Final answer:
Using the density of nitrogen gas and the stoichiometry of its production from sodium azide, 547.3 grams of sodium azide are required to produce 283 L of nitrogen gas for an automotive airbag.
Step-by-step explanation:
To calculate how many grams of sodium azide (NaN3) are required to produce 283 L of nitrogen gas for an automotive airbag, you need to use the provided density of nitrogen gas and the balanced chemical reaction for the decomposition of NaN3.
The balanced chemical equation for the decomposition of sodium azide is:
2NaN3 (s) → 2Na (s) + 3N2 (g)
Given the density of nitrogen gas as 1.25 g/L, you first calculate the mass of nitrogen gas needed:
Mass of N2 = 283 L × 1.25 g/L = 353.75 g
Next, using the molar mass of N2 (28 g/mol) and NaN3 (65 g/mol), and the stoichiometry of the reaction:
Moles of N2 required = 353.75 g / 28 g/mol = 12.63 mol
According to the reaction, 2 moles of NaN3 produce 3 moles of N2, so:
Moles of NaN3 required = (2/3) × 12.63 mol = 8.42 mol
Mass of NaN3 required = 8.42 mol × 65 g/mol = 547.3 g
Therefore, 547.3 grams of sodium azide are required to produce 283 L of nitrogen gas for an automotive airbag.