How many grams of Iron (Fe) can be obtained from 25.0 grams of FeSO4? I can't find out how to get the answer.

Answer 1

The molecular weight of FeSO4 = 151.9 g/mol
Of this, 55.8 g/mol is Fe. (atomic weight)

So, 55.8 g of Fe are contained in 151.9 g of FeSO4.
If x g of Fe are contained in 25.0 g of FeSO4, then :

x = 25.0 * 55.8 / 151.9

= 9.2

Answer 2

Get Fe's percentage of the molecular wieght of FeSO4.
Fe=55.85, O=16, S=32
molecule weight=55.85+32+(4*16)=151.85
55.85/151.85=36.78%

Your answer is 25g*0.3678=9.19g of Fe