Final answer:
To produce 10.0 cubic feet of nitrogen gas, we would theoretically need 547.74 grams of sodium azide (NaN3), using stoichiometry and the given reaction. The options provided do not match this calculation.
Step-by-step explanation:
The question is asking about the amount of sodium azide (NaN3) required to produce a specific volume of nitrogen gas (N2). To find the answer, we need to use the chemical equation for the decomposition of sodium azide:
2NaN3 (s) → 2Na (s) + 3N2 (g)
At standard temperature and pressure (STP), one mole of gas occupies 22.4 liters (or approximately 0.79 cubic feet). First, we convert 10.0 cubic feet to liters to find the moles of nitrogen gas and then use stoichiometry to calculate the mass of NaN3.
- 10.0 ft3 = 283.168 liters
- Moles of N2 = 283.168 L / 22.4 L/mol = 12.64 mol
- The molar ratio of NaN3 to N2 in the reaction is 2:3, so moles of NaN3 needed = (2/3) * 12.64 mol = 8.4267 mol
- Molar mass of NaN3 = 65 g/mol
- Mass of NaN3 = 8.4267 mol * 65 g/mol = 547.74 grams
However, since the options provided seem to be incomplete or incorrect, we cannot match the correct answer to the given choices.