MnO _4^{-}oxidises (i) oxalate ion in acidic | vedclass

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Question

(b)
$2 MnO _4^{-}+5 C _2 O _4^{2-} \stackrel{16 H ^{+}}{\longrightarrow}$
$2 Mn ^{2+}+10 CO _2+8 H _2 O$
As $2$ moles of $MnO _4^{-}$oxidises $5\,

Vedclass · Accepted Answer

$2: 5$ and $1: 8$

Vedclass · Answer

(b)
$2 MnO _4^{-}+5 C _2 O _4^{2-} \stackrel{16 H ^{+}}{\longrightarrow}$
$2 Mn ^{2+}+10 CO _2+8 H _2 O$
As $2$ moles of $MnO _4^{-}$oxidises $5\, moles$ of $C _2 O _4^{2-}$
$	herefore MnO _4^{-}: C _2 O _4^{2-}=2: 5$
$2 MnO _4^{-}+16 HCl \longrightarrow 2 MnCl _2+5 Cl _2+2Cl^-+8H_2O$
As $2\, moles$ of $MnO _4^{-}$oxidises $16\, moles$ of $HCl$
$	herefore MnO _4^{-}: HCl =2: 16=1: 8$

$MnO _4^{-}$oxidises $(i)$ oxalate ion in acidic medium at $333\, K$ and $(ii)$ $HCl$. For balanced chemical equations, the ratios $\left[ MnO _4^{-}: C _2 O _4^{2-}\right]$ in $(i)$ and $\left[ MnO _4^{-}: HCl \right]$ in $(ii)$ respectively are

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