7.83 ml of a solution of Fe2+(aq) is titrated with 24.9 ml of 0.375 M KMnO4...

Question

Question

science chemistry

Answer 1

Answer #1

Here:

M(MnO4-)=0.375 M

V(MnO4-)=24.9 mL

V(Fe2+)=7.83 mL

According to balanced reaction:

5*number of mol of MnO4- =1*number of mol of Fe2+

5*M(MnO4-)*V(MnO4-) =1*M(Fe2+)*V(Fe2+)

5*0.375*24.9 = 1*M(Fe2+)*7.83

M(Fe2+) = 5.96 M

Answer: 5.96 M

Answer 2

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