Question

How many grams of oxalic acid (H2C204) is required to completely reduce 25 ml of 0.1 M KMnO4 solution? (Molecular weight of H2C2O4 = 90) ​

Answer 1

Answer:

0.36 g.

I hope it will help you

Answer 2

we have to find the amount of oxalic acid required to completely reduce 25 ml of 0.1 M KMnO₄ solution.

solution : no of moles of potassium permanganate ( KMnO₄) = volume of solution in L × molarity of the solution

= $\frac{25}{1000}L\times0.1M$ = 2.5 × 10⁻³ mol

reaction of oxalic acid and potassium permanganate in acidic medium is given by

$2KMnO_4+5H_2C_2O_4+3H_2SO_4\implies2MnSO_4+10CO_2+8H_2O+K_2SO_4$

here you see, 5 moles of oxalic acid is required to completely reduce 2 moles of potassium permanganate.

∴ $\frac{5}{2}\times2.5\times10^{-3}$ = 6.25 × 10⁻³ mol of oxalic acid is required to completely reduce 2.5 × 10⁻³ mol of KMnO₄.

as it has given that molecular weight of oxalic acid, (H₂C₂O₄) = 90 g/mol

∴ the mass of the oxalic acid = no of moles of oxalic acid × molecular weight of oxalic acid

= 6.25 ×  10⁻³ × 90 = 0.5625 g

therefore 0.5625 g of oxalic acid is required to reduce 25 ml of 0.1 M potassium permanganate.