Given the redox reaction:X Na_2HAsO_3 + Y NaB | vedclass

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(C) The unbalanced equation is: $Na_2HAsO_3 + NaBrO_3 + HCl \longrightarrow NaBr + H_3AsO_4 + NaCl$$1$. Identify oxidation states:$As$ in $Na_2HAsO_3$

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$3, 1, 6$

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(C) The unbalanced equation is: $Na_2HAsO_3 + NaBrO_3 + HCl \longrightarrow NaBr + H_3AsO_4 + NaCl$$1$. Identify oxidation states:$As$ in $Na_2HAsO_3$ is $+3$,and in $H_3AsO_4$ is $+5$ (oxidation,change of $+2$).$Br$ in $NaBrO_3$ is $+5$,and in $NaBr$ is $-1$ (reduction,change of $-6$).$2$. Balance the change in oxidation states:To balance the electron transfer,multiply the $As$ half-reaction by $3$ and the $Br$ half-reaction by $1$.$3 Na_2HAsO_3 + 1 NaBrO_3 \longrightarrow 3 H_3AsO_4 + 1 NaBr$$3$. Balance the remaining atoms:Total $Na$ on the left is $3 	imes 2 + 1 = 7$. On the right,we have $1$ from $NaBr$ and $6$ from $NaCl$ to make $7$.Total $Cl$ on the right is $6$,so $Z = 6$.The balanced equation is: $3 Na_2HAsO_3 + 1 NaBrO_3 + 6 HCl \longrightarrow 1 NaBr + 3 H_3AsO_4 + 6 NaCl$Thus,$X = 3$,$Y = 1$,and $Z = 6$.

Given the redox reaction:
$X Na_2HAsO_3 + Y NaBrO_3 + Z HCl \longrightarrow NaBr + H_3AsO_4 + NaCl$
The values of $X$,$Y$,and $Z$ in the balanced equation are respectively:

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Given the redox reaction:$X Na_2HAsO_3 + Y NaBrO_3 + Z HCl \longrightarrow NaBr + H_3AsO_4 + NaCl$The values of $X$,$Y$,and $Z$ in the balanced equation are respectively: