$X - - M n_{2}^{7 +} + Y - - O_{7}^{2 -} \to Z - - M n^{4 +} + A - - O_{2}$
X,Y,Z and A are stoichiometric coefficients.
Here, X,Y,Z and A are respectively

20, 12, 56, 42

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12, 28, 42, 56

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44, 28, 12, 56

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None of these

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Solution

The correct option is A 20, 12, 56, 42
Let’s calculate how many electrons to be added to either side.
$+ 7 X 2 + 4 6 e^{-} + M n_{2} \to 2 M n$

i. $+ 7 X 20 20_{7} \to 70_{2} + 28 e^{-}$

ii. Multiply Eq. (i) by 28 and Eq. (ii) by and add them.
$28 / x 6 e^{-} + 28 M n_{2}^{7 +} \to 56 M n^{4 +} 120_{7}^{2 -} \to 420_{2} + 28 / x 6 e^{-} - ---------------------------- - 28 M n_{2}^{7 +} + 120_{7}^{2 -} \to 56 M n^{4 +} + 420_{2}$

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Similar questions

X - - M n_{2}^{7 +} + Y - - O_{7}^{2 -} \to Z - - M n^{4 +} + A - - O_{2}

X - - K M n O_{4} + Y - - K_{2} C_{2} O_{4} + Z - - H_{2} S O_{4} \to A - - M n 0_{4} + B - - C 0_{2} + 6 K_{2} S O_{4} + S - - H_{2} O

$x K M n O_{4} + y K_{2} C_{2} O_{4} + z H_{2} S O_{4} ⟶ a M n O_{4} + b C O_{2} + 6 K_{2} S O_{4} + 8 H_{2} O$

Here x,y,z, a and b are respectively

x N a_{2} H A s O_{3} + y N a B r O_{3} + z H C l \to N a B r + H_{3} A s O_{4} + N a C l

x, y, z

N a_{2} H A s O_{3}, N a B r O_{3}, H C l

x + y + z

X - - K M n O_{4} + Y - - K_{2} C_{2} O_{4} + Z - - H_{2} S O_{4} \to A - - M n 0_{4} + B - - C 0_{2} + 6 K_{2} S O_{4} + S - - H_{2} O

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