Question #8bddf

1 Answer
May 20, 2017

Answer:

#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