Monday, June 3, 2019

redox - Calculating n factor of SO2


I'm having a hard time trying to figure out n factor of SOX2 in the reaction


FeSX2+OX2FeX2OX3+SOX2(unbalanced)


To balance it, I saw that in 1 mole of FeSX2, it looses a net total of 11 moles of electrons, while OX2 gains 4 moles of electrons per mole of it. That means they will have to react in a 4:11 ratio, from this, the equation balances out to


4FeSX2+11OX22FeX2OX3+8SOX2


The question was (wrt to this equation), what will be the equivalent weight of SOX2, if the molecular weight of SOX2 is M?


This means I will need the n factor of it to divide by M,


I tried to reverse this reaction


2FeX2OX3+8SOX24FeSX2+11OX2



But the problem is that now it's hard to tell from which O atoms (of the now-products), the oxidation state is getting changed to ±0 (zero). It's now confusing me.


The answer is given as M/5. I don't know why, according to the solution they only concidered the oxidation number change of S while ignoring the others, I don't think it's right.




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