You visited us 1 times! Enjoying our articles? Unlock Full Access!

Reaction with Metals

In the follow...

Question

In the following reaction
$X e F_{2} + {B r O_{3}}^{-} + H_{2} O ⟶ X e + {B r O_{4}}^{-} + 2 H F$

$e q u i v a l e n t m a s s o f {B r O_{3}}^{-} = \frac{m o l a r m a s s}{. . . . . . . . . .}$

Open in App

Solution

In the following reaction:

$X e F_{2} + {B r O_{3}}^{-} + H_{2} O ⟶ X e + {B r O_{4}}^{-} + 2 H F$

$e q u i v a l e n t m a s s o f B r O_{3}^{-} = \frac{m o l a r m a s s}{2}$ .

The oxidation number of Br changes form +5 to +7. Hence, the net change in the oxidation number is 2.

Suggest Corrections

Similar questions

X e F_{2} + B r O_{3}^{-} + H_{2} O ⟶ B r O_{4}^{-} + X e + 2 H F

B r O_{3}^{-}

B r O_{4}^{-}

X e F_{2}

X e F_{2} m_{1} + B r O_{3}^{-} m_{2} + H_{2} O ⟶ B r O_{4}^{-} + X e + 2 H F

m_{1}

m_{2}

X e F_{2}

B r O_{3}^{-}

X e F_{2}

B r O_{3}^{-}

X e O_{3} + 6 H^{+} + 6 F e^{2 +} ⟶ X e + 3 H_{2} O + 6 F e^{2 +}

e q u i v a l e n t m a s s o f X e O_{3} = \frac{m o l a r m a s s}{. . . . . . . . . .}

B r O_{3}^{-}

B r O_{4}^{-}

X e F_{6} + H_{2} O ⇌ X e O F_{4} + 2 H F

= k_{1}, X e O_{4} + X e F_{6} ⇌ X e O F_{4} + X e O_{3} F_{2}

= K_{2}

X e O_{4} + 2 H F ⇌ X e O_{3} F_{2} + H_{2} O

Join BYJU'S Learning Program