If x is the water of crystallisation of the ore carnallite,

and y be the oxidation state of the central atom of the product (A), that is responsible for a green flame in the following reaction,

$B O_{3}^{3 -} + c o n c . H_{2} S O_{4} + C H_{3} - C H_{2} - O H i g n i t e - -- \to A$

If z is the oxidation state of iron in the brown ring complex,
$F e^{2 +} + N O_{3}^{-} + c o n c . H_{2} S O_{4} \to B r o w n r i n g$

Then x+y+z is:

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Solution

Carnallite is $K C l . M g C l_{2} .6 H_{2} O$

$B O_{3}^{3 -} + c o n c . H_{2} S O_{4} + C H_{3} - C H_{2} - O H i g n i t e - -- \to (C_{2} H_{5} O)_{3} B$

The oxidation state of Boron in $(C_{2} H_{5} O)_{3} B$ is +3

$F e^{2 +} + N O_{3}^{-} + c o n c . H_{2} S O_{4} \to [F e (H_{2} O)_{5} N O] S O_{4}$

The oxidation state of Iron would be +1

x=6, y=3 and z=1

x+y+z= 10

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Similar questions

\begin{matrix} B O_{3}^{3 -} + c o n c . H_{2} S O_{4} + C H_{3} - C H_{2} - O H i g n i t e - -- \to & (A) G r e e n f l a m e \end{matrix}

F e^{2 +} (a q) + N O_{3}^{-} (a q) + H_{2} S O_{4} (c o n c) ⟶

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

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

K_{3} C r O_{8}

C r O_{4}^{2 -} + H_{2} O_{2} + O H^{-} \to [C r O_{8}]^{3 -} + H_{2} O + O_{2}

C r

X

O

Y

C r

Z

X + Y + Z

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