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VSEPR Theory

5.5 g of a mi...

Question

$5.5 g$ of a mixture of $F e S O_{4} \cdot 7 H_{2} O$ and $F e_{2} (S O_{4})_{3} \cdot 9 H_{2} O$ required $5.4 m L$ of $0.1 N K M n O_{4}$ solution for complete oxidation. Calculate the gram mole of hydrated ferric sulphate in the mixture.
(At. mass $H = 1, O = 16, S = 32, F e = 56$ ).

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Solution

Only $F e S O_{4} \cdot 7 H_{2} O$ will be oxidised by $K M n O_{4}$ .
Mol. mass of $F e S O_{4} \cdot 7 H_{2} O = 278$
As the conversion involves one electron,
$F e^{2 +} \to F e^{3 +} + e^{-}$
The eq. mass of $F e S O_{4} \cdot 7 H_{2} O = \frac{278}{1} = 278$
$5.4 m L$ of $0.1 N K M n O_{4}$
$\equiv 5.4 m L$ of $0.1 N F e S O_{4} \cdot 7 H_{2} O$ solution
Amount of $F e S O_{4} \cdot 7 H_{2} O = \frac{0.1 \times 278}{1000} \times 5.4 = 0.15 g$
Amount of $F e_{2} (S O_{4})_{3} \cdot 9 H_{2} O = (5.5 - 0.15) = 5.35 g$
Mol. mass of $F e_{2} (S O_{4})_{3} \cdot 9 H_{2} O = 562$
No. of $g$ moles of $F e_{2} (S O_{4})_{3} \cdot 9 H_{2} O = \frac{M a s s}{M o l . m a s s}$
$= \frac{5.35}{562} = 0.00952$
$= 9.52 \times 10^{- 3}$ .

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