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The reaction produces 176.85 g of Na.
The balanced equation is
2NaN₃ → 2Na + 3N₂
You can use either the mass of NaN₃ or of N₂ to calculate the mass of Na.
Using Mass of NaN₃
You must make the following conversions:
mass of NaN₃ → moles of NaN₃ → moles of Na → mass of Na
500 g NaN₃ × ##(1″ mol NaN₃”)/(65.01″ g NaN₃”)## = 7.69 mol NaN₃
7.69 mol NaN₃ × ##(2 ” mol Na”)/(2 ” mol NaN₃”)## = 7.69 mol Na
7.69 mol Na × ##(22.99 ” g Na”)/(1″ mol Na”)## = 176.8 g Na
Using Mass of N₂
You must make the following conversions:
mass of N₂ → moles of N₂ → moles of Na → mass of Na
323.20 g Na × ##(2″ mol N₂”)/(28.01″ g N₂”)## = 11.539 mol N₂
11.539 mol N₂ × ##(2 ” mol Na”)/(3 ” mol N₂”)## = 7.692 mol Na
7.692 mol Na × ##(22.99 ” g Na”)/(1″ mol Na”)## = 176.85 g Na
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