25 ml of 0.50 M H2O2 solution is added to 50ml of 0.20 M KMnO4 in acidic solution.
Answers
Question:
25 ml of 0.50 M H₂O₂ solution is added to 50 ml of 0.20 M KMnO₄ is acid solution. Which of the following statements is true?
a) 0.010 mole of oxygen is liberated
b) 0.005 mole of KMnO₄ are left
c) 0.030 g atom of oxygen is liberated
d) 0.0025 mole of H₂O₂ does not react with KMnO₄
Answer:
b) 0.005 mole of KMnO₄ are left.
Explanation:
KMn⁺¹O₄ + 5e⁻ + H⁺ ==> Mn²⁺ Valency Factor = 5
(O⁻¹)₂ ==> O₂ + 2e⁻ Valency Factor = 2
These are the equations involved.
Mass Equivalent = Volume x Molarity x Valency Factor
Mass Equivalent of H₂O₂ = 25 × 0.5 × 2 = 25
Mass Equivalent of KMnO₄= 50 × 0.2 × 5= 50
Difference of Mass Equivalent = 50 - 25 = 25
25 Mass Equivalent of KMnO₄ is left = ( W * 1000 ) / ( M/Valency Factor)
M = 25 / (1000*5) = 5 milli-moles of KMnO₄
Therefore, 25 Mass Equivalent or 5 milli-mole of KMnO₄ or 0.005 mole of KMnO₄ are left.
b) 0.005 mole of KMnO₄ are left.
Explanation:
KMn⁺¹O₄ + 5e⁻ + H⁺ ==> Mn²⁺ Valency Factor = 5
(O⁻¹)₂ ==> O₂ + 2e⁻ Valency Factor = 2
▸These are the equations involved.
➦Mass Equivalent = Volume x Molarity x Valency Factor
Mass Equivalent of H₂O₂ = 25 × 0.5 × 2 = 25
➻Mass Equivalent of KMnO₄= 50 × 0.2 × 5= 50
Difference of Mass Equivalent = 50 - 25 = 25
25 Mass Equivalent of KMnO₄ is left = ( W * 1000 ) / ( M/Valency Factor)
M = 25 / (1000*5) = 5 milli-moles of KMnO₄
➦Therefore, 25 Mass Equivalent or 5 milli-mole of KMnO₄ or 0.005 mole of KMnO₄ are left.✔
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