AP chem stochiometry

Question

3.47 grams of hydrated ammonium iron (II) sulfate hexahydrate [FeSO4(NH4)2SO4*6H2O] was dissolved in 200. mL of water. Twenty milliliters (20.0 mL) of this had acid added to it and then reacted completely with 12.6 mL of a KMnO4 solution. Calculate the concentration of the KMnO4 solution given the complete, balanced redox equation below.

5 Fe^+2 + MnO4^-1 + 8 H^-1 --> 5 Fe^+3 + Mn^+2 + 4 H2O

Answer 1

Hey Margie,

Fortunately, you have been given the balanced equation for the reaction of Fe²⁺ with MnO4-. So, we can use the stoichiometry of this equation to find the mols of MnO4- used and then find its concentration. But before we can do that, we must first find the mols of Fe2+ that are present. We do that using the provided information as follows:

molar mass [FeSO₄(NH₄)₂SO₄*6H₂O] = 392 g / mol

3.47 g [FeSO₄(NH₄)₂SO₄*6H₂O] x 1 mol / 392 g = 0.008852 mols

0.008852 mols / 200 ml x 20 ml = 0.0008852 mols [FeSO₄(NH₄)₂SO₄*6H₂O] taken

0.0008852 mols [FeSO₄(NH₄)₂SO₄*6H₂O] x 1 mol Fe / mol [FeSO₄(NH₄)₂SO₄*6H₂O] = 0.0008852 mols Fe²⁺

mols MnO₄^- needed = 0.0008852 mols Fe²⁺ x 1 mol MnO₄^- / 5 mols Fe²⁺ = 0.004426 mols MnO₄^-

Concentration of MnO₄^- = 0.004426 mols / 0.0126 L = 0.351 M (3 sig. figs.)