3 KClO 3-3 H 2 SO 4- 3 KHSO 4- HClO 4-2 ClO 2- H 2 OEquivalent weight of KClO 3 is MA- 101-4B- 2 n-2C- M - M -2D- M -4- M -2

The correct option is C ( M+M2 )ClO_3^ → ClO_4^ + 2e^ (x=2)oxidation x=5, nbsp; nbsp;x=7 e^ + ClO_3^ → ClO_2(x=1)reduction x=5, nbsp; x=4 E_W=M+M2 ...

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Byju's Answer

Standard XII

Chemistry

Oxidising Reagent and Reducing Reagent

3 KClO 3+3 H ...

Question

$3 K C l O_{3} + 3 H_{2} S O_{4} \to 3 K H S O_{4} + H C l O_{4} + 2 C l O_{2} + H_{2} O$
Equivalent weight of $K C l O_{3}$ is

\frac{M}{4}

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\frac{M}{2}

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(M + \frac{M}{2})

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(\frac{M}{4} + \frac{M}{2})

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Solution

The correct option is C $(M + \frac{M}{2})$
$C l O_{3}^{-} \to C l O_{4}^{-} + 2 e^{-} (x = 2) oxidation$
$x = 5$ , $x = 7$
$e^{-} + C l O_{3}^{-} \to C l O_{2} (x = 1) reduction$
$x = 5$ , $x = 4$
$E_{W} = M + \frac{M}{2}$

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3 K C l O_{3} + 3 H_{2} S O_{4} \to 3 K H S O_{4} + H C l O_{4} + 2 C l O_{2} + H_{2} O

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3KClO3+3H2SO4→3KHSO4+HClO4+2ClO2+H2O Equivalent weight of KClO3 is

The correct option is C (M+M2)ClO−3→ClO−4+2e−(x=2)oxidation x=5, x=7 e−+ClO−3→ClO2(x=1)reduction x=5, x=4 EW=M+M2

$3 K C l O_{3} + 3 H_{2} S O_{4} \to 3 K H S O_{4} + H C l O_{4} + 2 C l O_{2} + H_{2} O$
Equivalent weight of $K C l O_{3}$ is

The correct option is C $(M + \frac{M}{2})$
$C l O_{3}^{-} \to C l O_{4}^{-} + 2 e^{-} (x = 2) oxidation$
$x = 5$ , $x = 7$
$e^{-} + C l O_{3}^{-} \to C l O_{2} (x = 1) reduction$
$x = 5$ , $x = 4$
$E_{W} = M + \frac{M}{2}$