A 458 g sample containing Mn3O4 was dissolved and all manganese was converted to Mn2+. In thepresence of fluoride ion, Mn2+ is titrated with 3 lit of KMnO4 solution (which was 1.25 N againstoxalate in acidic medium), both reactants being converted to a complex of Mn(III) , What was the %of Mn3O4 in the sample?
Question
A 458 g sample containing Mn3O4 was dissolved and all manganese was converted to Mn2+. In thepresence of fluoride ion, Mn2+ is titrated with 3 lit of KMnO4 solution (which was 1.25 N againstoxalate in acidic medium), both reactants being converted to a complex of Mn(III) , What was the %of Mn3O4 in the sample?
Solution
To solve this problem, we need to follow these steps:
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First, we need to understand the reaction between Mn2+ and KMnO4. In acidic medium, KMnO4 is reduced to Mn2+ while Mn2+ is oxidized to Mn3+. The balanced equation is:
2Mn2+ + 5KMnO4 + 16H+ → 2Mn3+ + 5Mn2+ + 8H2O
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From the balanced equation, we can see that 5 moles of KMnO4 react with 2 moles of Mn2+. Therefore, the molar ratio of KMnO4 to Mn2+ is 5:2.
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We know that the KMnO4 solution was 1.25 N, which means it contains 1.25 moles of KMnO4 per liter. Since we used 3 liters of this solution, we have 3*1.25 = 3.75 moles of KMnO4.
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Using the molar ratio from step 2, we can calculate the moles of Mn2+ in the sample. (3.75 moles KMnO4) * (2 moles Mn2+ / 5 moles KMnO4) = 1.5 moles Mn2+.
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The molar mass of Mn3O4 is 228.81 g/mol. Since Mn3O4 contains 3 moles of Mn for every mole of Mn3O4, the moles of Mn3O4 in the sample is (1.5 moles Mn2+) / 3 = 0.5 moles Mn3O4.
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The mass of Mn3O4 in the sample is (0.5 moles Mn3O4) * (228.81 g/mol) = 114.405 g.
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Finally, the percentage of Mn3O4 in the sample is (114.405 g / 458 g) * 100% = 24.98%.
So, the percentage of Mn3O4 in the sample is approximately 25%.
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