Approximately 21.45 grams of NaN₃ is required to fill the automobile airbag.
To determine the mass of NaN₃ required for an automobile airbag, we can use the ideal gas law and stoichiometry.
Step 1: Write the balanced chemical equation for the decomposition of NaN₃, which produces nitrogen gas (N₂) and sodium (Na) as products.
2 NaN₃(s) → 3 N₂(g) + 2 Na(s)
Step 2: Use the ideal gas law, PV = nRT, to find the number of moles (n) of nitrogen gas produced. The conditions in an automobile airbag are typically close to room temperature (around 298 K) and atmospheric pressure (around 1 atm).
V = 11.8 L
T = 298 K
R (gas constant) = 0.0821 L·atm/(mol·K)
P = 1 atm
PV = nRT
(1 atm) * (11.8 L) = n * (0.0821 L·atm/(mol·K)) * (298 K)
n ≈ 0.49 moles of N₂
Step 3: Use the stoichiometry of the balanced equation to relate moles of NaN₃ to moles of N₂. From the balanced equation, we know that 2 moles of NaN₃ produce 3 moles of N₂.
(0.49 moles of N₂) * (2 moles of NaN₃ / 3 moles of N₂) = 0.33 moles of NaN₃
Step 4: Calculate the molar mass of NaN₃.
Na: 23 g/mol
N: 14 g/mol
Molar mass of NaN₃ = 23 g/mol (Na) + 3 * 14 g/mol (N) = 65 g/mol
Step 5: Calculate the mass of NaN₃ required.
Mass = moles × molar mass
Mass = 0.33 moles × 65 g/mol ≈ 21.45 grams
So, approximately 21.45 grams of NaN₃ is required to fill the automobile airbag.