Question

Question #8bddf

Answer 1

`2.6 "g Fe"_2"O"_3`

Explanation:

The reaction for this problem is:

`4"Fe"_3"O"_4 + "O"_2 -> 6"Fe"_2"O"_3`

The molar mass of `"Fe"_3"O"_4` is 231.533 g/mol

So, you have `2.5 "g" times (1" mol")/(231.533"g") = 0.0108" mol " "Fe"_3"O"_4`

Now multiply this by the `6:4` ratio of `"Fe"_2"O"_3` to `"Fe"_3"O"_4` to get `0.0162 " mol ""Fe"_2"O"_3`.

Finally multiply this by the molar mass of `"Fe"_2"O"_3 - 159.69 " g/mol"`.

`0.0162 " mol" times 159.69 " g/mol" = 2.6 "g Fe"_2"O"_3`

Final Answer