Equivalent weight of FeS₂ in the half reaction, FeS₂ → Fe₂O₃ + SO₂ is :

${\mathrm{FeS}}_2 +{\mathrm{O}}_2 \to {\mathrm{Fe}}_2{\mathrm{O}}_3 +{\mathrm{SO}}_2$

Balanced reaction is

$4{\mathrm{FeS}}_2 +11{\mathrm{O}}_2\to 2{\mathrm{Fe}}_2{\mathrm{O}}_3 +8{\mathrm{SO}}_2$

1 mole of ${\mathrm{FeS}}_2$​=11/4 moles of  O₂

Each O− atom takes 2e⁻ for reduction

∴1 O_2​ molecule takes 4e−

 NO. of electrons gained by O_2​ =  $4\times \frac{11}{4}$ =11

∴ Number of e⁻ lost by FeS_2​=11

∴ Equivalent wt  of FeS_2​ in this reaction= $\frac{\mathrm{M}}{11}$