I'm having a hard time trying to figure out n factor of SOX2 in the reaction
FeSX2+OX2⟶FeX2OX3+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+11OX2⟶2FeX2OX3+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+8SOX2⟶4FeSX2+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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