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Sulphuric Acid

n factor for ...

Question

$n$ factor for $C u_{2} S$ in the reaction: $C u_{2} S + K M n O_{4} ⟶ C u^{2 +} + S O_{2} + M n^{2 +}$ , is:

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Solution

$C u_{2} S + K M n O_{4} ⟶ C u^{2 +} + S O_{2} + M n^{2 +}$
+1 -2 +2 +4
So, $n$ -factor of $C u_{2} S = 4 - (- 2) + 2 \times (2 - 1) = 2 + 6 = 8$ .

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