0.4g of silver powder ws boiled with excess of Fe₂(SO₄)₃ solution..the products are Ag₂SO₄ & FeSO₄ .calculate volume of 0.1N KMnO₄ in acidic medium needed to reoxidise FeSO₄ formed..

To tackle this problem, we need to break it down into a few manageable steps. First, we will determine how much silver reacts with the iron(III) sulfate, and then we will calculate the amount of iron(II) sulfate produced. Finally, we will find out how much potassium permanganate (KMnO4) is required to oxidize the iron(II) sulfate back to iron(III) sulfate in an acidic medium.

Step 1: Determine the moles of silver

The molar mass of silver (Ag) is approximately 107.87 g/mol. Given that we have 0.4 g of silver powder, we can calculate the number of moles of silver:

• Moles of Ag = Mass of Ag / Molar mass of Ag
• Moles of Ag = 0.4 g / 107.87 g/mol ≈ 0.00370 moles

Step 2: Reaction with iron(III) sulfate

The reaction between silver and iron(III) sulfate can be represented as follows:

• 3 Ag + 2 Fe2(SO4)3 → 3 Ag2SO4 + 2 FeSO4

From the balanced equation, we see that 3 moles of silver react with 2 moles of iron(III) sulfate to produce 2 moles of iron(II) sulfate. Therefore, we need to calculate how many moles of iron(II) sulfate are produced from the moles of silver we have:

• Using the stoichiometry from the equation, we find:
• Moles of FeSO4 produced = (2 moles FeSO4 / 3 moles Ag) × moles of Ag
• Moles of FeSO4 = (2/3) × 0.00370 ≈ 0.00247 moles

Step 3: Oxidation of iron(II) sulfate

In acidic medium, potassium permanganate (KMnO4) acts as an oxidizing agent. The balanced reaction for the oxidation of iron(II) sulfate to iron(III) sulfate is:

• 5 FeSO4 + KMnO4 + 8 H2SO4 → 5 Fe2(SO4)3 + K2SO4 + 8 H2O + MnSO4

From this equation, we can see that 1 mole of KMnO4 reacts with 5 moles of FeSO4. Therefore, we can calculate the moles of KMnO4 needed to oxidize the iron(II) sulfate produced:

• Moles of KMnO4 = (1 mole KMnO4 / 5 moles FeSO4) × moles of FeSO4
• Moles of KMnO4 = (1/5) × 0.00247 ≈ 0.000494 moles

Step 4: Calculate the volume of KMnO4 solution

Now that we know the moles of KMnO4 required, we can find the volume of a 0.1 N KMnO4 solution needed. Since KMnO4 is a strong oxidizing agent, its normality is equivalent to its molarity in this context. Therefore, we can use the formula:

• Volume (L) = Moles / Normality
• Volume (L) = 0.000494 moles / 0.1 N = 0.00494 L

To convert this volume into milliliters:

• Volume (mL) = 0.00494 L × 1000 mL/L = 4.94 mL

Final Result

In summary, approximately 4.94 mL of 0.1 N KMnO4 solution is needed to reoxidize the iron(II) sulfate formed in the reaction. This calculation illustrates the stoichiometric relationships in chemical reactions and the importance of understanding molarity and normality in titrations.