Answer: The volume of the sample after the reaction takes place is 29.25 L.

Explanation:

The given reaction equation is as follows.

[tex]OF_{2}(g) + H_{2}O(g) \rightarrow O_{2}(g) + 2HF(g)[/tex]

So, moles of product formed are calculated as follows.

[tex]\frac{3}{2} \times 0.17 mol \\= 0.255 mol[/tex]

Hence, the given data is as follows.

[tex]n_{1}[/tex] = 0.17 mol,      [tex]n_{2}[/tex] = 0.255 mol

[tex]V_{1}[/tex] = 19.5 L,         [tex]V_{2} = ?[/tex]

As the temperature and pressure are constant. Hence, formula used to calculate the volume of sample after the reaction is as follows.

[tex]\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}[/tex]

Substitute the values into above formula as follows.

[tex]\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{19.5 L}{0.17 mol} = \frac{V_{2}}{0.255 mol}\\V_{2} = \frac{19.5 L \times 0.255 mol}{0.17 mol}\\= \frac{4.9725}{0.17} L\\= 29.25 L[/tex]

Thus, we can conclude that the volume of the sample after the reaction takes place is 29.25 L.