Answer:
28.01 percent
Explanation:
Iron (III) sulphate, [tex]Fe_2(SO_4)_3[/tex], has 2 atoms of Fe.
One atom of Fe has a molar weight of 56. Hence, 2 atoms of Fe would have a total molar weight of 56 x 2 = 112 g/mol
Molar weight of [tex]Fe_2(SO_4)_3[/tex] = 399.88 g/mol
Percentage Fe in [tex]Fe_2(SO_4)_3[/tex] = 112/399.88 x 100%
= 28.01%
The percentage of iron is iron (III) sulphate is, therefore, 28.01 percent.
