Question

A sample of gas contains 0.1700 mol of OF2(g) and 0.1700 mol of H2O(g) and occupies a volume of 19.5 L. The following reaction takes place: OF2(g) + H2O(g)O2(g) + 2HF(g) Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant. L

Answer 1

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

Step-by-step explanation:

The given reaction equation is as follows.

`OF_(2)(g) + H_(2)O(g) \rightarrow O_(2)(g) + 2HF(g)`

So, moles of product formed are calculated as follows.

`(3)/(2) * 0.17 mol \\= 0.255 mol`

Hence, the given data is as follows.

`n_(1)` = 0.17 mol,
`n_(2)` = 0.255 mol

`V_(1)` = 19.5 L,
`V_(2) = ?`

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

`(V_(1))/(n_(1)) = (V_(2))/(n_(2))`

Substitute the values into above formula as follows.

`(V_(1))/(n_(1)) = (V_(2))/(n_(2))\\(19.5 L)/(0.17 mol) = (V_(2))/(0.255 mol)\\V_(2) = (19.5 L * 0.255 mol)/(0.17 mol)\\= (4.9725)/(0.17) L\\= 29.25 L`

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