After balancing the equation,C_2O_4^{2-} + H^ | vedclass

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(C) The redox reaction is: $a C_2O_4^{2-} + b MnO_4^- + H^+ \longrightarrow p CO_2 + q Mn^{2+} + H_2O$Step $1$: Determine the $n$-factors.For $C_2O_4^

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$10$

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(C) The redox reaction is: $a C_2O_4^{2-} + b MnO_4^- + H^+ \longrightarrow p CO_2 + q Mn^{2+} + H_2O$Step $1$: Determine the $n$-factors.For $C_2O_4^{2-}$,the oxidation state of $C$ changes from $+3$ to $+4$. The $n$-factor is $(4-3) 	imes 2 = 2$.For $MnO_4^-$,the oxidation state of $Mn$ changes from $+7$ to $+2$. The $n$-factor is $(7-2) 	imes 1 = 5$.Step $2$: Balance the $n$-factors by cross-multiplying.We take $5$ moles of oxalate $(a=5)$ and $2$ moles of permanganate $(b=2)$.Step $3$: Balance the atoms.For $C$-atoms: $2a = p \implies p = 2 	imes 5 = 10$.For $Mn$-atoms: $b = q \implies q = 2$.For $O$-atoms: $5(4) + 2(4) = 10(2) + 2(1) + H_2O \implies 28 = 22 + H_2O \implies H_2O = 6$.For $H$-atoms: $H^+ = 2 	imes 6 = 12$.The balanced equation is $5C_2O_4^{2-} + 2MnO_4^- + 16H^+ \longrightarrow 10CO_2 + 2Mn^{2+} + 8H_2O$.Thus,the coefficient of $CO_2$ is $10$.

After balancing the equation,$C_2O_4^{2-} + H^+ + MnO_4^- \longrightarrow CO_2 + Mn^{2+} + H_2O$,the coefficient of $CO_2$ is:

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