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Stoichiometric Calculations

A polyvalent ...

Question

A polyvalent metal weighing $0.1$ g and having atomic weight $51$ g reacted with dil. $H_{2} S O_{4}$ to give $43.9$ mL of $H_{2}$ at STP. This solution containing the metal in lower oxidation state is found to require $58.8$ mL of $0.1$ N $K M n O_{4}$ for complete oxidation. What is the oxidation state of the metal in reaction?

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Solution

The number of equivalents of $K M n O_{4}$ used $= 0.1 \times 0.0588 = 0.00585$ eq.
The number of moles of metal $= \frac{0.1}{51} = 0.00196$ moles
The number of electrons transferred in the reaction is $\frac{0.00585}{0.00196} = 3$ (Since, moles x number of electrons = Equivalents)
Hence, the oxidation state of the metal is $+ 3$ .

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