I have 50 gram of feed in which 28% manganese (Mn) and 25% iron (Fe) is present. I mixed this feed in 500ml water with 1M H2SO4 (27.7ml) to make MnSO4 and FeSO4 solution (aqueous form), then I added Oxalic acid dihydrated (H2C2O4.2H2O) (powder form). How much oxalic acid I need to precipitate all the Mn and Fe ions from the solution?

Possible reaction could be:

C2H2O4.2H2O(aq) + MnSO4.H2O(aq) = MnC2O4.3H2O(aq) + H2SO4(aq)

C2H2O4.2H2O(aq) + FeSO4.H2O(aq) = FeC2O4.3H2O(aq) + H2SO4(aq)

To determine how much oxalic acid you need to precipitate all the manganese (Mn) and iron (Fe) ions from your solution, we first need to calculate the amounts of Mn and Fe present in your feed and then find out how much oxalic acid is required for the precipitation reactions.

Step 1: Calculate the Amounts of Mn and Fe

Your feed contains 50 grams with 28% manganese and 25% iron. Let's calculate the mass of each metal:

• Manganese: 50 g × 0.28 = 14 g of Mn
• Iron: 50 g × 0.25 = 12.5 g of Fe

Step 2: Convert Mass to Moles

Next, we need to convert these masses into moles using the molar masses of Mn and Fe:

• Molar mass of Mn: 54.94 g/mol
• Molar mass of Fe: 55.85 g/mol

Now, calculate the moles:

• Moles of Mn: 14 g ÷ 54.94 g/mol ≈ 0.254 moles
• Moles of Fe: 12.5 g ÷ 55.85 g/mol ≈ 0.224 moles

Step 3: Stoichiometry of the Precipitation Reactions

The reactions for the precipitation of manganese and iron with oxalic acid are as follows:

• Manganese reaction: C2H2O4.2H2O + MnSO4.H2O → MnC2O4.3H2O + H2SO4
• Iron reaction: C2H2O4.2H2O + FeSO4.H2O → FeC2O4.3H2O + H2SO4

From these reactions, we can see that one mole of oxalic acid reacts with one mole of Mn and one mole of Fe. Therefore, the total moles of oxalic acid required will be the sum of the moles of Mn and Fe:

• Total moles of oxalic acid: 0.254 moles (Mn) + 0.224 moles (Fe) = 0.478 moles

Step 4: Calculate the Mass of Oxalic Acid Needed

Now, we need to find out how much mass of oxalic acid dihydrate (H2C2O4.2H2O) is required. The molar mass of oxalic acid dihydrate is:

• Molar mass of H2C2O4.2H2O: 90.03 g/mol

Using the number of moles calculated earlier, we can find the mass:

• Mass of oxalic acid: 0.478 moles × 90.03 g/mol ≈ 43.0 g

Final Result

To precipitate all the manganese and iron ions from your solution, you will need approximately 43.0 grams of oxalic acid dihydrate. This amount will ensure that all the Mn and Fe ions are effectively precipitated as their respective oxalates.