For the reaction, Fe₂S₃ + 5O₂$\to$2FeSO₄ + SO₂, the equivalent weight of Fe₂S₃ is

${\mathrm{Fe}}_2{\stackrel{-2}{\mathrm{S}}}_3+5{\mathrm{O}}_2\to 2\mathrm{Fe}\stackrel{+6}{\mathrm{S}}{\mathrm{O}}_4+\stackrel{+4}{\mathrm{S}}{\mathrm{O}}_2$

Total number of electrons gained (n-factor) is calculated as:

$$\begin{gathered} 2\times 2+2\times 6+1\times 4=20 \\ \mathrm{Equivalent} \mathrm{Weight}=\frac{\mathrm{Molecular} \mathrm{weight}}{\mathrm{n}-\mathrm{factor}} \\ \mathrm{Equivalent} \mathrm{Weight}=\frac{\mathrm{M}}{20} \end{gathered}$$