Question

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15.8 g KMnO4 is completely treated with 0.2 M HCl solution According to reaction
2KMnO4 + 16HCI = 2KCl + 2MnCl2 + 5C12 + 8H20
Volume of HC1 in litre is-​

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

2KMnO₄ + 16HCl ---> 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O

Equivalent weight of KMnO₄ = 158/5 = 31.6

31.6 g of KMnO₄ reacts with 1000 ml of 1N HCl

15.8 g of KMnO₄ reacts with 500 ml of 1N HCl

Since equivalents of KMnO₄ = equivalents of HCl

N₁V₁ = N₂V₂

15.8 g of KMnO₄ reacts with 2500 ml of 0.2N HCl

For HCl, n factor = 1

Therefore Molarity = Normality

15.8 g of KMnO₄ reacts with 2.5 litres of 0.2 M HCl

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