An ore contains fe3o4 and no other iron. the iron in a 51.9-gram sample of the ore is all converted by a series of chemical reactions to fe2o3. the mass of fe2o3 is measured to be 18.7 g. what was the mass of fe3o4 in the sample of ore? answer in units of g.

First let us convert the mass of Fe2O3 to moles by dividing the mass of Fe2O3 by its molecular mass, 159.69 g/mol. 

moles Fe2O3 = 18.7 / 159.69 = 0.117 mol

Then we know that there are 2 moles of Fe for every 1 mole of Fe2O3. So,

moles Fe = 2 * 0.117 mol = 0.234 mol

Now to make Fe3O4 requires 3 moles of Fe, therefore:

moles Fe3O4 = 0.234 mol (1 / 3) = 0.078 mol

Converting to mass by multiplying with the molar mass of Fe3O4, 231.533 g/mol:

it's filtration, but this is an assumption so my apologies if i got this

2b2t

explanation:

2b2t