Rust contains iron (III) oxide Fe2 O3 .Then, how many moles of iron (III) ions present in 12 g of rust?

Answer 1

Convert:
grams <--> mol using molar mass.

Molar mass of Fe2O3 = 159.69 g/mol

12g Fe2O3 (1mol Fe2O3 /159.69 g Fe2O3) = 0.075 mol Fe2O3

0.075 mol Fe2O3 (2mol Fe3+/1mol Fe2O3) = 0.15 mol Fe3+

0.15 mol of Iron (III) ions.

Answer 2

1. Add up the molecular weight of one molecule of Fe2 O3. The answere will be in grams per mole (g/mol).

2. To find out how many moles of rust you have, divide 12 g by your answer from #1

3. OK, so from the molecular formula, you see that there are 2 atoms of Fe (3+) for every 1 molecule of rust, which translates to 2 moles of Fe for every 1 mole of rust molecules. So the equation would look like this:

moles of Fe2 O3 x 2 moles of Fe (3+)/ 1 mole of Fe2 O3 = # of moles of Fe (3+)

So you just have to double your answer from step 2.

Gotta love Gen Chem.

Cheers.

Answer 3

The above answer is partly correct. After you have worked out how many moles of iron oxide there is, you need to multiply this by two, since there are two iron ions in each iron oxide particle (Fe**2** O3).