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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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