A 25.2 g 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 grams. What was
the percent iron in the sample of ore?
Answer in units of %

Mass Fe in sample = (12 x 2 x atomic mass Fe/molar mass Fe2O3) = ?

%Fe = (mass Fe/mass sample)*100 = ?

To find the percent of iron in the sample of ore, we need to determine the mass of iron in the Fe2O3 and then calculate the percentage.

1. Calculate the molar mass of Fe2O3 by adding the atomic masses of each element:
Fe: 55.85 g/mol
O: 16.00 g/mol (x3 since there are 3 oxygen atoms)
Molar mass of Fe2O3 = (55.85 g/mol) + (16.00 g/mol x 3) = 159.85 g/mol

2. Use the molar mass of Fe2O3 to convert the mass of Fe2O3 to moles:
Moles of Fe2O3 = 12 g / 159.85 g/mol = 0.075 mol

3. Since the chemical reaction converts all the iron in the sample to Fe2O3, the moles of iron is the same as the moles of Fe2O3.

4. Calculate the mass of iron in the sample using the moles of Fe2O3:
Mass of iron = Moles of Fe2O3 x Molar mass of Fe = 0.075 mol x 55.85 g/mol = 4.19 g

5. Finally, calculate the percent of iron in the sample of ore:
Percent of iron = (Mass of iron / Mass of sample) x 100%
= (4.19 g / 25.2 g) x 100%
= 16.63%

Therefore, the percent of iron in the sample of ore is approximately 16.63%.

To find the percent of iron in the sample of ore, you need to calculate the mass percent of iron in the Fe2O3.

First, calculate the molar mass of Fe2O3:
- The molar mass of Fe is 55.845 g/mol.
- The molar mass of O is 15.999 g/mol.
Since Fe2O3 has two Fe atoms and three O atoms, the molar mass of Fe2O3 is:
(2 * 55.845 g/mol) + (3 * 15.999 g/mol) = 159.688 g/mol.

Next, calculate the mass percent of iron in Fe2O3:
- The mass of Fe2O3 is given as 12 grams.
- The molar mass of Fe2O3 is 159.688 g/mol.
The mass percent of iron is:
(55.845 g/mol / 159.688 g/mol) * 100% ≈ 35.0%.

Therefore, the percent of iron in the sample of ore is approximately 35.0%.