Gas Laws: Air Bags (desperate for help)

Question

The air bags in your vehicle work using the following sets of reactions:

 

2NaN3-->2Na+3N2

 

10Na+2KNO3-->K2O+5NA2O+N2

 

When your vehicle detects a collision, the sodium azide (NaN3) is activated (heated up) to start the decomposition reaction. This produces nitrogen gas and sodium metal. The sodium metal produced in the first reaction is then used in the second reaction to produce more nitrogen gas to inflate your airbag in your vehicle.

 

Say a typical airbag contains 130. g of NaN3 and the temperature inside your airbag are around 40 degrees Celsius with a normal air pressure of 1.0 atm. What volume of N2 gas is produced when your air bag is deployed?

Answer 1

We need to use PV=nRT twice. We are given 130 grams of NaN₃. Let's convert this to moles:

130 g NaN₃ x (1 mole)/(65.0 g) = 2.00 moles NaN₃

 

We need a volume of N₂. Therefore, we must convert moles of NaN₃ to moles of N₂ using the first equation:

2.00 moles NaN₃ x (1 mole N₂)/(2 moles NaN₃) = 1.00 mole N₂.

 

We use 1.00 mole N₂ as "n" in the ideal gas law. Remember that T must be in Kelvin:

V=nRT/P= [(1 mole) x (0.0821 L⋅atm/K⋅mol) x (313.15 K)]/1.0 atm = 25.7 L N₂

 

N₂ is also a product of the second reaction. We convert moles of NaN₃ to moles of Na to get the amount of sodium produced in the first reaction. Then, using the second equation, we convert moles of Na to moles of N₂ using the second balanced equation:

 

2.00 moles NaN₃ x (2 moles Na)/(2 moles NaN₃) = 2.00 moles Na

2.00 moles Na x (1 mole N₂)/(10 moles Na) = 0.200 moles N₂

 

Now we use PV=nRT again with the new moles:

V=nRT/P= [(0.200 mole) x (0.0821 L⋅atm/K⋅mol) x (313.15 K)/1 atm = 5.14 L

 

The total volume is simply the sum of the two volumes:

V_N2 = 5.14 L + 25.7 L = 30.8 L