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Equivalent Mass

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Question

Find the number of moles of $K M n O_{4}$ needed to oxidise one mole $C u_{2} S$ in acidic medium. The reaction is $K M n O_{4} + C u_{2} S \to M n^{2 +} + C u^{2 +} + S O_{2}$

0.4

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1.6

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Solution

The correct option is B 1.6
$K + 7 M n O_{4} + + 1 C u_{2} - 2 S \to M n^{2 +} + C u^{2 +} + + 4 S O_{2}$
Formula used for the n-factor calculation,
$n_{f} = (| O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} |) \times number of atoms$
$n_{f} of K M n O_{4} = | (+ 7) - (+ 2) | \times 1 = 5$

If in a reaction, two elements of same reactant molecule are undergoing either oxidation or reduction then the n-factor of the molecule will be the sum of the individual n-factors of both the elements.
$n_{f} of C u_{2} S = (| (+ 1) - (+ 2) | \times 2) + (| (- 2) - (+ 4) | \times 1) = 8$

From law of equivalence

$Equivalents of C u_{2} S = Equivalents of K M n O_{4}$
$Moles of C u_{2} S \times n_{f} = Moles of K M n O_{4} \times n_{f}$
$Moles of K M n O_{4} needed per mole of C u_{2} S = \frac{1 \times 8}{5} = 1.6$

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K M n O_{4}

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