Acidic solution of $F e^{2 +}$ ion gives a brown ring when it comes in contact with $N O_{3}^{-}$ ion. Complete the following reactions of this process.
$[F e (H_{2} O)_{6}]^{2 +} +$ _ $\to$ $+ H_{2} O$ .

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Solution

This is the common nitrate test used to detect and to measure of
$[F e (H_{2} O)_{6}]^{2 +} + {N O_{3}}^{-} ⟶ {[F e (H_{2} O)_{5} (N O)] B r o w n R i n g}^{2 +} + H_{2} O$

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Similar questions

F e^{2 +}

N O^{-} 3

[F e (H_{2} O)_{6}]^{2 +} + N O_{3}^{-} + H^{+} \to

+ [F e (H_{2} O)_{6}]^{3 +} + H_{2} O

[F e (H_{2} O)_{5} N O]^{2 +}

N O_{3}^{-}

{N O}_{2}^{-}

{N O}_{3}^{-}

{N O}_{2}^{-}

{N O}_{2}^{-}

B r O_{3}^{-}

B r^{-}

B r_{2}

K B r O_{3}

H_{2} O_{2} (a q) + F e^{2 +} (a q) \to F e^{3 +} (a q) + H_{2} O (l)

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