Question

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

Answer 1

Reaction:
$\underset{2As=+3\ast 2=+63S=-2\ast 3}{{As}_2{S}_3}⟶\underset{2\times 5=10}{As{O}_4^{3-}}+\underset{3\times 6=18}{3S{O}_4^{3-}}$
net change in oxidation of $A{s}_2{S}_3=18+4+6=28$
Thus, equivalent mass of $A_2{S}_3=\frac{\text{Molar mass}}{\text{change in oxidation}}=\frac{M}{28}$