Problem:
What volume of 0.1N oxalic acid solution can be reduced by 250g of an 8 percent by weight KMnO4 solution?

Solution:

To find the volume of the oxalic acid solution that can be reduced by the given amount of KMnO4 solution, we need to determine the stoichiometry between the two compounds.

Stoichiometry:
The balanced chemical equation for the reaction between oxalic acid (C2H2O4) and potassium permanganate (KMnO4) is as follows:

5C2H2O4 + 2KMnO4 + 3H2SO4 → 10CO2 + 2MnSO4 + K2SO4 + 8H2O

From the balanced equation, we can see that 5 moles of oxalic acid react with 2 moles of potassium permanganate.

Calculating Moles of KMnO4:
To calculate the moles of KMnO4, we first need to determine the molecular weight of KMnO4.

Molecular weight of KMnO4 = (39.1 + 54.9 + 16.0*4) g/mol = 158.04 g/mol

Given that the weight of KMnO4 is 250g and the percentage by weight is 8%, we can calculate the moles of KMnO4 as follows:

Moles of KMnO4 = (250g * 0.08) / 158.04 g/mol = 0.126 moles

Calculating Moles of Oxalic Acid:
Since the stoichiometric ratio between KMnO4 and C2H2O4 is 2:5, the moles of oxalic acid can be calculated as follows:

Moles of C2H2O4 = (0.126 moles * 5) / 2 = 0.315 moles

Calculating Volume of Oxalic Acid Solution:
Now, using the given concentration of 0.1N (equivalent to 0.1 moles per liter), we can calculate the volume of the oxalic acid solution as follows:

Volume of Oxalic Acid Solution = (0.315 moles) / (0.1 moles/L) = 3.15 liters

Therefore, the volume of 0.1N oxalic acid solution that can be reduced by 250g of an 8% KMnO4 solution is 3.15 liters.