Calculate the Percentage of MnO2 in a mineral specimen.?

Calculate the Percentage of MnO2 in a mineral specimen in the I2 liberated by a 0.1344 g sample in the net reaction:

MnO2(s) +4H+ + 2I- -%26gt; Mn2+ + I2 + 2H2O

required 32.30mL of 0.07220 M Na2S2O3.

I don't understand what the steps of the problem are.Calculate the Percentage of MnO2 in a mineral specimen.?
The number of moles of I2 is the same as the number of moles of MnO2.



The I2 liberated reacts with 6 moles thiosulfate so you work out the number of moles



of thiosulfate and divide by 6 to get moles MnO2.



Moles thiosulfate = 0.0323 L x 0.07220 M = 0.00233 moles.



0.00233 / 6 = 0.0003887 moles MnO2.



Mass MnO2 = 0.0003887 moles x 86.94 g/mole = 0.0338 g.



( 0.0338 / 0.1344 ) x 100 = 0.3% MnO2.Calculate the Percentage of MnO2 in a mineral specimen.?
I can't explain the detail, you may search the reference. It's an iodometric titration. An iodide, usually potassium iodide is added to a solution of MnO2, it reacts based on the reaction you wrote above:
MnO2(s) +4H+ + 2I- ---%26gt; Mn2+ + I2 + 2H2O

The amount of I2 that's released by the titration is titrated with sodium thiosulphate:
I2 + 2S2O3^2- ---%26gt; S4O6^2- + 2I-

Pay attention that 2 iodide molecule reacts with one MnO2 to produce one I2. In this case I2 = MnO2. And one I2 is needed to react with 2 molecule of thiosulphate. Or one thiosulphate = I2/2 = MnO2/2. Now you can determine the precentage of MnO2.

In equivalence point:
MV = MV
Where M is molarity and V is volume.

I prefer this to be solved in normality rather than molarity, but solve the problem as you wish.