Answer :
Final answer:
Around 125.53 grams of sodium azide are required to fill a 70.0 L airbag at 295 K and 101.2 kPa, using the ideal gas law and the stoichiometry from the decomposition of sodium azide to produce nitrogen gas.
Explanation:
To determine what mass of sodium azide (NaN3) would be required to completely fill an airbag with a volume of 70.0 liters at a temperature of 295 K and atmospheric pressure of 101.2 kPa, we first use the ideal gas law (PV = nRT) to calculate the number of moles of nitrogen gas (N2) needed.
Then we use the stoichiometry of the decomposition reaction of sodium azide, which is 2 NaN3(s)
arr 2 Na(s) + 3 N2(g), to find the mass of NaN3 needed.
Let's calculate the number of moles (n) of N2 using the ideal gas law where P = pressure (101.2 kPa), V = volume (70.0 L), R = the ideal gas constant (8.314 J/mol·K or 0.0821 L·atm/mol·K), and T = temperature (295 K):
PV = nRT arr n = PV / RT
Converting pressure to atmospheres (1 atm = 101.325 kPa), we have:
P = 101.2 kPa / 101.325 kPa/atm
= approximately 1 atm
Now:
n = (1 atm arr (70.0 L)) / (0.0821 L·atm/mol·K arr 295 K)
= 2.896 moles of N2
From the balanced chemical equation, it takes 2 moles of NaN3 to produce 3 moles of N2. To find the moles of NaN3 needed:
(2.896 moles of N2) arr ((2 moles of NaN3) / (3 moles of N2)) = 1.931 moles of NaN3
The molar mass of NaN3 is approximately 65.01 g/mol, so:
Mass of NaN3 = 1.931 moles arr 65.01 g/mol
= 125.53 g
Therefore, 125.53 grams of sodium azide would be required to inflate the airbag completely under the given conditions.
