How many mL. of aqueous solution of `KMnO_(4)` containing `158g/L` must be used to complete the conversation of `75.0g` of `KI` to iodine by the reaction
`KMnO_(4)+KI+H_(2)SO_(4)rarrK_(2)SO_(4)+MnSO_(4)+I_(2)+6H_(2)O`

Explanation

To solve the problem, we need to determine how many mL of an aqueous solution of KMnO₄ containing 158 g/L is required to completely convert 75.0 g of KI to iodine (I₂) using the given reaction:

\[ \text{KMnO}_4 + \text{KI} + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + \text{MnSO}_4 + \text{I}_2 + 6\text{H}_2\text{O} \]

### Step 1: Calculate the number of moles of KI
First, we need to calculate the moles of KI present in 75.0 g.

- Molar mass of KI = K (39.1 g/mol) + I (126.9 g/mol) = 166 g/mol
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