How many litres of `SO_(2)` taken at NTP have to be passed through a solution of `HCIO_(3)` to reduce `16.9` g of it to HCI?

Explanation

To solve the problem of how many liters of `SO2` are required to reduce `16.9 g` of `HClO3` to `HCl`, we can follow these steps:

### Step 1: Determine the equivalent weight of `HClO3`
The equivalent weight of a substance can be calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{n} \]
where \( n \) is the number of electrons transferred per molecule in the reaction.
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