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Normality

How many mill...

Question

How many millimoles of $H S C N$ will react with iodine, so that the produced $I^{-}$ will react with $10 ml$ of neutral $0.2 M K M n O_{4}$ ?

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Solution

$K I + K M n O_{4} + H_{2} O ⟶ K I O_{3} + K O H + M n O_{2}$
$(n_{f - K I} = 6) (n_{f - K M n O_{4}} = 3)$
Equivalents of $K I = K M n O_{4}$
Number of equivalents of $K I = 3 \times 0.2 \times 10 \times 10^{- 3} = 6 m e q$
Moles = equivalents $\div$ n - factor
Moles of $K I = 6 \div 6 = 1 m m o l$
$S^{-} - C \equiv N + I_{2} + H^{+} ⟶ S O_{4}^{2 -} + H C N^{-} + I^{-} + H_{2} O$
$(n_{f - I_{2}} = 1) (n_{f - S^{-} C N} = 8)$
Equivalents of $S^{-} C N = I_{2} = 1 m e q$
Moles of $S^{-} C N = 1 \div 8 = 0.125 m m o l$

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