please answer fast15.8 g KMnO4 is completely treated with 0.2 M HCl solution According to reaction2KMnO4 + 16HCI = 2KCl + 2MnCl2 + 5C12 + 8H20Volume of HC1 in litre is- About the author Bella
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 Do mark as brainliest if it helped! Reply
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
Do mark as brainliest if it helped!
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