Question

A sample of gas contains 0.1300 mol of N2(g) and 0.2600 mol of O2(g) and occupies a volume of 23.9 L. The following reaction takes place: N2(g) + 2O2(g)2NO2(g) Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.

Answer 1

Answer : The volume of the sample after the reaction takes place is, 15.93 liters.

Explanation : Given,

Moles of
`N_2` = 0.13 mole

Moles of
`O_2` = 0.26 mole

Initial volume of gas = 23.9 L

First we have to calculate the moles of
`NO_2` gas.

The balanced chemical reaction is :

`N_2(g)+2O_2(g)\rightarrow 2NO_2(g)`

From the balanced reaction, we conclude that

As, 1 mole of
`N_2` react with 2 moles of
`O_2` to give 2 moles of
`NO_2`.

So, 0.13 mole of
`N_2` react with
`2* 0.13=0.26` moles of
`O_2` to give
`2* 0.13=0.26` moles of
`NO_2`.

According to the Avogadro's Law, the volume of the gas is directly proportional to the number of moles of the gas at constant pressure and temperature.

`V\propto n`

or,

`(V_1)/(V_2)=(n_1)/(n_2)`

where,

`V_1` = initial volume of gas = 23.9 L

`V_2` = final volume of gas = ?

`n_1` = initial moles of gas = 0.13 + 0.26 = 0.39 mole

`n_2` = final moles of gas = 0.26 mole

Now put all the given values in the above formula, we get the final temperature of the gas.

`(23.9L)/(V_2)=(0.39mole)/(0.26mole)`

`V_2=15.93L`

Therefore, the volume of the sample after the reaction takes place is, 15.93 liters.