In the following equation:
$F e S_{2} + O_{2} \to F e_{2} O_{3} + S O_{2}$
Calculate the volume of air (in L) measured at STP used in oxidation assuming that air contains
$20 %$ of oxygen by mole.
(Given moles of $F e S_{2}$ is 10)

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Solution

$F e S_{2} + O_{2} \to F e_{2} O_{3} + S O_{2}$
Here initial mole of $F e S_{2}$ is 10.
Applying POAC on $O$ atom,
$2 (n_{O_{2}}) = 3 (n_{F e_{2} O_{3}}) + 2 (n_{S O_{2}}) . . . (i)$
Applying POAC on $F e$ atom,
$1 (n_{F e S_{2}}) = 2 (n_{F e_{2} O_{3}}) . . . (i i)$
Using (i) and (ii)
$2 (n_{O_{2}}) = 3 (\frac{n_{F e S_{2}}}{2}) + 2 (n_{S O_{2}})$
$2 (n_{O_{2}}) = 3 (\frac{10}{2}) + 2 (20)$
$2 (n_{O_{2}}) = \frac{55}{2}$

Now, $moles of air = \frac{100}{20} \times moles of O_{2}$
$Volume = \frac{100}{20} \times \frac{55}{2} \times 22.4 = 3080 L$

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