What volume of 0.1 N oxalic acid solution can be reduced by 250 gram o...
Calculation of Volume of Oxalic Acid Solution Reduced by KMnO4
Given Information
- Concentration of oxalic acid solution = 0.1 N
- Weight of KMnO4 solution = 250 g
- Percent weight of KMnO4 solution = 8%
Calculations
Step 1: Calculation of moles of KMnO4
Molar mass of KMnO4 = 158.03 g/mol
Percentage weight of KMnO4 solution = 8%
Therefore, weight of KMnO4 in 250 g solution = 0.08 x 250 g = 20 g
Number of moles of KMnO4 = 20 g / 158.03 g/mol = 0.1265 moles
Step 2: Calculation of moles of oxalic acid
Balanced chemical equation for the reaction between KMnO4 and oxalic acid:
5C2H2O4 + 2KMnO4 + 3H2SO4 → 10CO2 + 2MnSO4 + K2SO4 + 8H2O
From the equation, we can see that 2 moles of KMnO4 react with 5 moles of oxalic acid.
Therefore, number of moles of oxalic acid = 0.1265 moles x (5/2) = 0.31625 moles
Step 3: Calculation of volume of oxalic acid solution
Molarity of oxalic acid solution = 0.1 N
Number of moles of oxalic acid = 0.31625 moles
Therefore, volume of oxalic acid solution = 0.31625 moles / 0.1 N = 3.1625 liters or 3162.5 ml
Answer
The volume of 0.1 N oxalic acid solution that can be reduced by 250 g of an 8% by weight KMnO4 solution is 3162.5 ml.