Chemistry lab question

Question

The permanganate ion reacts with diarsenic trioxide in an acidic solution to produce the manganese(II) and arsenate ions. The solution containing the per manganate ion was prepared by dissolving 0.321 g of NaMnO 4 in enough water to produce 100.0 mL of solution . This solution was then placed in a buret . It required 24.70 mL of this solution to react with 0.792 g of an impure sample of As 2 O 3 that was dissolved in water and acid . 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 4 (MM=141.9) solution ?

i. How many grams of pure As 2 O 3 (MM=197.8) were present in the sample ?

j. What is the % As 2 O s , in the sample ?

i just need help on I.

Answer 1

a. As₂O₃ ----------- AsO₄^3-

MnO₄^- ---------- Mn²⁺

b. As₂O₃ + 5H₂O ----------- 2 AsO₄^3- + 10 H⁺ + 4 e^- Oxidation half reaction

MnO₄^- + 8 H⁺ + 5 e^- ---------- Mn²⁺ +4 H₂O Reduction half reaction

c. Balanced equation

5 As₂O₃ + 4MnO₄^- +9 H₂O ----------10 AsO₄^3- + 4Mn²⁺ + 18 H⁺

d. MnO₄^- is reduced

e. As₂O₃ is oxidized

f.MnO₄^- is the oxidizing agent

g.As₂O₃ it is the reducing agent.

h. 0.321 grams NaMnO₄ is 0.0023 moles and the molarity is obtained by dividing with the volume 0.1 liter(100 mL), therefore the molarity is 0.023M

24.70 mL of NaMnO₄ 0.023 M is required to react with AS₂O₃

Multiplying the volume of NaMnO₄ ( 24.70 mL which is 0.02470 liter) with it's molarity 0.023M gives the moles of NaMnO₄ which is the moles of MnO₄^-

Therefore the moles of MnO₄^- = 0.02470 Liters X 0.023M = 0.0005681 moles of MnO₄^-

^{0.0005681 moles} of MnO₄^- reacts with AS₂O₃

But as per the equation we balanced in the answer for c, MnO₄^- and AS₂O₃ react with each other in 4:5 mole ratio

0.0005681 moles of MnO₄^- x 5 moles of AS₂O₃ / 4 moles of MnO₄^- -= It gives 7.10 X 10 ^-4 moles of AS₂O₃

_{0.0005681 moles of permanaganate ion reacts with 0.00071moles of Arsenic trioxide}

_{i) 0.00071 moles of} AS₂O₃ is converted into grams by multiplying with its molar mass (197.8 g/mol) which gives 0.14 grams of AS₂O₃

j) Percentage of AS₂O₃ in the sample is = 0.14/ 0.792 X100 gives the percentage of AS₂O₃ in the sample which is 17.73%