$a N_{2} H_{4} + b {B r O}_{3}^{-} ⟶ a N_{2} + b B r^{-} + 6 H_{2} O$ ,

What are the values of $a$ and $b$ in the above reaction?

3, 2

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2, 3

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4, 6

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6, 4

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Solution

The correct option is C 3, 2
The balanced reaction is given below:

$3 N_{2} H_{4} + 2 {B r O}_{3}^{-} ⟶ 3 N_{2} + 2 B r^{-} + 6 H_{2} O$
Where a and b are stoichiometric co-efficients.

Hence, the value of $a$ and $b$ are $3$ and $2$ respectively.

Suggest Corrections

Similar questions

a N_{2} H_{4} + b {B r O}_{3}^{-} ⟶ a N_{2} + b B r^{-} + 6 H_{2} O

K B r O_{3}

a N_{2} H_{4} + b {B r O}_{3}^{-} ⟶ a N_{2} + b B r^{-} + 6 H_{2} O

N_{2} H_{4}

H g S O_{4} / H_{2} S O_{4} / H_{2} O - ------------ \to

M g + N_{2} \to A (s)

A + 6 H_{2} O \to M g (O H)_{2} + B (g)

A

B

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