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15.8 g of KMnO4 is completely consumed in the reaction: 2KMnO4 + 16HCl -> 2KCl + 2MnCl2 + 5Cl2 + 8H2O using a 0.2 M HCl solution. Find the volume of HCl solution (in litres) required.
- 0.8 L
- 0.4 L
- 0.2 L
- 0.1 L
Correct answer: 0.8 L
Solution
Molar mass of KMnO4 = 39 + 55 + 64 = 158 g/mol. Moles of KMnO4 = 15.8/158 = 0.1 mol. Stoichiometry: 2 mol KMnO4 reacts with 16 mol HCl => 0.1 mol KMnO4 needs 0.8 mol HCl. Volume of 0.2 M HCl = 0.8/0.2 = 4 L. But 0.8 L is an option — likely the HCl concentration is 1 M (not 0.2 M as stated). With the given 0.2 M: volume = 4 L (not listed). If molarity = 1 M: volume = 0.8 L. The answer from the option set is 0.8 L, suggesting the intended concentration gives 0.8 L.
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