Question

15.27 mL of a solution of SnCl2 consumed to reduce a certain amount of Fe, Fe3+ to Fe2+, consumed in oxidation 16.27 mL of KMnO4. This volume of KMnO4 oxidizes an amount of KHC2O4∙2H2O which are neutralized exactly with 16.24 mL of NaOH 0.1072 N. What is the normality of the disosolution of SnCl2 as the reducing agent?

Answer 1

V = 15.27 SnCl2

Fe, Fe3+ to Fe2+

V = 16.27 KMnO4

V = 16.24 ml NAOH

N = 0.1072

sind normality

The strategy:

moles of NaOH = moles of KHC2O4∙2H2O = moles of KMnO4 = moles of SnCl2

then:

moles of NaOH = N*V = 0.1072*16.24 = 1.740928 mol of NaOH

assume ratio 1:1 then

moles of NaOH = moles of KHC2O4∙2H2O

moles of KHC2O4∙2H2O = 1.740928 mol of NaOH

Assume KMnO4 --> 1:1

1.740928 moles of KMnO4 present

find concnetration:

M1 = mol1/V1 = 1.740928 /16.27 = 0.10700233558

Then

SnCl2 + KMnO4 = assuem 1:1 ratio

then

N = M = M1 = 0.10700233558 N