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Consider the following reaction:

$xMnO_{4}^{-}+yC_{2}O_{4}^{2-}+zH^{+}\rightarrow xMn^{2+}+2yCO_{2}+\frac{z}{2}H_{2}O$ The values of $x$ , $y$ and $z$ in the reaction are, respectively:
Option: 1
5, 2 and 8

Option: 2
5, 2 and 16

Option: 3
2, 5 and 8

Option: 4
2, 5 and 16

Answers (1)

best_answer

The given reaction is:

$xMnO_{4}^{-}+yC_{2}O_{4}^{2-}+zH^{+}\rightarrow xMn^{2+}+2yCO_{2}+\frac{z}{2}H_{2}O$

Balancing the charges on both sides, we got

$-x-2y+z=2x$

$\Rightarrow -2y+z=3x\:\:\cdot \cdot \cdot (1)$

Balancing the O atoms on both sides, we got

$4x+4y=4y+\frac{z}{2}$

$\Rightarrow z=8x$

$\Rightarrow x=\frac{z}{8}\:\cdot \cdot \cdot (2)$

Substituting the value of z in (1)

$\Rightarrow -2y+8x=3x$

$\Rightarrow -2y=-5x$

$\Rightarrow y=\frac{5x}{2}\:\cdot \cdot \cdot (3)$

Now, substitute the value of y and z in terms of x in the original reaction:

$xMnO^{-}_{4}+\frac{5}{2}xC_{2}O_{4}^{2-}+8xH^{+}\rightarrow xMn^{2+}+5xCO_{2}+4xH_{2}O$

Solving further we get,

$2MnO_{4}^{-}+5C_{2}O_{4}^{2-}+16H^{+}\rightarrow 2Mn^{2+}+10CO_{2}+8H_{2}O$

$\therefore$ by correlating, we get

x=2, y=5, z=16

Therefore,option(4) is correct

Posted by

Ajit Kumar Dubey

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