(redox balancing, titrations and stoichiometry) 3. A sample of calcium oxalate (
ID: 854662 • Letter: #
Question
(redox balancing, titrations and stoichiometry) 3. A sample of calcium oxalate (CaC2O4) is dissolved in excess acid, followed by its titration with permanganate according to this reaction (unbalanced): If 37.68 mL of 0.1019 M MnO4 was used to completely titrate the oxalate, what is the mass calcium oxalate in the original sample? (balancing redox reactions) Balance following redox reactions, taking place in acidic conditions. In each case, indicate which element is reduced, which element is oxidized.Explanation / Answer
2MnO4- +5 C2O42-+ 16H+? 2Mn2++ 10CO2 + 8H2O
Number of moles of KMnO4 present the given solution = 37.68mL*0.1019/1000
=3.84*10^-3 moles
According to above equation 2 moles ofKMnO4 reacts with 5 moles of Calicium oxalate
So 3.84*10^-3 moles of KMnO4 =(5/2)* 3.84*10^-3=9.6*10^-3 moles of Calicium Oxalate
1 mole of CaC2O4 = 128.1gms
Therefore 9.6*10^-3 moles of CaC2O4= 9.6*10^-3 *128.1=1.2297= 1.23gms
Weight of the CaC2O4= 1.23gms
Reaction-1: 2S2O32- +I3- --------------> S4O62- + 3I- , Here I get oxidised and S get reduced.
Reaction-2: NO3- + 4Zn + 10 H+ = 4 Zn2+ + NH4++ 3 H2O, Here Zn get oxidised, N get reduced
Reaction-3: C2H5OH + 2 Cr2O72-+16 H+ ? 4 Cr3+ + 2 CO2 + 11 H2O
Here Cr+6 get reduced and Alcohol get oxidised .
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