Please show all work on how to complete the question\'s different parts as they

Question

Please show all work on how to complete the question's different parts as they are all needed steps to completing the lab. Thank you very much!

I.   Prelab Questions
a.   Write balanced equations for the oxidation of sodium oxalate and the reduction of the permanganate ion in acidic solutions.

b.   Calculate the equivalent weight of the oxalate ion.

c.   If 0.5468 g of sodium oxalate, Na2C2O4, requires 35.43 mL of a KMnO4 solution to reach the end point, what is the normality of the KMnO4 solution?

d.   What volume of 0.100 N KMnO4 would be required to titrate 0.56 g of K2[Cu(C2O4)2]•2H2O?

e.   Define a catalyst and explain how it is used in a chemical reaction.

f.   Calcium is determined in a limestone sample by precipitation CaC2O4, dissolving the precipitate in H2SO4, and titrating with standard KMnO4 . The precipitate from a limestone sample weighting 0.4463 g requires 32.17 mL of 0.1336 N KMnO4 for titration. Calculate the percentage of CaO in the sample.

Explanation / Answer

C2O4(2-) -> 2CO2 + 2e-

MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O

Overall : 2MnO4- + 5C2O4(2-) + 16H+ -> 2Mn2+ 10CO2 + 2Mn2+ + 8H2O

b) Equivalent weight of oxalate ion = 88/2 =44 ( two replacable sodium ions)

c) Overall reaction is

2KMnO4 + Na2C2O4 --> 2 NaMnO4 + K2C2O4

Molecular weight of Na2C2O4= 134, moles of Na2C2O4= 0.5468/134=0.004081

1mole of Na2C2O4 requires 2 moles of KMnO4

0.004081 moles of Na2C2O4 requires 2*0.004081=0.008161

Molartiy = moles/ Volume (L)= 0.008161*1000/35.43=0.230347

d)

The oxalate ion in the K2 [Cu(C2O4)2].2H2O gets oxidised by the KMnO4:

So 5e- + 8H+ + MnO4- ---> Mn2+ + 4H2O   (1)

C2O42- ---> 2CO2 +2e-   (2)

Eq.1*2 + Eq.2 *5

16H++2MnO4- +5C2O4-2 --------> 2Mn+2 +8H2O+ 5CO2

5 moles of C2O4 requires 2 mole of KMnO4

Molecular weight of K2 [Cu (C2O4)2].2H2O= 353.8

Moles of complex= 0.56/353.8=0.001583

1 mole of Complex contains 2 mole of C2O4

moles of C2O4=2*0.001583=0.003166 moles

5 moles of C2O4 requires 2 moles of C2O4

0.003166 moles of C2O4 requires 2*0.003166/5=0.001266

Voume of KmnO4= 0.001266/0.1 L= 0.01266L= 12.66 ml

e) A catalyst reduces the activation energy of reaction by allowing the reaction to proceed in an alternative path.

In the formation of NH3 by N2+3H2-------> 2NH3

Iron is catalyst used. The iron does not participate in the reaction. The quantity of catalyst required is very less.

f) The reaction is

KMnO4 + CaC2O4+ H2SO4 ----> MnSO4+ K2SO4+ CaSO4+ CO2 +H2O

moles of KMnO4= 0.1336*32.17/1000=0.004298

Moles of CaC2O4 = 0.004298

moles of Ca in the sample ( same as CaO)= 0.004298)

Molecular weight of CaO=56

mass of CaO= 0.004298*56=.24 gm

Percentage of CaO in the sample =100*0.24/0.4463=53.77%

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.