calculate volume n ml of KMnO₄ in acid medium needed to oxidise 1g of FeC₂O₄.

To determine the volume of potassium permanganate (KMnO4) in milliliters required to oxidize 1 gram of ferrous oxalate (FeC2O4) in an acidic medium, we first need to understand the stoichiometry of the reaction involved. The oxidation of ferrous oxalate by KMnO4 occurs in an acidic environment, where KMnO4 acts as an oxidizing agent. Let's break down the steps to find the answer.

Step 1: Identify the Reaction

The balanced chemical equation for the reaction between KMnO4 and FeC2O4 in acidic medium can be represented as follows:

5 FeC2O4 + 2 KMnO4 + 6 H2SO4 → 5 Fe2(SO4)3 + 2 MnSO4 + 8 H2O + 10 CO2

From this equation, we see that 2 moles of KMnO4 react with 5 moles of FeC2O4.

Step 2: Calculate Moles of FeC2O4

Next, we need to calculate the number of moles of FeC2O4 in 1 gram. The molar mass of FeC2O4 is calculated as follows:

• Iron (Fe): 55.85 g/mol
• Carbon (C): 12.01 g/mol (2 atoms)
• Oxygen (O): 16.00 g/mol (4 atoms)

The molar mass of FeC2O4 = 55.85 + (2 × 12.01) + (4 × 16.00) = 55.85 + 24.02 + 64.00 = 143.87 g/mol.

Now, we can calculate the moles of FeC2O4 in 1 gram:

Moles of FeC2O4 = mass (g) / molar mass (g/mol) = 1 g / 143.87 g/mol ≈ 0.00695 moles.

Step 3: Determine Moles of KMnO4 Required

Using the stoichiometry from the balanced equation, we find out how many moles of KMnO4 are needed to react with the moles of FeC2O4:

From the equation, 5 moles of FeC2O4 require 2 moles of KMnO4. Therefore, the moles of KMnO4 needed can be calculated as follows:

Moles of KMnO4 = (2 moles KMnO4 / 5 moles FeC2O4) × moles of FeC2O4

Moles of KMnO4 = (2/5) × 0.00695 ≈ 0.00278 moles.

Step 4: Calculate Volume of KMnO4 Solution

To find the volume of KMnO4 solution needed, we need to know the concentration of the KMnO4 solution. Let's assume we have a 0.02 M (molar) solution of KMnO4. The volume can be calculated using the formula:

Volume (L) = moles / concentration (mol/L)

Volume (L) = 0.00278 moles / 0.02 mol/L = 0.139 L.

To convert this to milliliters:

Volume (mL) = 0.139 L × 1000 mL/L = 139 mL.

Final Calculation

Thus, approximately 139 mL of a 0.02 M KMnO4 solution is required to oxidize 1 gram of FeC2O4 in an acidic medium. If the concentration of your KMnO4 solution differs, you can adjust the calculations accordingly using the same method.