How many grams of iron can be recovered from 11.8 kg Fe3O4?
I tried to multiply 11.8 kg by the percent composition of iron, but that didn't work.
It should.
11.8 kg x (3*55.847/231.539) = ?
Did you use 0.7236 for the factor (72.36%)?
Yes, but I'm doing my chemistry online hw, and it said to round to sig figs. I put 8.54 g, and it said it was wrong. Thank you so much for responding I'm super confused.
Most books now show Fe3O4 as FeO*Fe2O3 (a mixture of Fe(II) and Fe(III) but that shouldn't change any of the calculations. I would have reported 8.54 kg also.
To calculate the grams of iron that can be recovered from 11.8 kg of Fe3O4, you need to consider the molar mass and the percent composition of iron in Fe3O4.
First, let's determine the molecular formula mass of Fe3O4, which is the sum of the atomic masses of each element in the formula:
Molar mass of Fe = 55.85 g/mol
Molar mass of O = 16.00 g/mol
Thus, the molecular formula mass of Fe3O4 is:
(3 * Molar mass of Fe) + (4 * Molar mass of O)
(3 * 55.85 g/mol) + (4 * 16.00 g/mol) = 159.69 g/mol
Next, find the percent composition of iron in Fe3O4:
Percent composition of iron = (Molar mass of Fe / Molecular formula mass of Fe3O4) * 100
= (55.85 g/mol / 159.69 g/mol) * 100
= 34.98%
Now, to calculate the grams of iron in 11.8 kg of Fe3O4, you can use the following equation:
Grams of iron = (Percent composition of iron / 100) x Mass of Fe3O4
= (34.98 / 100) x 11.8 kg
= 4.131 g
Therefore, approximately 4.131 grams of iron can be recovered from 11.8 kg of Fe3O4.