![2. (2 points) Calculate the percentage of Fe in pure iron(II) ammonium sulfate hexahydrate: (NH4)2[Fe(SO4)2]• 6 H20.](http://img.homeworklib.com/questions/84ce9960-70e1-11ea-aeed-fbde19c26f1d.png?x-oss-process=image/resize,w_560)
Homework Answers
1)
No. of moles of MnO4- required = molarity x volume
= 0.0244 x 36.44
= 0.889 millimoles
change in oxidation state = 5
change in oxidation state = 1
Thus no. of moles of Fe present = 5 / 1 x 0.889
= 4.45 millimoles
Mass of iron present = 4.45 / 1000 x 56 g / mol
= 0.249 g
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