Question

In the following reaction (unbalanced), equivalent weight of $A{s}_2{S}_3$ is related to molecular weight $M$ by :
${As}_2{S}_3+H^{+}+{NO}_3^{-}⟶NO+H_{2}O+As{O}_4^{3-}+S{O}_4^{2-}$

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

Answer 1

Answer: (C)

$\frac{M}{28}$

Reaction:
$\underset{2As=+3\ast 2=+63S=-2\ast 3=-6}{{As}_2{S}_3}⟶\underset{2\times 5=10}{As{O}_4^{3-}}+\underset{3\times 6=18}{3S{O}_4^{2-}}$
Net change in oxidation of $A{s}_2{S}_3=(10-6)+(18-(-6\left)\right)=28$
Thus, equivalent mass of $A_2{S}_3=\left(Molarmass\right)/\left(changeinoxidation\right)=\frac{M}{28}$