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:

Assign $A, B, C, D$ from the given type of reaction.
$NH_4Cl + NaNO_2 \xrightarrow{\Delta} N_2 \uparrow + NaCl + 2H_2O$

$ClO_2$ oxidizes $H_2O_2$ to $O_2$ in basic medium and is itself reduced to $Cl^-$. How many moles of $H_2O_2$ will be oxidized by one mole of $ClO_2$ (in $.5$)?

Which product is obtained during the reaction of $MnO_4^-$ with $I^-$ in a faintly alkaline condition?

The equivalent weight of $MnSO_4$ is half of its molecular weight when it is converted to

Consider the following redox reaction taking place in acidic medium: $BH_{4}^{-}(aq) + ClO_{3}^{-}(aq) \rightarrow H_{2}BO_{3}^{-}(aq) + Cl^{-}(aq)$. If the Nernst equation for the above balanced reaction is $E_{cell} = E_{cell}^{o} - \frac{RT}{nF} \ln Q$, then the value of $n$ is . . . . . . . (Nearest integer)