Question

Automobile air bags use the decomposition of sodium azide as their sources of gas for rapid inflation, represented in the reaction below. What mass (in grams) of NaN3 is required to provide 40.0 L of N2 at 25°C and 763 torr?

Answer 1

Answer : The mass of $NaN_3$ required is 71.175 grams.

Explanation :

To calculate the moles of nitrogen gas, we use the equation given by ideal gas :

PV = nRT

where,

P = Pressure of nitrogen gas = 763 torr

V = Volume of the nitrogen gas = 40.0 L

n = number of moles of gas = ?

R = Gas constant = $62.364\text{ L.torr }mol^{-1}K^{-1}$

T = Temperature of helium gas = $25^oC=273+25=298K$

Putting values in above equation, we get:

$763torr\times 40.0L=n\times 62.364\text{ L.torr }mol^{-1}K^{-1}\times 298K\\\\n=1.642mol$

Now we have to calculate the moles of $NaN_3$.

The balanced chemical reaction is:

$2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)$

From the balanced chemical reaction, we conclude that

As, 3 moles of $N_2$ produced from 2 moles $NaN_3$

So, 1.642 moles of $N_2$ produced from $\frac{2}{3}\times 1.642=1.095$ moles $NaN_3$

Now we have to calculate the mass of $NaN_3$.

Molar mass of $NaN_3$ = 65 g/mol

$\text{Mass of }NaN_3=\text{Moles of }NaN_3\times \text{Molar mass of }NaN_3$

$\text{Mass of }NaN_3=1.095mole\times 65g/mole=71.175g$

Therefore, the mass of $NaN_3$ required is 71.175 grams.