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An important step in the preparation of sulfuric acid involves the production of sulfur trioxide ($\ce{SO3}$) from sulfur dioxide ($\ce{SO2}$) and oxygen.

$$\ce{2 SO2 (g) + O2 (g) -> 2 SO3 (g)}$$

If all volumes are measured at the same temperature and pressure, calculate the volume of sulfur trioxide produced when $\pu{25 mL}$ of sulfur dioxide is placed in a container with $\pu{8 mL}$ of oxygen. What will be the total volume of all gases remaining in the container?

I seem to keep getting the wrong answer for the question, and I'm not totally sure why. I assumed I needed to use Avogadro's law, given that temperature and pressure are constant. So I used:

$$\frac{V_1}{n_1} = \frac{V_2}{n_2}$$

I determined $V_1$, $n_1$ the following way:

$$V_1 = \pu{8 mL} + \pu{25 mL} = \pu{33 mL}$$ $$n_1 = \pu{2 mol} + \pu{1 mol} = \pu{3 mol}$$

From the equation I assumed $n_2 = \pu{2 mol}$. This gave me

$$\frac{\pu{33 mL}}{\pu{3 mol}} = \frac{V_2}{\pu{2 mol}} \implies V_2 = \pu{22 mL}$$

The provided answer says $\pu{16 mL}$, and I am just unsure of how to alter the method.

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    $\begingroup$ Here's a hint: if I increased the volume of sulfur dioxide to 1000mL, but kept the amount of oxygen the same (8mL), would you expect there to be much more sulfur trioxide produced? Why or why not? $\endgroup$ – chipbuster Nov 11 '17 at 3:03
  • $\begingroup$ @chipbuster I would assume that not much more would be produced, because it would be limited by the smaller amount of oxygen?? $\endgroup$ – uyujc Nov 11 '17 at 3:09
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    $\begingroup$ @uyujc Absolutely, so why don't you determine the limiting reagent first, and then apply the same law again? $$\frac{V_i}{n_i} = \text{const}$$ $\endgroup$ – andselisk Nov 11 '17 at 3:16
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    $\begingroup$ @andselisk Ah! Thank you!! I got the first bit. Would you be able to help with how to go about finding the total volume of all gases remaining?? I know there would be 16 mL of sulfur trioxide (produced from the reaction), and would it be right to assume all the oxygen has been used up?? $\endgroup$ – uyujc Nov 11 '17 at 3:23
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    $\begingroup$ @uyujc Well, assuming the reaction is complete and irreversible, in the system you have some leftovers of sulfur dioxide $\ce{SO2}$ and the product, sulfur trioxide $\ce{SO3}$ (as you mentioned, all oxygen is used). All you have to do is to find how much $\ce{SO2}$ is left and add $\pu{16 mL}$ of $\ce{SO3}$ to it. Keep in mind you can post an answer to your own question, this is considered a good practice here:) $\endgroup$ – andselisk Nov 11 '17 at 3:28

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