140 LABORATORY EXPERIMENTS IN GENERAL CHEMISTRY 3. The permanganate ion reacts w

Question

140 LABORATORY EXPERIMENTS IN GENERAL CHEMISTRY 3. The permanganate ion reacts with diarsenic trioxide in an acidic solution to produce the manganese (ID and arsenate ions. The solution containing the per- manganate ion was prepared by dissolving 0.321 gof NaMno4 in enough water to produce 100.0 mL of solution. This solution was then placed in a buret. It required this solution to react with g impure sample of that was dissolved in Given this information, answer the following questions a. Write the two half equations. b. Balance each half equation. c. Write the balanced redox equation. d. What is being reduced? e. What is being oxidized? f What is the oxidizing agent? g. What is the reducing agent? h. What is the molarity of the NaMno (MM 141.9) solution? i. How many grams of pure As2oa (MM 197.8) were present in the sample? j. What is the As20, in the sample?

Explanation / Answer

a,b,c] 5 As2O3 + 4 MnO4^- + 12 H^+ + 9H2O --> 10 H3AsO4 + 4 Mn^2+

d and e ) Because... each As in As2O3 at +3, loses 2 electrons in each As in H3AsO4 So, the oxidation: As2O3 --> 2H3AsO4 + 4 e- lost and because... each Mn in (MnO4)^-1 at +7, and takes 5 electrons and become Mn2 +

So As is being reduced and Mn is being oxidized.

f and g) Specie that gains electrons and is reduced is oxidizing agent that is As

And Mn loose electrons and get oxidized,it is reducing agent.

h ) Molarity of Na MnO4= Moles/Litre

Moles =0.321g/141.9

Volume=100 mL=0.1 L

Molarity=0.321/141.9 * 1/0.1

=0.023 M

i) So there were 0.0023 Moles of Na + and MnO4- ions

So 4 moles of MnO4- ions react with 5 moles of As2O3

Then 0.0023 moles of MnO4- will react with 5/4 * 0.0023

=0.0029 moles of As2O3

Grams of pure As2O3= 0.0029 * 197.8

=0.57 gram As2O3

j ) Percentage of As2O3 in sample=0.57(Mass of pure As2O3) /0.792 (mass of whole sample) * 100

0.719 6*100

=71.96

rOUNDING OFF

=72 %

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