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 $2\times 0.13=0.26$ moles of $O_2$ to give $2\times 0.13=0.26$ 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,

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

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.

$\frac{23.9L}{V_2}=\frac{0.39mole}{0.26mole}$

$V_2=15.93L$

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