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

Step-by-step explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number
`6.023* 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}}=(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 =
`(2)/(3)* 1.09=0.73moles` of
`NaN_3`

Mass of
`NaN_3=moles* {\text {Molar mass}}=0.73moles* 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