Final answer:
To calculate the percentage of MnO₂ in a mineral sample, the moles of Na₂S₂O₃ reacting with I₂ liberated by MnO₂ are determined and converted to moles of MnO₂, which are then used to calculate the mass and percent of MnO₂ in the sample. However, the percentage obtained exceeds 100%, suggesting an error in the data or calculations.
Step-by-step explanation:
To calculate the percentage of MnO₂ in the mineral specimen, we need to first determine the moles of Na₂S₂O₃ that reacted. Using the titration data provided, we find that:
32.30 mL of 0.07220 M Na₂S₂O₃ = 32.30 mL x (1 L / 1000 mL) x 0.07220 mol/L = 0.00233166 mol of Na₂S₂O₃
The stoichiometry of the reaction between Na₂S₂O₃ and the iodine (I₂) which was liberated by MnO₂ is 1:1. Therefore, 0.00233166 mol I₂ was produced by MnO₂.
Next, the molar mass of MnO₂ is 86.94 g/mol. The moles of MnO₂ can be found using the moles of I₂ (since the moles are equal due to the stoichiometry of the reaction): 0.00233166 mol MnO₂. To find the mass of MnO₂ that produced this amount of I₂, we multiply the moles by the molar mass of MnO₂:
0.00233166 mol x 86.94 g/mol = 0.20275 g of MnO₂
Finally, to find the percentage of MnO₂ in the sample, we use the mass of MnO₂ and the original sample mass:
(0.20275 g MnO₂ / 0.1344 g sample) x 100 = 150.86%
However, it is not possible to have a mass percentage over 100%. This indicates a possible mistake in either the provided experimental data or in the calculation process. It should be carefully reviewed.