A 0.5 g sample containing `MnO_(2)` is treated with HCl liberating `Cl_(2)` is passed into a solution of KI and 30.0 " mL of " 0.1 M `Na_(2)S_(2)O_(3)` are required to titrate the liberated iodine. Calculate the percentage of `MnO_(2)` is the sample.

Explanation

To solve the problem, we need to determine the percentage of \( \text{MnO}_2 \) in a 0.5 g sample based on the titration results with \( \text{Na}_2\text{S}_2\text{O}_3 \). Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the moles of \( \text{Na}_2\text{S}_2\text{O}_3 \)
We are given that 30.0 mL of 0.1 M \( \text{Na}_2\text{S}_2\text{O}_3 \) is required for titration.

\[ \text{Moles of } \text{Na}_2\text{S}_2\text{O}_3 = \text{Molarity} \times \text{Volume (L)} \]
...