Question

(a) How many milliliters of 0.1490 M KMnO4 are needed to react with 131.0 mL of 0.1490 M oxalic acid?

______ mL

(b) How many milliliters of 0.1490 M oxalic acid are required to
react with 131.0 mL of 0.1490 *M* KMnO_{4}?

_______ mL

Answer #1

a) 2KMnO4 + 3C2H2O4 -----------------> K2CO3 + 2MnO4 + 3H2O + 5CO2

millimoles of oxalic acid = 131 x 0.1490 = 19.519

3 millimoles oxalic acid reacts with 2 millimoles KMnO4

19.519 millimoles oxalic acid reacts with 19.519 x 2 / 3 = 13.0 millimoles of KMnO4

13 = V x 0.149

V = 87.25 mL

volume of KMnO4 required = 87.25 mL

b) millimoles of KMnO4 = 131 x 0.149 = 19.519

2 millimoles KMnO4 reacts with 3 millimoles of oxalic acid

19.519 millimoles KMnO4 reacts with 19.519 x 3 / 2 = 29.28 millimoles oxallic acid

29.28 = V x 0.149

V = 196.58 mL

volume of oxallic acid require = 196.58 mL

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