The amount of sodium azide required will be 123.72 grams
From the equation of the reaction below:
[tex]2NaN_3 --- > 2Na + 3N_2[/tex]
The mole ratio of sodium azide to nitrogen gas is 2:3
Mole of 80.0 g nitrogen = 80/28.02 = 2.855 moles
Equivalent mole of sodium azide = 2/3 x 2.855 = 1.903 moles
Mass of 1.903 moles sodium azide = 1.903 x 65 = 123.72 grams
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