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Question

In the following reaction,
xMn2O7yMnO2+zO2 x, y, and z respectively are:


A

2, 4, 3

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B
2,3,4
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C

4,3,2

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D

None of these

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Solution

The correct option is A

2, 4, 3


Mn2O7 is splitted into two parts ( +7×2Mn2 and 2×7O7 )

Let the following equation be numbered ...(i)

(i). 6e+Mn22Mn
2x=14 2x=8
so change in oxidation state require =148=6e

Let the following equation be numbered ...(ii)

(ii) 2O7 7O2+28e
14x=28 14x=0
so change in oxidation state require 0(28)=28e

Now, multiply equation (i) by 28 and equation (ii) by 6 and add them as

28×6e+28Mn7+256Mn4+
12O2742O2+28×6e
overall reaction: 28Mn7+2+12O2756Mn4++42O2
So,
28Mn2O7+12Mn2O756MnO2+42O2 ...(iii)

In the above equation (iii), balance the Mn atoms disregarding O atoms.

40Mn2O756MnO2+24MnO2+42O2 ...(iv)

In the above equation, now balance the O atoms to get

40Mn2O780MnO2+42O2+18O2

The final, balanced equation is:

2Mn2O74MnO2+3O2


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