To 100 " mL of " `KMnO_4` solution containing 0.632 g of `KMnO_4` 200 " mL of " `SnCl_2` containing 2.4g is added in presence of HCl. To the resulting solution, an excess of `HgCl_2` is added at once. How many grams of `Hg_2Cl_2` will be precipitated? (molarcular mass of `KMnO_4` is 158, `SnCl_2` is 95, and `Hg_2Cl_2` is 471 g `mol^(-1))`

Explanation

To solve the problem step by step, we will follow the stoichiometric principles and calculations involved in the reactions between the given compounds.

### Step 1: Calculate the number of moles of KMnO4

Given:
- Mass of KMnO4 = 0.632 g
- Molar mass of KMnO4 = 158 g/mol

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