Question

Determine the weight of Fe(NH4)2(SO4)2(H20)6.which is required to react completely with 20.0 mL of 0.004 M KMnO4

Answer 1

Ionic equation for reaction KMnO4 and Mohr's salt is

MnO4(-)   + 5Fe(2+)    +   8H+ -> Mn(2+) + 5Fe(3+) + 4H2O

Hence, 1 mol KMnO4 requires 5 moles of Fe(NH4)₂(SO4)₂(H2O)₆

  Moles of KMnO4 = Molarity x volume in Litres = 0.004 M x 0.02 L= 0.00008 mol

Hence,

MnO4(-)   +           5Fe(2+)             +      8H+ -----------> Mn(2+) + 5Fe(3+) + 4H2O

1 mol                5 mol = 5 x 392 g

0.00008 mol               ?

            ? = ( 0.00008 mol/1 mol ) x 5 x 392 g of Mohr's salt

                 = 0.1568 g of Mohr's salt

Therefore,

weight of Fe(NH4)2(SO4)2(H20)6.which is required to react completely with 20.0 mL of 0.004 M KMnO4

         = 0.1568 grams