A 7.6 g sample of iron ore is treated as follows.

The iron in the sample is all converted by a
series of chemical reactions to Fe2O3. The
mass of Fe2O3 is measured to be 12.9 g. What
was the mass of iron in the sample of ore?
Answer in units of g

Didn't I do this for you (or one similar to it) a couple of days ago.

12.9g x (1 mol Fe2O3/molar mass Fe2O3) x (2 mols Fe/1 mol Fe2O3) x (atomic mass Fe/1 mol Fe) = ?

There is an easier way to do this but chemical factors are not taught in most schools now.
12.9 x (2*Fe/Fe2O3) = 12.9*2*55.85/159.7 = 9.02 g.
Check your post. There is a problem because the mass of Fe CAN'T be more than the 7.6g you started with.

To find the mass of iron in the sample of ore, we'll use the information given. We know that the iron in the sample is converted to Fe2O3, and the mass of Fe2O3 is given as 12.9 g.

Now, let's calculate the molar mass of Fe2O3:
Mass of Fe = atomic mass of Fe = 55.845 g/mol
Mass of O = atomic mass of O = 16.00 g/mol
Total molar mass of Fe2O3 = (2 * mass of Fe) + (3 * mass of O)
= (2 * 55.845 g/mol) + (3 * 16.00 g/mol)
= 159.69 g/mol

Now, we'll use the molar mass of Fe2O3 to calculate the number of moles of Fe2O3 present in the sample:
Number of moles of Fe2O3 = mass of Fe2O3 / molar mass of Fe2O3
= 12.9 g / 159.69 g/mol

Since we converted all the iron in the sample to Fe2O3, the number of moles of Fe in the sample is the same as the number of moles of Fe2O3.

Now, let's calculate the mass of iron in the sample:
Mass of iron = number of moles of Fe * molar mass of Fe
= (12.9 g / 159.69 g/mol) * 55.845 g/mol

Calculating this expression, we get the mass of iron in the sample of ore.