mol" = 2.6 "g Fe"_2"O}}_{3}$