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Stoichiometric Calculations

25 mL of a so...

Question

$25$ mL of a solution of ferric alum, $F e_{2} (S O_{4})_{3} . (N H_{4})_{2} . S O_{4} .24 H_{2} O$ containing $1.25$ g of the salt was boiled with iron when the reaction occurs as:

$F e + F e_{2} (S O_{4})_{3} ⟶ 3 F e S O_{4}$

The unreacted iron was filtered off and solution treated with $0.107 N$ $K M n O_{4}$ in acid medium.
What is titre value of $F e$ in mL as nearest integer?

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Solution

The reaction is as follows :

$F e + F e_{2} (S O_{4})_{3} ⟶ 3 F e S O_{4}$

$964$ g $F e_{2} (S O_{4})_{3} . (N H_{4})_{2} S O_{4} .24 H_{2} O = 3 \times 152$ g $F e S O_{4}$

$∴ 1.25$ g $F e_{2} (S O_{4})_{3} . (N H_{4})_{2} S O_{4} .24 H_{2} O$ gives $= \frac{3 \times 152 \times 1.25}{964} = 0.5912$ g $F e S O_{4}$

This $F e S O_{4}$ is oxidized by $K M n O_{4}$

$∴$ Meq. of $F e S O_{4}$ $=$ Meq. of $K M n O_{4}$

$\frac{0.5912}{(\frac{152}{1})} \times 1000 = 0.107 \times V$

$∴ V = 36.36$ mL

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