Question

Please HELP!!! Automotive airbags inflate when sodium azide decomposes explosively to its constituent elements. How many grams of sodium azide are required to produce 24.4 L of nitrogen gas at standard temperature and pressure? 2NaN3 --> 2Na + 3N2
47.2 g of sodium azide


106.2 g of sodium azide


1.63 g of sodium azide


0.726 g of sodium azide

Answer 1

Answer: 47.2 g of sodium azide

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles of nitrogen}=\frac{\text{Given volume}}{\text {Molar volume}}=\frac{24.4L}{22.4L}=1.09moles$

The balanced chemical reaction is:

$2NaN_3\rightarrow 2Na+3N_2$  

According to stoichiometry :  

3 moles of $N_2$ are produced by = 2 moles of $NaN_3$

Thus 1.09 moles of $N_2$ are produced by =$\frac{2}{3}\times 1.09=0.73moles$  of $NaN_3$

Mass of $NaN_3=moles\times {\text {Molar mass}}=0.73moles\times 65g/mol=47.2g$

Thus 47.2 g of sodium azide are required to produce 24.4 L of nitrogen gas at standard temperature and pressure