Consider the following reaction:xMnO_4^{-} + | vedclass

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(C) The half-equations of the reaction are:$MnO_4^{-} \rightarrow Mn^{2+}$$C_2O_4^{2-} \rightarrow CO_2$The balanced half-equations are:$(MnO_4^{-} +

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$2, 5$ and $16$

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(C) The half-equations of the reaction are:$MnO_4^{-} ightarrow Mn^{2+}$$C_2O_4^{2-} ightarrow CO_2$The balanced half-equations are:$(MnO_4^{-} + 8H^{+} + 5e^{-} ightarrow Mn^{2+} + 4H_2O) 	imes 2$$(C_2O_4^{2-} ightarrow 2CO_2 + 2e^{-}) 	imes 5$Equating the number of electrons,we get:$2MnO_4^{-} + 16H^{+} + 10e^{-} ightarrow 2Mn^{2+} + 8H_2O$$5C_2O_4^{2-} ightarrow 10CO_2 + 10e^{-}$On adding both the equations,we get:$2MnO_4^{-} + 5C_2O_4^{2-} + 16H^{+} ightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$Thus,the values of $x, y$ and $z$ are $2, 5$ and $16,$ respectively.

Consider the following reaction:
$xMnO_4^{-} + yC_2O_4^{2-} + zH^{+} \rightarrow xMn^{2+} + 2yCO_2 + \frac{z}{2}H_2O$
The values of $x, y$ and $z$ in the reaction are,respectively:

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Consider the following reaction:$xMnO_4^{-} + yC_2O_4^{2-} + zH^{+} \rightarrow xMn^{2+} + 2yCO_2 + \frac{z}{2}H_2O$The values of $x, y$ and $z$ in the reaction are,respectively: