How many moles of NaN3 must be extracted from 65 grams of N2?

Question

2 moles of Sodium Azide (NaN3) produces about 3 moles of Nitrogen (N2) in a car's airbag.

How many moles of NaN3 are needed to fill a 65 gram airbag? Assume the gas is produced at a temperature of 30 °C and 2.5 atm. (How many moles of NaN3 must be extracted from 65 grams of N2?)

Please include a detailed description of each step if you can please! :)

Answer 1

2NaN₃ ==> 2Na + 3N₂ ... balanced equation

I think you wrote the question incorrectly, and I will assume that you meant the air bag is 65 liters, not 65 grams. The problem then becomes how many moles of NaN3 are needed to produce enough N2 gas to occupy 65 liters. If this is NOT what you meant, then please respond in the comment section.

We will treat N2 gas as an ideal gas, and use the ideal gas law to calculate how many moles of N2 is equivalent to 6 L of N2. Once we have moles of N2, we can go back to the balance equation (above), use the stoichiometry (3 N₂ : 2 NaN₃) to find moles of NaN₃.

Ideal gas law: PV = nRT

P = pressure in atm = 2.5 atm

V = volume in L = 65 L

n = moles of N2 = ?

R = gas constant = 0.0821 Latm/Kmol

T = temperature in Kelvin = 30ºC + 273 = 303K

Solving for n (moles) we have ...

n = PV/RT = (2.5)(65) / (0.0821)(303)

n = 6.53 moles N₂ needed

Now we use the mole ratio to find moles of NaN₃ that would produce 6.53 mols N₂:

6.53 mols N₂ x 2 mols NaN₃ / 3 mols N₂ = 4.35 moles NaN₃