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As apparent, the molar ratio of sulfur to oxygen in the reactants is $1:2$, which would make sulfur the limiting reagent and ultimately 1 part of $\ce{SO2}$ would form, amounting to $20~\pu{mL}$ only. That tells us that we need to consider sulfur in its $\beta$-octasulfur state. It might be wrongly assumed that octasulfur ($\ce{S8}$) exists only in the solid state, but the Wikipedia page for sulfur tells us otherwise:

The structure of the $\ce{S8}$ ring is virtually unchanged by this [solid to gas (sublimation)] phase change, which affects the intermolecular interactions.

But the question says "to form $40~\pu{mL}$ of sulfur dioxide gas", which clearly wouldn't be the case if we used the first equation, apparently producing only $20~\pu{mL}$ of $\ce{SO2}$ gas.

So it would be fitting to use this equation instead:

$$\ce{\underset{\pu{20 mL}}{\small\frac{1}{8}S8(g)} + \underset{\pu{40 mL}}{O2(g)} -> \underset{\pu{40 mL}}{SO2(g)}}$$

Now for the question:

Determine the number of atoms in the molecule of sulfur in the vapour state at this temperature.

The molar ratio, now, is $1:8$. That would mean the reaction consumes $40\times(1:8)~\pu{mL}$ of $\ce{S8}$. That gives us $40:320$ or $5:40$. So $5~\pu{mL}$ of $\ce{S8}$ is consumed.

So the reaction upon completion must leave behind $20-5=15~\pu{mL}$ of $\ce{S8}$.

Now, it is quite clear from the question that we are not at standard state (aka STP). We'll have to use the universal gas law to find out the amount of substance in moles of $\ce{S8}$ left behind.

As usual (according to Apoorv Potnis, in this case),


But we realize that we are missing the $P$ term. In good humor I assume it to be $1~ \pu{atm}$.

Temperature is $1000+273.15=1273.15~\pu{K}$

$$1~\pu{atm}\times15\cdot10^{-3}~\pu{L}= n\times0.082057~\pu{L~atm~mol-1K-1}\times1273.15~\pu{K}$$

This gives us $n=0.0001435$.

Number of molecules of $\ce{S8}$ would be $\text{N}_{\small{A}}\cdot~n=6.02214\times~10^{23}\cdot0.0001435$

Atoms is no trouble– just multiply by eight, only to get $86417709000000000000$ or ~$8.64\times10^{19}$.