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Question: Answered & Verified by Expert
\( 5.5 \mathrm{~g} \) of a mixture of \( \mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O} \) and \( \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O} \) requires \( 5.4 \mathrm{~mL} \) of \( 0.1 \mathrm{~N} \mathrm{KMnO}_{4} \) solution for complete oxidation. Calculate the number of moles of hydrated ferric sulphate in the mixture.
ChemistryRedox ReactionsJEE Main
Solution:
1860 Upvotes Verified Answer
The correct answer is: 0.0095

FeSO4·7H2O will oxidise as Fe is present in +2 oxidation state.

In Fe2SO43·9H2OFe is already present in +3 oxidation state, so it will not oxidise.

Hence, milliequivalents of FeSO4·7H2O= milliequivalents of KMnO4 

N1V1 FeSO4·7H2O=N2V2 KMnO4x278×1000=0.1×5.4x=0.1501 g

Amount of hydrated ferric sulphate =5.5 g0.1501 g=5.34 g

Number of moles = Mass / Molar mass =5.34/562=0.0095 mol

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