How many moles of acidified KMnO4 are required to oxidise a sample containing 10 moles of KI, 20 moles of H2O2, 3040g of FeSO4 and 30 moles of 2FeC2O4.3SnC2O4.

To determine how many moles of acidified potassium permanganate (KMnO4) are required to oxidize the given substances, we need to analyze each component's oxidation reaction and stoichiometry. Let's break this down step by step.

Understanding the Reactions

KMnO4 is a strong oxidizing agent, especially in acidic conditions. It can oxidize various substances, and each reaction has a specific stoichiometric ratio. Here’s how we can approach the problem:

1. Oxidation of KI

When potassium iodide (KI) is oxidized by KMnO4, iodine (I2) is produced. The balanced reaction is:

• 2 KMnO4 + 16 H+ + 10 KI → 2 Mn2+ + 8 H2O + 5 I2

From this equation, we see that 2 moles of KMnO4 react with 10 moles of KI. Therefore, for 10 moles of KI, we need:

• 10 moles KI → 2 moles KMnO4

2. Oxidation of H2O2

Hydrogen peroxide (H2O2) can also be oxidized by KMnO4. The balanced reaction is:

• 2 KMnO4 + 3 H2O2 + 4 H+ → 2 Mn2+ + 3 O2 + 6 H2O

From this equation, 2 moles of KMnO4 react with 3 moles of H2O2. Thus, for 20 moles of H2O2, we need:

• 20 moles H2O2 → (20/3) * 2 = 13.33 moles KMnO4

3. Oxidation of FeSO4

Iron(II) sulfate (FeSO4) is oxidized to iron(III) sulfate (Fe2(SO4)3) by KMnO4. The balanced reaction is:

• 5 KMnO4 + 8 H+ + 5 Fe2+ → 5 Mn2+ + 4 H2O + 5 Fe3+

Here, 5 moles of KMnO4 react with 5 moles of Fe2+. For 3040 g of FeSO4, we first need to find the number of moles:

• Molar mass of FeSO4 = 55.85 (Fe) + 32.07 (S) + 4*16 (O) = 151.91 g/mol
• Moles of FeSO4 = 3040 g / 151.91 g/mol ≈ 20 moles

Thus, for 20 moles of FeSO4, we need:

• 20 moles Fe2+ → 20 moles KMnO4

4. Oxidation of 2FeC2O4.3SnC2O4

For the oxalate ion (C2O4^2-), the reaction with KMnO4 is:

• 2 KMnO4 + 5 C2O4^2- + 16 H+ → 2 Mn2+ + 10 CO2 + 8 H2O

From this, 2 moles of KMnO4 react with 5 moles of oxalate. For 30 moles of 2FeC2O4.3SnC2O4, we need to calculate the moles of oxalate:

• 30 moles of 2FeC2O4.3SnC2O4 contains 30 moles of C2O4^2-.

Thus, for 30 moles of oxalate, we need:

• 30 moles C2O4^2- → (30/5) * 2 = 12 moles KMnO4

Calculating Total KMnO4 Required

Now, let’s sum up the moles of KMnO4 required for each component:

• From KI: 2 moles
• From H2O2: 13.33 moles
• From FeSO4: 20 moles
• From 2FeC2O4.3SnC2O4: 12 moles

Adding these together gives:

• Total KMnO4 = 2 + 13.33 + 20 + 12 = 47.33 moles

Final Answer

Therefore, approximately 47.33 moles of acidified KMnO4 are required to oxidize the sample containing the specified amounts of KI, H2O2, FeSO4, and 2FeC2O4.3SnC2O4.