# In the following reaction, 2NaN₃ decomposes to form 2Na plus 3N2. If 500 g of NaN3 decomposes to form 323.20 g of N2, how much Na is produced?

May 22, 2014

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 × $\left(1 \text{ mol NaN₃")/(65.01" g NaN₃}\right)$ = 7.69 mol NaN₃

7.69 mol NaN₃ × $\left(2 \text{ mol Na")/(2 " mol NaN₃}\right)$ = 7.69 mol Na

7.69 mol Na × $\left(22.99 \text{ g Na")/(1" mol Na}\right)$ = 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 × $\left(2 \text{ mol N₂")/(28.01" g N₂}\right)$ = 11.539 mol N₂

11.539 mol N₂ × $\left(2 \text{ mol Na")/(3 " mol N₂}\right)$ = 7.692 mol Na

7.692 mol Na × $\left(22.99 \text{ g Na")/(1" mol Na}\right)$ = 176.85 g Na