Equivalent weight of $Fe{S}_2$  in the half reaction    ${\mathrm{FeS}}_2\to {\mathrm{Fe}}_2{\mathrm{O}}_3+{\mathrm{SO}}_2$  is 

Balanced reaction is

4FeS₂​+11O_2​⟶2Fe₂​O₃​+8SO_2​

1 mole of FeS₂​=11/4 moles of O_2​

Each O− atom takes 2e− for reduction

∴1O2​ molecule takes 4e−

∴ Number of e− gained by O₂​=11/4​×4=11

∴ Number of e− lost by FeS₂​=11

∴ Equivalent weight of FeS₂​ in this reaction= M/11