15 mL of 0.5 M H2O2 reacts with 10 mL of 0.5 M KMnO4 in excess of sulphuric acid, according to reaction2KMnO4 + 3H2SO4 + 5H2O2→ K2SO4 + 2MnSO4 + 8H2O + 5O2The volume of O2 produced at STP is

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Grade 11Mechanics

15 mL of 0.5 M H₂O₂ reacts with 10 mL of 0.5 M KMnO₄ in excess of sulphuric acid, according to reaction
2KMnO₄ + 3H₂SO₄ + 5H₂O₂
→ K₂SO₄ + 2MnSO₄ + 8H₂O + 5O₂
The volume of O₂ produced at STP is

Devendra Verma

8 Years agoGrade 11

1 Answer

Shailendra Kumar Sharma

8 Years ago

2KMnO₄ + 3H₂SO₄ + 5H₂O₂→ K₂SO₄ + 2MnSO₄ + 8H₂O + 5O₂

_{15 ml of .5 M H2O2 i.e. 7.5 mmol of total H2O2 we have}

_{10 ml of .5 KmnO4 i.e. total of 5 mmol of KMnO4 we have}

_{Now stoichiometry, says we need 5 moles of H2O2 and 2 moles of KMnO4 which clears that H2O2 is limiting reagent.
and other thing is with 5 mol of H2O2 gives 5 mole of O2 so 7.5 mmol of H2O2 will give 7.5mmol of O2 = 7.5 *22.4 mL =168mL}

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15 mL of 0.5 M H2O2 reacts with 10 mL of 0.5 M KMnO4 in excess of sulphuric acid, according to reaction2KMnO4 + 3H2SO4 + 5H2O2→ K2SO4 + 2MnSO4 + 8H2O + 5O2The volume of O2 produced at STP is

Devendra Verma

1 Answer

Shailendra Kumar Sharma

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15 mL of 0.5 M H₂O₂ reacts with 10 mL of 0.5 M KMnO₄ in excess of sulphuric acid, according to reaction
2KMnO₄ + 3H₂SO₄ + 5H₂O₂
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