How many grams of sodium azide are needed to provide sufficient nitrogen gas to fill a 50.0 × 50.0 × 25.0 cm bag to a pressure of 1.11 atm at 13.0 °C?

Question

How many grams of sodium azide are needed to provide sufficient nitrogen gas to fill a 50.0 × 50.0 × 25.0 cm bag to a pressure of 1.11 atm at 13.0 °C? 

     g sodium azide   

Answer 1

PV = nRT where:

1. P = pressure in atm
2. P = 1.11 atm
3. V = volume in L
4. 1 cm³ = 1 mL
5. 50 * 50 * 25 = 62500 cm³ = 62500 mL
6. 62500 mL * 1 L / 1000 mL = 62.5 L
7. n = moles of nitrogen
8. R = gas constant 0.08206 L*atm/mol * K
9. T = temperature in K
10. 13 + 273 = 286 K

Putting it all together, we are solving for n moles of nitrogen

n = PV/RT so n = (1.11 * 62.5) / (0.08206 * 286) --> n=2.956 moles of nitrogen.

However, we are looking for mass of sodium azide. This means finding the moles and them multiplying by the molar mass

2NaN₃(s) ---> 2Na(s) + 3N₂(g)

2.956 moles of N₂ * ( 2 moles NaN₃ / 3 moles N₂) = 1.971 moles NaN₃ * 65.0099 g/mol = 128. grams

with only THREE significant figures