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In the following equation: 4FeS(2) + 11O...

In the following equation: `4FeS_(2) + 11O_(2) to 2Fe2O3 + 8SO2`. The volume of air measured at 1 atm and 273 K, used in oxidation of 6g `FeS_(2)` assuming that air contains 20% of oxygen by volume is __________ L. [Atomic mass Fe = 56, S = 32] Use molar volume 22.4L at 1 atm & 273 K.

Text Solution

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(15.4 L)
Molecular weight of `FeS_(2)` is 120 g.
4(120)gm `to` 11 mol `O_(2)`
6 gm `Fes_(2) to `?
Solving `(11)/(80)` mol of `O_(2)`
Volume of `O_(2) ((11)/(80) xx 22.4)) = 3.08`
Volume of air = `((100)/(20) xx 3.08) = 15.4L`
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