2 NaN3(s) ---> 2 Na(s) + 3 N2(g) Nitrogen gas has a density of 0.629 g/mL at 30°C. How many grams of NaN3 are required to make 4.0 gal of Nitrogen gas.

Question

2 NaN3(s) ---> 2 Na(s) + 3 N2(g)

Nitrogen gas has a density of 0.629 g/mL at 30°C. How many grams of NaN3 are required to make 4.0 gal of Nitrogen gas.

 

Answer 1

2NaN₃(s) ---> 2Na(s) + 3N₂(g) ... balanced equation

Let us first convert 4.0 gal N₂ gas to liters.

1 gal = 3.785 liters

4 gal = 4 gal x 3.785 L / gal = 15.14 L

So, the question boils down to how many grams of NaN₃ are required to make 15.14 liters of N₂ gas.

From the balanced equation we see that it takes 2 moles NaN₃ to make 3 moles N₂ gas.

So we will now convert 15.14 L of N₂ gas (what we want to make) to moles.

15.14 L N₂ x 1000 ml / L x 0.629 g / ml = 9523 g N₂ x 1 mol N₂ / 28 g = 340 moles N₂

Next we'll find the moles of NaN₃ needed:

340 mols N₂ x 2 mols NaN₃ / 3 mols N₂ = 227 moles NaN₃ needed

Finally, convert this to grams of NaN₃:

227 mols NaN₃ x 65.0 g NaN₃ / mol = 14,755 g NaN₃ needed = 15,000 g NaN₃ (to 2 sig. figs.)