The easy way to do it is as follows:
32.8g x (2*molar mass Fe3O4/3*molar mass Fe2O3) = g Fe3O4.
Then (g Fe3O4/51.98)*100 = %Fe3O4
An ore contains Fe3O4 and no other iron. The
iron in a 51.98 gram sample of the ore is all
converted by a series of chemical reactions to
Fe2O3. The mass of Fe2O3 is measured to be
32.8 grams. What was the percent Fe3O4 in
the sample of ore?
Answer in units of %.
1 answer
1 answer