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Stoichiometric Calculations

5.5 g of a mi...

Question

$5.5$ g of a mixture of $F e S O_{4} .7 H_{2} O$ and $F e_{2} (S O_{4})_{3} .9 H_{2} O$ required $5.4$ mL of $0.1 N K M n O_{4}$ solution for complete oxidation. Moles of hydrated ferric sulphate in mixture are $9.52 \times 10^{- x}$ moles. The value of $x$ is :

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Solution

The redox changes are:
$M n^{7 +} + 5 e ⟶ M n^{2 +}$
$F e^{2 +} ⟶ F e^{3 +} + e$
Only $F e S O_{4} \cdot 7 H_{2} O$ present in mixture reacts with $K M n O_{4}$ and therefore
$M e q . o f F e S O_{4} \cdot 7 H_{2} O = M e q . o f K M n O_{4}$
$(\frac{w}{E}) \times 1000 = 5.4 \times 0.1$
or
$⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{w}{(\frac{278}{1})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ \times 1000 = 0.54$
$(∵ E_{F e S O_{4} \cdot 7 H_{2} O} = \frac{M}{I})$
or $w_{F e S O_{4} \cdot 7 H_{2} O} = 0.150 g$
$∴ m a s s o f F e_{2} (S O_{4})_{3} \cdot 9 H_{2} O = (5.5 - 0.150) g$
$= 5.350 g$
$∵ M o l a r m a s s o f F e_{2} (S O_{4})_{3} \cdot 9 H_{2} O = 562$
$∴ M o l e o f F e_{2} (S O_{4})_{3} \cdot 9 H_{2} O = \frac{5.35}{562}$
$= 9.52 \times 10^{- 3} m o l e s$
so value of x is $3$ .

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