Hydrogen disulfide heterogenous equilibrium


> 
> Hydrogen sulfide can be removed from natural gas by the reaction
> 
> 
> $$\ce{2 H2S(g) + SO2(g) -> 3 S(s) + 2 H2O(g)}$$
> 
> 
> **a**) Calculate $ΔG^\circ$ and $K$ for this reaction at $\pu{298 K}.$  
> 
> **b**) What is the mass of sulfur at the equilibrium (at $\pu{298 K})$ when $\pu{1.00 mol}$ sulfur dioxide are sealed in a vacuumed 1 liter vessel?
> 
> 
> $$
> \begin{array}{lrrr}
> \hline
>  & Δ\_\mathrm fH^\circ/\pu{kJ mol-1} & Δ\_\mathrm fG^\circ/\pu{kJ mol-1} & S^\circ/\pu{J mol-1 K-1} \\
> \hline
> \ce{H2S(g)} & -20.6 & -33.4 & 205.8 \\
> \ce{SO2(g)} & -296.8 & -300.1 & 248.2 \\
> \ce{S(s)} & 0 & 0 & 32.1 \\
> \ce{H2O(g)} & -241.8 & -228.6 & 188.8 \\
> \hline
> \end{array}
> $$
> 
> 
> 


I was able to calculate the Gibbs free energy and the equilibrium constant, using the formula


$$ΔG = -RT\ln K$$


and got the following answers, though I am not sure whether they are correct:


$$
\begin{align}
ΔG &≈ \pu{-90.2 kJ mol-1} \\
K &≈ \pu{6.28E15}
\end{align}
$$


Next, I tried to solve for the mass of sulfur but was unable to do so. I tried to solve it in the following way, which led to no solutions:


$$\begin{array} {c c c c c c c}
\ce{&2 H2S(g) &+ &SO2(g) &-> &3 S(s) &+ &2 H2O(g)} \\
c\_\mathrm{alku}/\pu{mol dm-3}&1.00 && 1.00 && 0 && 0 \\
c\_\mathrm{loppu}/\pu{mol dm-3}&1.00 - 2x && 1.00 - x && 3x && 2x
\end{array}$$


$$\pu{6.28E15} = \frac{(2x)^2}{(1.00 - 2x)^2(1.00 - x)}$$


I would like to know whether my answers for step **a** are incorrect and how I should approach step **b**.


