$F e^{3 +}$ and $C u^{2 +}$ Can be separated by Excess of $N H_{3} (a q)$ .

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Solution

Fe3+ and Cu2+ can be separated by using excess of NH3(aq), as follows:
Excess of $N H_{3} (a q)$ percipitates $F e^{3 +}$ as $F e (O H)_{3}$
while $C u^{2 +}$ remains soluble as $[C u (N H_{3})_{4}]^{2 +}$ ,
White ppt. with $B a c l_{2} \Rightarrow S O_{4}^{2 -}$ ,
Choclate coloured ppt with $K_{4} [F e (C N)_{6}] \Rightarrow C u^{2 +}$ ,
Thus, hydrated salts is : $C u S O_{4} \cdot x H_{2} O$
$C u S O_{4} \cdot x H_{2} O ⟶ B a S O_{4}$
159.5+18x 233
2.495 2.33
$∴ x = 5$

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Fe3+ and Cu2+ Can be separated by Excess ofNH3(aq).

$F e^{3 +}$ and $C u^{2 +}$ Can be separated by Excess of $N H_{3} (a q)$ .