Answer:
sodium azide required = 142.4 g
Explanation:
Decomposition of sodium azide .
2NaN₃ = 2Na + 3 N₂
2 mole 3 mole of nitrogen
2 moles of sodium azide gives 3 moles of nitrogen
mol weight of sodium azide = 65
mol weight of nitrogen = 28
2 x 65 g of sodium azide gives 3 x 28 g of nitrogen .
130 g of sodium azide gives 84 g of nitrogen .
mass of nitrogen required = 1.25 x 73.6
= 92 g
84 g of nitrogen requires decomposition of 130 g of sodium azide
92 g of nitrogen requires decomposition of 130x 92 / 84 g of sodium azide
sodium azide required = 130x 92 / 84 = 142.4 g
