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 following:
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 and 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 after the completion of the reaction.
Useful Hint: Hydrogen peroxide is a strong oxidizing agent and considered to be a molecular compound.
1.) To write down the chemical equation for this reaction, we need to identify the reactants and the products involved. The given information tells us that potassium permanganate (KMnO4) and hydrogen peroxide (H2O2) are the reactants. The products are potassium sulfate (K2SO4), manganese (II) sulfate (MnSO4), and water (H2O).
The chemical equation can be written as:
2KMnO4(aq) + 3H2O2(aq) + 2H2SO4(aq) → K2SO4(aq) + 2MnSO4(aq) + 3O2(g) + 4H2O(l)
2.) To balance the chemical equation, we need to ensure that the number of atoms of each element is equal on both sides of the equation.
Balanced chemical equation:
2KMnO4(aq) + 3H2O2(aq) + 2H2SO4(aq) → K2SO4(aq) + 2MnSO4(aq) + 3O2(g) + 4H2O(l)
3.) The type of reaction observed in this case is a redox reaction (oxidation-reduction reaction). Hydrogen peroxide (H2O2) acts as an oxidizing agent, and potassium permanganate (KMnO4) acts as a reducing agent.
The color change observed is due to the reduction of purple-colored permanganate ions (MnO4-) to colorless manganese (II) ions (Mn2+). The pink color of the solution is due to the presence of the permanganate ions. As they are reduced, the color fades and eventually becomes colorless.
4.) Before the reaction:
The solution contains potassium permanganate ions (KMnO4-), hydrogen peroxide ions (H2O2), and hydrogen sulfate ions (HSO4-).
After the reaction:
The solution contains potassium sulfate ions (K2SO4), manganese (II) sulfate ions (MnSO4), and water molecules (H2O).
5.) The balanced equation shows that 1 mole of KMnO4 produces 2 moles of MnSO4. The molar mass of MnSO4 is 151.0019 g/mol, so for 5.65 g of KMnO4, we need to calculate the number of moles and then multiply by the molar ratio.
To calculate the number of moles of KMnO4:
Number of moles = Mass / Molar mass
Number of moles = 5.65 g / 158.034 g/mol (molar mass of KMnO4)
Next, we use the molar ratio from the balanced equation:
1 mole of KMnO4 produces 2 moles of MnSO4
Total moles of MnSO4 = Moles of KMnO4 * (2 moles of MnSO4 / 1 mole of KMnO4)
Finally, substitute the values into the equation:
Total moles of MnSO4 = (5.65 g / 158.034 g/mol) * (2 moles of MnSO4 / 1 mole of KMnO4)
Calculate this expression to find the total number of moles of MnSO4.