A_2O_x is oxidized to AO_3^- by MnO_4^- in ac | vedclass

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(B) In the reaction,$A$ in $A_2O_x$ is oxidized to $AO_3^-$. Let the oxidation state of $A$ in $A_2O_x$ be $n$. Then $2n + x(-2) = 0$,so $n = x$. In $

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The value of $x = 3$

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(B) In the reaction,$A$ in $A_2O_x$ is oxidized to $AO_3^-$. Let the oxidation state of $A$ in $A_2O_x$ be $n$. Then $2n + x(-2) = 0$,so $n = x$. In $AO_3^-$,the oxidation state of $A$ is $y$. Then $y + 3(-2) = -1$,so $y = +5$. The change in oxidation state per atom of $A$ is $(5 - x)$. Since there are $2$ atoms of $A$ in $A_2O_x$,the total change is $2(5 - x) = 10 - 2x$. For $MnO_4^-$ in acidic medium,$Mn^{+7} + 5e^- ightarrow Mn^{+2}$,so the change in oxidation state is $5$. Equating the equivalents: $n_f 	imes 	ext{moles of } A_2O_x = n_f 	imes 	ext{moles of } KMnO_4$. $(10 - 2x) 	imes (1.5 	imes 10^{-3}) = 5 	imes (0.03 	imes 0.040)$. $(10 - 2x) 	imes 1.5 	imes 10^{-3} = 5 	imes 1.2 	imes 10^{-3}$. $10 - 2x = \frac{6}{1.5} = 4$. $2x = 6$,so $x = 3$. Thus,the value of $x$ is $3$ and the oxide is $A_2O_3$.

$A_2O_x$ is oxidized to $AO_3^-$ by $MnO_4^-$ in acidic medium. If $1.5 \times 10^{-3} \text{ mole}$ of $A_2O_x$ requires $40 \text{ mL}$ of $0.03 \text{ M } KMnO_4$ solution in acidic medium,which of the following statement$(s)$ is/are correct?

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