5H_2O_2 + XClO_2 + 2OH^- to XCl^- + YO_2 + 6H | vedclass

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(B) To balance the reaction $5H_2O_2 + XClO_2 + 2OH^- 	o XCl^- + YO_2 + 6H_2O$:$1$. Balance Chlorine $(Cl)$: There are $X$ atoms of $Cl$ on both side

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$X = 2$ and $Y = 5$

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(B) To balance the reaction $5H_2O_2 + XClO_2 + 2OH^- 	o XCl^- + YO_2 + 6H_2O$:$1$. Balance Chlorine $(Cl)$: There are $X$ atoms of $Cl$ on both sides.$2$. Balance Oxygen $(O)$: On the left side,total $O = 5(2) + 2(X) + 2 = 12 + 2X$. On the right side,total $O = 2Y + 6(1) = 2Y + 6$.$3$. Balance Hydrogen $(H)$: On the left side,total $H = 5(2) + 2 = 12$. On the right side,total $H = 6(2) = 12$. (Already balanced).$4$. Balance Charge: Left side charge = $-2$. Right side charge = $-X$. Thus,$X = 2$.$5$. Substitute $X = 2$ into the Oxygen balance equation: $12 + 2(2) = 2Y + 6 \implies 16 = 2Y + 6 \implies 2Y = 10 \implies Y = 5$.Therefore,$X = 2$ and $Y = 5$.

$5H_2O_2 + XClO_2 + 2OH^- \to XCl^- + YO_2 + 6H_2O$
In the above balanced reaction,the reaction is balanced when,

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$5H_2O_2 + XClO_2 + 2OH^- \to XCl^- + YO_2 + 6H_2O$In the above balanced reaction,the reaction is balanced when,