Balance the following reactions by oxidation number method :
$F e S_{2} + O_{2} \to F e_{2} O_{3} + S O_{2}$

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Solution

$F e^{+ 2} \to F e^{+ 3} 1 ↑$
$S_{2}^{- 1} \to 2 S^{+ 4} 10 ↑$
both have total $11 ↑ * 4$

$F e^{+ 2} \to F e^{+ 3}; o n e l o s s o f e l e c t r o n$
$S_{2}^{- 1} \to 2 S^{+ 4}; 2 * 5 = 10 e l e c t r o n l o s s e d$
both have total 11 electron.
And $O_{2} ⟶ 2 O^{2 -} (2 \times 2 = 4$ $electron gained)$
$4 [F e S_{2} + 11 O_{2} ⟶ 2 F e^{3 +} + 8 S^{+ 4} + 22 O^{- 2}]$
or $3 F e S_{2} + 11 O_{2} ⟶ 2 F e_{2} O_{3} + 8 S O_{2}$

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