Question

$O_2+Fe{S}_2\to FeO+S{O}_2$
The number of moles of $Fe{S}_2$ required to produce $20\text{mol}$ electrons to reduce $O_2$ in the above reaction is:

• A. $2$
• B. $10$
• C. $\frac{10}{3}$
• D. $\frac{5}{3}$

Answer 1

The correct option is A $2$

$O_2+Fe{S}_2\to FeO+S{O}_2$
Oxidation half reaction:
$S_2^{2-}\to 2S^{4+}+10e^{-}$
Reduction half reaction:
$4e^{-}+O_2\to 2O^{2-}$
So, on doing $2\times \left(i\right)+5\times \left(ii\right)$
$\therefore 2S_2^{2-}+6O_2\to 4S^{4+}+10O^{2-}2Fe{S}_2+5O_2\to 4S{O}_2+2FeO$
Since, one mole of $Fe{S}_2$ produces $10$ electrons.
So, $2$ moles of $Fe{S}_2$ are required to produce $20\text{mol}$ electrons.