Question

5 H2C2O4 + 2MnO4^- + 6H2O ----> 10 CO2 + 2 Mn^2+ + 8 H2O

How many milliliters of 0.1650 M KMnO4 are needed to react with 108.0 mL of 0.2550 M oxalic acid?

There are 2 moles of MnO4^- to 5 moles of oxalic acid.

my method was 0.2550 M oxalic acid * .108 liters = 0.02754 moles oxalic acid.

0.02754 moles oxalic acid ( 2 moles MnO4^-/5 moles oxalic acid) = 0.011016 moles MnO4^-

Molarity of MnO4^- at .108 L = 0.102 M MnO4-

M1V1 = M2V2

0.102(0.108) = .1650 (v)

v= 7.210 L or 0.007210 mL

is this correct??

Answer 1

you are correct up to final stage but you did small mistake dear . go through my solution

solution :

2 mol MnO4- --------------------> 5 mol oxalic acid

x mol MnO4- -------------------> 0.02754 mol oxalic acid

moles of MnO4- = x = 2 x 0.02754 / 5 = 0.0110

molarity = moles /volume

0.1650 = 0.0110 / volume

volume = 0.06676 L = 66.76 mL

so    volume = 66.76 mL

ANOTHER SIMPLE METHOD FOR YOU

5 H2C2O4 + 2MnO4^- + 6H2O ----> 10 CO2 + 2 Mn^2+ + 8 H2O

M1 = molarity of KMnO4 = 0.1650 M

V1 = volume of KMnO4 = ??

n1 = number of moles of KMnO4 from equation = 2

M2 = molarity of H2C2O4 = 0.2550 M

V2 = volume of H2C2O4 = 108.0 mL

n2 = number of moles of H2C2O4 from equation = 5

formula :

M1 V1 / n1 = M2 V2 / n2

0.1650 x V1 / 2 = 0.2550 x 108.0 / 5

V1 = 66.76 mL