$K M n O_{4} + C_{2} O_{4} H_{2} + H_{2} S O_{4} \to K_{2} S O_{4} + M n S O_{4} + C O_{2} + H_{2} O$
Balance by oxidation number method and ion electron method?

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Solution

$^{+ 1} K M^{+ 7} n O_{4}^{- 2} +^{+ 3} C_{2} O_{4}^{- 2} H_{2}^{+ 1} +^{1} H_{2} 6 S O_{4}^{- 2} ⇌^{1} K_{2} 6 S O_{4}^{- 2} +^{2} M n S^{8} O_{4}^{- 2} +^{4} C O_{2}^{- 2} +^{1} H_{2} O^{- 2}$

Oxidation: $^{3} C_{2} O_{4}^{- 2} H_{2}^{1} ⟶ 2 6 C O_{2}^{- 2} + 2 e^{-}$
Reduction: $K M n O_{4} ⟶ M n^{+ 2} 6 S O_{4}^{- 2}$

$\Rightarrow$ O: $C_{2} O_{4} H_{2} ⟶ 2 C O_{2} + 2 e^{-} + 2 H^{+}$
R: $2 K M n O_{4} + 3 H_{2} S O_{4} + 10 e^{-} + 10 H^{+} ⟶ 2 M n S O_{4} + K_{2} S O_{4} + 8 H_{2} O$

$\Rightarrow$ O: $5 C_{2} O_{4} H_{2} ⟶ 10 C O_{2} + 10 e^{-} + 10 H^{+}$
R: $2 K M n O_{4} + 3 H_{2} S O_{4} + 10 e^{-} + 10 H^{+} ⟶ 2 M n S O_{4} + K_{2} S O_{4} + 8 H_{2} O$

$∴ 5 C_{2} O_{4} H_{2} + 2 K M n O_{4} + 3 H_{2} S O_{4} ⟶ 2 M n S O_{4} + 10 C O_{2} + K_{2} S O_{4} + 8 H_{2} O$ .

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