Final answer:
To inflate an airbag with a volume of 26 L at 1.15 atm and 260C, approximately 27 grams of sodium azide is required. This is calculated using the ideal gas law and stoichiometry, considering the molar mass of NaN₃ and the balanced decomposition reaction.
Step-by-step explanation:
To calculate the amount of sodium azide (NaN3) required to inflate an airbag with nitrogen gas, we need to apply the ideal gas law and stoichiometry. The ideal gas law is PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is temperature in Kelvin. First, we convert the airbag volume to moles of nitrogen gas required using the ideal gas law.
Given:
- Volume (V) = 26 L
- Pressure (P) = 1.15 atm
- Temperature (T) = 260 C = 533 K (temperature must be converted to Kelvin by adding 273 to the Celsius temperature)
Using the ideal gas law (PV = nRT) and the fact that R = 0.0821 L*atm/(K*mol) for the ideal gas constant, we can solve for n (number of moles of N₂):
PV = nRT 1.15 atm * 26 L = n * 0.0821 L*atm/(K*mol) * 533 K n = 0.623 moles of N₂
Next, we use stoichiometry. From the balanced chemical equation 2 NaN₃ → 3 N₂ + 2 Na, we see that 2 moles of NaN₃ produces 3 moles of N2. Therefore, we can find the moles of NaN3 needed:
(0.623 moles N₂) (2 moles NaN₃ / 3 moles N₂) = 0.415 moles NaN₃
Finally, we convert moles of NaN3 to grams. The molar mass of NaN₃ is approximately 65.01 g/mol.
0.415 moles NaN₃ * 65.01 g/mol = 26.979 grams of NaN₃
Therefore, approximately 27 grams of sodium azide must be decomposed to inflate the airbag.