In the following balanced chemical reaction, $a, b, c$ and $d$ , respectively, correspond to :
$I O_{3}^{⊖} + a I^{⊖} + b H^{\oplus} ⟶ c H_{2} O + d I_{2}$

5, 6, 3, 3

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5, 3, 6, 3

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3, 5, 3, 6

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5, 6, 5, 5

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Solution

The correct option is A 5, 6, 3, 3
Steps for balancing:
$12 H^{\oplus} + 10 e^{-} + 2 I O_{3}^{⊖} ⟶ I_{2} + 6 H_{2} O$
$10 I^{⊖} ⟶ 5 I_{2} + 10 e^{-}$
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$10 I^{⊖} + 12 H^{\oplus} + 2 I O_{3}^{⊖} ⟶ 6 H_{2} O + 6 I_{2}$
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or $5 I^{⊖} + 6 H^{\oplus} + I O_{3}^{⊖} ⟶ 3 H_{2} O + 3 I_{2}$

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Similar questions

I O_{3}^{-} + a I^{-} + b H^{+} \to c H_{2} O + d I_{2}

I O_{3}^{-} + a I^{-} + b H^{+} \to c H_{2} O + d I_{2} a, b, c, d repectively correspond to :

6 \frac{2}{3} + 3 \frac{3}{5} + 3 \frac{5}{6} = ?

C_{6} H_{5} C O C H_{3} S e O_{2} - -- \to

A \times (B \cup C)

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