The "air bags" that are currently installed in automobiles to prevent injuries in the event of a crash are equipped with sodium azide, NaN3, which decomposes when activated by an electronic igniter to produce nitrogen gas that fills the bag. How many liters of nitrogen, measured at 25 °C and 1.00 atm, will be produced by 100.0 g of NaN3?

56.51L

Explanation:

Firstly, we need to write the equation of reaction. This means we are writing equation of reaction for the decomposition of sodium azide.

2NaN3 - -> 3N2. + 2Na

We can see from the equation that 2 moles of the side gives 3 moles of nitrogen.

We need to get the number of moles of the azide actually in the reaction.

This is gotten by dividing the mass of the azide by the molar mass. The molar mass of the azide is 23 + 3 (14) = 23 + 42 = 65g/mol

The number of moles is 100g/65 = 1.54 moles

From here, we get the number of moles of nitrogen produced after decomposition.

From what we can see:

2 moles of azide = 3 nitrogen moles

1.54 moles of azide = x nitrogen moles

x = (1.54 * 3) / 2 = 2.31 moles of nitrogen

To get the volume produced, we use the ideal gas equation

PV = nRT

P = 1 atm

V=?

n = 2.31 moles

R = 0.08205 L atm mol^-1 K^-1

T = 25 + 273.15 = 298.15K

V = nRT/P = (2.31 * 0.08205 * 298.15) / 1 = 56.51L