A car's airbag needs to be filled with 140 liters of nitrogen gas. Using 2NaN3 → 2Na+3N2 , determine how many grams of sodium azide (NaN3) will be required to produce the 140 liters of nitrogen gas.

Answer 1

Hello, Joselyn,

Since no information is provided on the temperature and pressure of the N₂ in the 140L airbag, I will assume conditions of STP. At STP, all gases occupy 22.4 liters per mole of gas. This means we can convert the 140L directly into moles N₂:

(140L)/(22.4L/mole N₂) = 6.25 moles N₂

If temperature and pressure had been provided, the best approach would be to convert the 140L into the volume at STP, or else use the ideal gas law to calculate moles N₂. My preference is to convert the volume to STP since the ideal gas law requires the value of the gas constant, which is a perfectly fine approach, but slows things down looking for the value that has the correct units for the data provided (e.g., atm, KPa, mmHg, etc).

Now that we established we have 6.25 moles of N₂ in the 140 liters, we can see from the balanced equation that we get 3 moles of N₂ for every 2 moles of sodium azide, NaN₃. That's a molar ratio of 3/2, so that means we only need 2 moles NaN₃ for every 3 moles of N₂ produced. Use this ratio to find the moles of sodium azide required:

(6.25 moles N₂)*((2 moles NaN₃)/(3 moles of N₂) = 4.17 moles of NaN₃

Multiply 4.17 moles NaN₃ by its molar mass of 65.01 g/mole NaN₃ to find the mass of NaN₃ required to produce 140 liters of N₂ at STP. I get 271.1 grams, which is 271 grams at 3 sig figs. However 140 liters is really only 2 sig figs. I hesitated giving up my hard-fought 1 gram if I went to 2 sig figs, so I'll let you decide the correct number. (Too painful for me to round down).

Bob