Calculating n factor of SO2 I'm having a hard time trying to figure out n factor of $\ce{SO2}$ in the reaction

$$\ce{FeS2 + O2 → Fe2O3 + SO2}\qquad \text{(unbalanced)}$$

To balance it, I saw that in 1 mole of $\ce{FeS2}$, it looses a net total of 11 moles of electrons, while $\ce{O2}$ 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

$$\ce{4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2}$$

The question was (wrt to this equation), what will be the equivalent weight of $\ce{SO2}$, if the molecular weight of $\ce{SO2}$ is $M$?

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

I tried to reverse this reaction

$$\ce{2 Fe2O3 + 8 SO2 → 4 FeS2 + 11 O2}$$

But the problem is that now it's hard to tell from which $\ce{O}$ atoms (of the now-products), the oxidation state is getting changed to $\pm0$ (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 $\ce S$ while ignoring the others, I don't think it's right.
As this is clearly a homework question, I will only point you in the right direction.

Balance the oxidation reaction:

$$\ce{S2^2- -&gt; 2 SO2 + $n$e^-}$$

Balancing the charges will give you $n$. (I'll leave the balancing to the OP, as this is homework).

The equivalent mass of $\ce{SO2}$ in this reaction is then:

$$N=\frac{2M}{n}$$

The factor $2$ is needed because the electronically balanced reaction has two molecules of sulphur dioxide in the equation.