contestada

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.

Respuesta :

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

Explanation : Given,

Moles of [tex]N_2[/tex] = 0.13 mole

Moles of [tex]O_2[/tex] = 0.26 mole

Initial volume of gas = 23.9 L

First we have to calculate the moles of [tex]NO_2[/tex] gas.

The balanced chemical reaction is :

[tex]N_2(g)+2O_2(g)\rightarrow 2NO_2(g)[/tex]

From the balanced reaction, we conclude that

As, 1 mole of [tex]N_2[/tex] react with 2 moles of [tex]O_2[/tex] to give 2 moles of [tex]NO_2[/tex].

So, 0.13 mole of [tex]N_2[/tex] react with [tex]2\times 0.13=0.26[/tex] moles of [tex]O_2[/tex] to give [tex]2\times 0.13=0.26[/tex] moles of [tex]NO_2[/tex].

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.

[tex]V\propto n[/tex]

or,

[tex]\frac{V_1}{V_2}=\frac{n_1}{n_2}[/tex]

where,

[tex]V_1[/tex] = initial volume of gas = 23.9 L

[tex]V_2[/tex] = final volume of gas = ?

[tex]n_1[/tex] = initial moles of gas = 0.13 + 0.26 = 0.39 mole

[tex]n_2[/tex] = 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.

[tex]\frac{23.9L}{V_2}=\frac{0.39mole}{0.26mole}[/tex]

[tex]V_2=15.93L[/tex]

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