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For the following reaction, what volume of $\ce{SO3}$ can be produced from $4.2\ \mathrm{L}$ of $\ce{O2}$ (measured at the same temperature and pressure), assuming an excess of $\ce{SO2}$? $$\ce{2SO2(g) + O2(g) -> 2SO3 (g)}$$

I demonstrate what I tried here but I have no idea what to try here, anyone have any suggestions on what to do?

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TL;DR There is twice the amount of $\ce{SO3}$ produced compared to oxygen, thus the volume will be twice as much as oxygen. Therefore $8.4$ litres of $\ce{SO3}$ will be produced.

Since $\ce{SO2}$ is in excess, the amount of $\ce{SO3}$ produced will be twice as the amount of moles of oxygen. So if there is originally x moles of oxygen gas, 2x moles of $\ce{SO2}$ is produced. It is important to note that temperature and pressure remain constant throughout the reaction.

Note one of gas laws state, the same number of moles of any gaseous substance will have the same volume at constant temperature and pressure. So x moles of $\ce{SO3}$ will hold the same volume as x moles of oxygen gas, so 4.2 litres.

Now there is also another gas law which states that the volume of a gas is directly proportional to the number of moles of the substance. So 2x moles of $\ce{SO3}$ will have twice the volume of x moles of $\ce{SO3}$. Therefore the volume of 2x moles of $\ce{SO3}$ will be 8.4 moles.

Therefore the answer is $8.4$ litres

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