A 38 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 11.1 g. What
was the mass of iron in the sample of ore?
Answer in units of g

Another stoichiometry problem BUT there is a short cut that saves a lot of time; however, the shortcut is not taught, to my knowledge, anymore.

Here is the short cut.
grams Fe in the ore = grams Fe2O3 x (2*atomic mass Fe/molar mass Fe2O3) = ?

not quite sure how to do this how do i get an answer through this thing

6.93755

To find the mass of iron in the sample of ore, we need to determine the difference in mass before and after the chemical reactions.

1. Start by subtracting the mass of Fe2O3 (11.1 g) from the initial mass of the sample (38 g).

38 g - 11.1 g = 26.9 g

The mass difference (26.9 g) represents the mass of oxygen that was present in the initial iron ore sample.

2. Now, we need to find the molar mass of iron (Fe) and Fe2O3 to determine the stoichiometric ratio between them.

The molar mass of iron (Fe) is 55.845 g/mol.
The molar mass of Fe2O3 is calculated as (55.845 g/mol x 2) + (16.00 g/mol x 3) = 159.69 g/mol.

3. Next, we can set up a proportion using the stoichiometric ratio between Fe2O3 and Fe:

(159.69 g Fe2O3 / 1 mol Fe2O3) = (26.9 g Fe / x mol Fe)

4. Cross multiply the ratio and solve for x:

(159.69 g Fe2O3) * (x mol Fe) = (26.9 g Fe) * (1 mol Fe2O3)

x = (26.9 g Fe) * (1 mol Fe2O3) / (159.69 g Fe2O3)

x ≈ 0.1688 mol Fe

5. Finally, convert the mol of Fe to grams by multiplying by the molar mass of Fe:

Mass of Fe = (0.1688 mol Fe) * (55.845 g/mol Fe)

Mass of Fe ≈ 9.43 g

Therefore, the mass of iron in the sample of ore is approximately 9.43 g.