Question

Which of the following statements is/are correct in the following reaction.
$A{s}_2{S}_5+N{O}_3^{⊝}\to As{O}_4^{3-}+N{O}_2+S{O}_4^{2-}$

• A. The equivalent weight of $A{s}_2{S}_5$ is $\frac{M}{40}$ ($M$= molecular weight of $A{s}_2{S}_5$).
• B. The equivalent weight of $N{O}_3^{⊝}$ is $dfracM3$ ($M$= molecular weight of $N{O}_3^{⊝}$ ion).
• C. $n$-factor for the conversion of $A{s}_2{S}_5$ to $As{O}_4^{3-}$ is zero.
• D. $n$-factor for the conversion of $A{s}_2{S}_5$ to $S{O}_4^{2-}$ is $30$.

Answer 1

Answer: (A), (B), (C)

The equivalent weight of $A{s}_2{S}_5$ is $\frac{M}{40}$ ($M$= molecular weight of $A{s}_2{S}_5$).
The equivalent weight of $N{O}_3^{⊝}$ is $dfracM3$ ($M$= molecular weight of $N{O}_3^{⊝}$ ion).
$n$-factor for the conversion of $A{s}_2{S}_5$ to $As{O}_4^{3-}$ is zero.

In $\left(\begin{array}{cc}\stackrel{+5\times 2}{A{s}_2}& \stackrel{-2\times 5}{S_5}\end{array}\right)$,
$\left[\begin{array}{c}\stackrel{+5}{\underset{2x=10}{A{s}_2}}\stackrel{no}{\underset{redox}{\to }}\underset{\underset{2x=10}{2x-16=-6}}{2As{O}_4^{3-}}\end{array}\right]n=0$ ....(i)
$\underset{5x=-10}{S_5^{-2}}\to \underset{\underset{5x=30}{5x-40=-10}}{5S{O}_4^{2-}}+40e^{-}]$ ......(ii)
Add equations (i) and (ii), we get
$A{s}_2{S}_5\to 2As{O}_4^{3-}+5S{O}_4^{2-}+40e^{-}$ $(n=40)$
$Ew=\frac{M}{40}$
Reduction:
$\left[\begin{array}{c}3e^{-}+\underset{\underset{x=5}{x-6=-1}}{N{O}_3^{⊝}}\to \underset{\underset{x=2}{x-4=2}}{N{O}_2}\end{array}\right]n=3$