Stoichiometry of Gaseous Reactions
I never really learned the method of solving this stuff with Avogadro's Law, but here's an almost identical way.

Your balanced equation is as follows...

$\ce{2H2S + 3O2 -&gt; 2SO2 + 2H2O}$, and I think you agree on that by looking at your calculations.

At a given temperature and pressure, $1 mol = x L$. For example, at STP (standard temperature and pressure) this is $1 mol = 22.4 L$. It wasn't exactly defined what temperature and pressure are, but they are constant, so we can use a variable $x$ to denote some conversion (which cancels out anyway, I assume giving rise to Avogadro's Law).

$12.0L_{\ce{SO2}} * \frac{1mol_{\ce{SO2}}}{xL_{\ce{SO2}}} * \frac{2mol_{\ce{H2S}}}{2mol_{\ce{SO2}}} * \frac{xL_{\ce{H2S}}}{1mol_{\ce{H2S}}} = 12.0L_{\ce{H2S}}$

But this is assuming $100\%$ yield. I see you divided by $0.885$ above, which is correct, but I've gotten a metric ton of these wrong before, so I prefer the proportions way.

Regardless, $\frac{12L_{\ce{H2S}}}{0.885} = 13.6L_{\ce{H2S}}$

I think you definitely overcomplicated things by assuming a density of $1g/mL$ among other things. Your dimensional analysis also seems strange near the end of the left hand side where you somehow cancel $g$ with $mol_{\ce{SO2}}$.

If I were to do it with Avogadro's Law (taking a guess here), I'd still get $12.0L$ because the mole ratio is the same. And then I'd divide like you did.