Question

When a pink aqueous solution of potassium permanganate, faintly acidified with dilute sulfuric acid was treated with 10% aq. hydrogen peroxide, the reaction took place with the evolution of gas bubbles, and the pink solution was turned colorless. Further chemical analysis revealed that the evolved gas was oxygen, and the resulting solution contains potassium sulfate and manganese (II) sulfate; water was also formed during the same reaction. Please answer the followings: 1) Write down the proper chemical equation for this reaction. 2) Balance the chemical equation. 3) Define the type/types of the reaction, and offer an explanation for the color change. 4) Write down the ions (and define types) present in the solution before & after the reaction. 5) If initially 5.65 g of potassium permanganate was taken for the reaction, calculate the total number of moles of manganese (II) sulfate present in the solution after the completion of the reaction

Answer 1

1) proper chemical equation

KMnO_4(aq) + H₂SO_4(aq)     K₂SO_4(aq) + MnSO_4(aq)  + H₂O_(l)  + O_2(g)

2) Balanced chemical equation

4KMnO_4(aq) + 6H₂SO_4(aq)    2 K₂SO_4(aq) + 4MnSO_4(aq)  + 6H₂O_(l)  + 5O_2(g)

3) Type of reaction is oxidation- reduction reaction or redox reaction

Mn have +7 oxidation state in KMnO₄  but after reaction it get reduced to +2 oxidation state in MnSO₄ therefore pink colour disappear.

4) Ion present before reaction are

4K⁺, 4Mn⁷⁺, 16O^2-, 12H⁺, 6SO₄^2-

Ion present after reaction are

4K⁺, 4Mn²⁺, 6SO₄^2-

5) molar mass of possium permagnate = 158.034 gm/mole

then 5.65 gm = 5.65 / 158.034 = 0.0357518 mole of possium permagnate

According to reaction 4 mole of KMnO₄ produce 4 mole of MnSO₄ then 0.0357518 mole of KMnO₄ produce 0.0357518 mole of MnSO₄

in 1 mole of MnSO₄ present 1 mole of Mn(II) thus, in 0.0357518  mole of MnSO₄ present 0.0357518 mole of Mn(II)