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The number of moles of KMnO4 that will be needed to react with one mole of ferrous sulphite in acidic solution is
- a)0.4
- b)0.6
- c)0.8
- d)1.0
Correct answer is option 'B'. Can you explain this answer?
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To determine the number of moles of KMnO4 needed to react with one mole of ferrous sulphite in acidic solution, we need to balance the chemical equation and use stoichiometry.
The balanced chemical equation for the reaction between KMnO4 and ferrous sulphite in acidic solution is as follows:
5FeSO3 + 2KMnO4 + 8H2SO4 → 5Fe2(SO4)3 + 2MnSO4 + K2SO4 + 8H2O
From the balanced equation, we can see that 2 moles of KMnO4 react with 5 moles of FeSO3. Therefore, the stoichiometric ratio is 2:5.
Now, let's calculate the number of moles of KMnO4 needed.
Given that we have 1 mole of FeSO3, we can set up the following proportion:
2 moles KMnO4 / 5 moles FeSO3 = x moles KMnO4 / 1 mole FeSO3
Cross-multiplying and solving for x, we get:
2 moles KMnO4 = 5 moles FeSO3 * x moles KMnO4
x moles KMnO4 = (2 moles KMnO4 * 1 mole FeSO3) / 5 moles FeSO3
x moles KMnO4 = 0.4 moles KMnO4
Therefore, the number of moles of KMnO4 that will be needed to react with one mole of ferrous sulphite in acidic solution is 0.4 moles, which corresponds to option B.
The balanced chemical equation for the reaction between KMnO4 and ferrous sulphite in acidic solution is as follows:
5FeSO3 + 2KMnO4 + 8H2SO4 → 5Fe2(SO4)3 + 2MnSO4 + K2SO4 + 8H2O
From the balanced equation, we can see that 2 moles of KMnO4 react with 5 moles of FeSO3. Therefore, the stoichiometric ratio is 2:5.
Now, let's calculate the number of moles of KMnO4 needed.
Given that we have 1 mole of FeSO3, we can set up the following proportion:
2 moles KMnO4 / 5 moles FeSO3 = x moles KMnO4 / 1 mole FeSO3
Cross-multiplying and solving for x, we get:
2 moles KMnO4 = 5 moles FeSO3 * x moles KMnO4
x moles KMnO4 = (2 moles KMnO4 * 1 mole FeSO3) / 5 moles FeSO3
x moles KMnO4 = 0.4 moles KMnO4
Therefore, the number of moles of KMnO4 that will be needed to react with one mole of ferrous sulphite in acidic solution is 0.4 moles, which corresponds to option B.
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