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Triss [41]
3 years ago
14

Please HELP!!! Automotive airbags inflate when sodium azide decomposes explosively to its constituent elements. How many grams o

f 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
Chemistry
1 answer:
Firdavs [7]3 years ago
8 0

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

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