What mass of iron can be obtained from 500.0g of Fe3O4?
I am having a hard time getting started on this question. I think I need to find the percent of iron in the compound and then multiply that by 500.0 g. But, when I tried that I got a really wonky answer.
Fe 3 55.85g/mol = 167.55/mol
O 4 16.00g/mol = 64.00g/mol
-----------
231.55g/mol
231.55g/mol = total molar mass
167.55g/mol Fe
-------------- x 100% = 72.36%
231.55g/mol
.7236% Fe x 500.0g = 361.8g
Did I get this right or do I need some major help?
looks fine
To find the mass of iron (Fe) that can be obtained from 500.0g of Fe3O4 (iron(II,III) oxide), you are correct in first finding the percentage of iron in the compound. However, it seems you made a small error in your calculation.
To calculate the percentage of iron in Fe3O4, you need to divide the molar mass of iron by the molar mass of Fe3O4 and then multiply by 100%. The correct calculation is:
167.55g/mol Fe / 231.55g/mol Fe3O4 x 100% = 72.36%
So, the correct percentage of iron in Fe3O4 is 72.36%.
To find the mass of iron, you need to multiply the 500.0g by the percentage of iron in Fe3O4:
500.0g x 72.36% = 361.8g
Therefore, the mass of iron that can be obtained from 500.0g of Fe3O4 is 361.8g.
You were on the right track, but there was just a small calculation error in your initial attempt. I hope this helps!