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Potassium Dichromate

Inorganic sal...

Question

Inorganic salt (A) which is hydrated and light green in colour and gives following reactions
(i) (A) decolourizes acidified $M n O_{4}^{-}$ and turns acidified $C r_{2} O_{7}^{2 -}$ green.
(ii) (A) in aqueous solution gives light green turbidity which dissolves in dil.acid.
(iii) (A) gives white ppt with $B a C l_{2}$ solution insoluble in conc. HCl. 2.78 g of (A) gave 2.33 g of white ppt.
(iv) (A) $△ \to (B) (b l a c k i s h b r o w n) + (C) + (D)      o x i d e s o f s u l p h u r$
$(C) o x i d a t i o n - ------ \to (D)$
(C) turns acidified $K_{2} C r_{2} O_{7}$ solution green
(B) + HCl $△ \to (E) d e e p y e l l o w$
(E) + KCNS $⟶$ red colour
Inorganic salt (A) is identified as $F e S O_{4} . 7 H_{2} O$ .
If true enter 1, else enter 0.

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Solution

By (iii) , it is clear that (A) has $S O_{4}^{2 -}$
(A) decolourises $M n O_{4}^{-} / H^{+}$
(A) turns $C r_{2} O_{7}^{2 -} / H^{+}$ green
$\Rightarrow$ (A) has $F e^{2 +}$
$\Rightarrow$ (A) is $F e S O_{4} . x H_{2} O$
(iii) $F e S O_{4} . x H_{2} O (152 + 18 x) (2.78) + B a C l_{2} ⟶ B a S O_{4} 233 (2.33)$
$\frac{152 + 18 x}{233}$ = $\frac{2.78}{2.33}$
$\Rightarrow$ x= 7
$\Rightarrow$ (A) is $F e S O_{4} . 7 H_{2} O$
Explanation :
(i) $M n O_{4}^{-} + 8 H^{+} 5 F e^{2 +} ⟶ 5 F e^{3 +} + M n^{2 +} c o l o u r l e s s + 4 H_{2} O$
(ii) $C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 F e^{2 +} ⟶ 6 F e^{3 +} + 2 C r 3 + g r e e n + 7 H_{2} O$
(iii) $F e S O_{4} + 2 H_{2} O ⟶ F e (O H)_{2} g r e e n t u r b i d i t y ↓ + H_{2} S O_{4}$
$F e (O H)_{2} + H_{2} S O_{4} ⟶ F e S O_{4} s o l u b l e + 2 H_{2} O$
(iv) $2 F e S O_{4} △ ⟶ F e_{2} O_{3} (B) + S O_{2} (C) + S O_{3} (D)$
$2 S O_{2} (C) + O_{2} ⟶ 2 S O_{3} (D)$
$C r_{2} O_{7}^{2 -} + 3 S O_{2} + 2 H^{+} ⟶ 2 C r^{3 +} g r e e n + 3 S O_{4}^{2 -} + H_{2} O$
$F e_{2} O_{3} + 6 H C l ⟶ 2 F e C l_{3} y e l l o w + 3 H_{2} O$
$F e C l_{3} + K C N S ⟶ [F e (C N S)] C l_{2} r e d + K C l$

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If true enter 1, else enter 0.

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