Answer:
Number of moles of nitrogen produced from 167.7 g of sodium azide are 3.87.
Step-by-step explanation:
Given data:
Mass of sodium azide = 167.7 g
Moles of nitrogen gas produced = ?
Solution:
First of all we will write the balance chemical equation of decomposition of sodium azide.
Chemical equation:
2NaN₃ → 3N₂ + 2Na
Number of moles of sodium azide:
Number of moles of sodium azide = mass / molar mass
Number of moles of sodium azide = 167.7 g/ 65 g/mol
Number of moles of sodium azide = 2.58 mol
Now we compare the moles of sodium azide with nitrogen from balance chemical equation.
NaN₃ : N₂
2 : 3
2.58 : 3/2×2.58 = 3.87
Number of moles of nitrogen produced from 167.7 g of sodium azide are 3.87.
Mass of nitrogen produced = moles × molar mass
Mass of nitrogen produced = 3.87 mol × 14 g/mol
Mass of nitrogen produced = 54.18 g.