Question

Potassium permanganate(KMnO4) reacts with oxalic acid (H2C2O4) in aqueous sulfuric acid according to: 2KMnO4 5H2C2O4 3H2SO4->2MnSO4 10 CO2 * H2O K2SO4 How many milliliters of a .250 M KMnO4 solution are needed to react completely with 3.225 G of oxalic acid

Answer 1

57.3mL of 0.250M KMnO4 are required

Step-by-step explanation:

3.225g of oxalic acid are:

3.225g×(1mol / 90.03g) = 0.03582mol of oxalic acid

Based on the reaction, 5 moles of oxalic acid react with 2 moles of KMnO4, thus, for a complete reaction of oxalic acid you need:

0.03582mol of oxalic acid × (2mol KMnO4 / 5mol Oxalic Acid) = 0.01433mol of KMnO4

If concentration of KMnO4 is 0.250M, liters in 0.01433mol are:

0.01433mol × (1L /0.250mol) = 0.0573L ≡ 57.3mL of 0.250M KMnO4 are required

I hope it helps!