In the reaction
` x FeS_2 + yO_2 rarr Fe_2 O_3 + SO_2`

Options

A. ` x=4, y=11`
B. ` x=3, y=10`
C. ` x=2, y=5`
D. ` x=4, y=13`

Answer

A

Explanation

This is an example of a reaction which occurs in the absence of acids and bases.
`overset(+2-1)(FeS_(2))+overset(0)(O_(2))rarroverset(+3)(Fe_(2))overset(-2)(O_(3))+overset(+4)(S)overset(-2)(O_(2))`
The oxidation number (O.N.) of Fe increases from ` +2` to ` +3` and that so S increases from ` -1` to ` O.N`. of O decreases from 0 to ` -2` . Thus, both Fe and S are oxidized while O is reduced. The change in the O.N. of Fe and S must be considered together as they must maintain their atomic rato of `1:2`.
The O.N. of `Fe` increases (+2 to +3) by one unti while the O.N. of two S atoms increases (-2 to +8) by ten units, Thus , total increase of O.N. is eleven units. The O.N. of two O atoms decreases (0 to -4) by four units. To balance the total increase and decrease in oxidtion numbers, multiply `FeS_2` by ` 4` and ` O_2 ` by `11` to give
` 4FeS_2 + 11O_2 rarr Fe_2O_3 +SO_2`
This implies that we must have ` 4 Fe, 8 S`, and ` 22O` on the ` RHS`. Since O dies not occur independently, the ` 22O` atoms must be factorized in such a way that they become parts of ` Fe_2O_3` and ` SO_2`:
`ubrace(4Fe + 6O)_(2Fe_2 O_3)" "ubrace(8S + 16O)_(8SO_2)`
Thus , the balanced equation is
` 4 FeS_2 + 11O_2 rarr 2Fe_2O_3 + 8SO_2`
Note that balancing of O atoms cannot be done by the addition fo ` H_2O` molecules as this is carried in the absence of acids and bases .