Question

$28{NO}_3^{-}+3A{s}_2{S}_3+4H_{2}O\to 6As{O}_3^{3-}+28NO+9{SO}_4^{2-}+8H^{+}$
What will be the equivalent mass of ${As}_2{S}_3$ in above reaction$?($Molecular mass of ${As}_2{S}_3=M)$

• A. $\frac{M}{2}$
• B. $\frac{M}{4}$
• C. $\frac{M}{24}$
• D. $\frac{M}{28}$

Answer 1

Answer: (D)

$\frac{M}{28}$

$28{NO}_3^{-}+3{As}_2{S}_3+4H_{2}O⟶6{AsO}_4^{3-}+28NO+{9SO}_4^{2-}+8H^{+}$

Oxidation state of $As$ before & after the reaction is $+3$ & $+5$ respectively.

Oxidation state of $S$ before & after the reaction is $-2$ & $+6$ respectively.

$\therefore n={(2\times 2)}_{forAs}+{(8\times 3)}_{forS}=28$

$\therefore$ Equivalenr mass$=\frac{Molecularmass}{n}$

$=\frac{M}{28}$