Balance the following reaction of oxidation number and ion electron method:
$K M n O_{4} + H_{2} S O_{4} + K_{2} C_{3} O_{4} \to M n S O_{4} + C O_{2} + K_{2} S O_{4}$ .

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Solution

By gon-electron method
$(K M n O_{4} + s e^{-} \to M n S O_{4}) \times 6$
$(K_{2} C_{3} O_{4} \to 3 C O_{2} + b e^{-}) \times 5$
$6 K M n O_{4} + 3 o e^{-} + 5 K_{2} C_{3} O_{4} + 4 H_{2} S O_{4} \to 6 M n S O_{4} + 15 C O_{2} + 3 o e^{-} + 8 K_{2} S O_{4} + 78_{H}^{+}$
By oxidation no method
$K M N O_{4} + H_{2} S O_{4} + K_{2} C_{3} O_{4} \to M n S O_{4} + 3 C O_{2} + K_{2} S O_{4}$
reduction by 5
oxidation by (6)
So using the coefficient
$6 K M n O_{4} + H_{2} S O_{4} + 5 K_{2} C_{3} O_{4} \to 6 M n S O_{4} + 15 C O_{2} + 8 K_{2} S O_{4} + 78 H + 38 H_{2} O$

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Similar questions

K M n O_{4} + C_{2} O_{4} H_{2} + H_{2} S O_{4} \to K_{2} S O_{4} + M n S O_{4} + C O_{2} + H_{2} O

A l + K M n O_{4} + H_{2} S O_{4} \to A l_{2} (S O_{4})_{3} + K_{2} S O_{4} + M n S O_{4} + H_{2} O

F e C_{2} O_{4} + K M n O_{4} + H_{2} S O_{4} \to F e_{2} (S O_{4})_{3} + C O_{2} + M n S O_{4} + K_{2} S O_{4} + H_{2} O

F e S O_{4} + K M n O_{4} + H_{2} S O_{4} \to F e_{2} (S O_{4})_{3} + M n S O_{4} + H_{2} O + K_{2} S O_{4}

K M n O_{4} + H_{2} S O_{4} + H C l \to K_{2} S O_{4} + M n S O_{4} + C l_{2} + H_{2} O

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