Question

Sodium azide (NaN₃) can decompose at 300 ^o C to produce sodium metal (Na) and nitrogen gas (N₂). This is one of the key reactions in an automobile air bag. The balanced equation for this reaction is:

2NaN₃(s) →2Na (s) + 3 N₂(g)

Part A

If 0.189 mol mole of NaN₃ is completely decomposed at 300.0 ^o   C and 1.00 atm, what volume (in L) of nitrogen gas would be produced?

Express your answer using the correct number of significant figures.

Part B

How many g of NaN₃ would be needed to produce 2.61 L of nitrogen gas at the same temperature and pressure? Use the correct number of significant figures in your answer.

Part C

How many g of NaN₃ would be need to produce 1.00 quart of nitrogen gas at the same temperature and pressure? (1.057 quart = 1.00 L) Enter your answer with the correct number of significant figures.

Answer 1

Ans. Part A: According to the stoichiometry of balanced reaction, 2 mol NaN₃ produces 3 mol N₂ gas.

So,

            Moles of N₂ produced = (3 / 2) x moles of NaN₃ taken

                                                = (3/ 2) x 0.189 mol

                                                = 0.2835 mol

# Given,

            Pressure, P = 1.00 atm

            Temperature, T = 300.0⁰C = 573.15 K

            Moles of N₂ = 0.2835 mol - as calculated above

# Using Ideal gas equation:            PV = nRT      - equation 1

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol^-1K^-1

            T = absolute temperature (in K) = (⁰C + 273.15) K

# Putting the values in equation 1 -

            1.00 atm x V = 0.2835 mol x (0.0821 atm L mol^-1K^-1) x 573.15 K

            Or, V = 13.3402668525 atm L / 1.00 atm = 13.34 L

Therefore, volume of N₂ gas produced = 13.34 L

# Part B: Putting the values in equation 1 to calculate moles of N₂ -

            1.00 atm x 2.61 L = n x (0.0821 atm L mol^-1K^-1) x 573.15 K

            Or, n = 2.61 atm L / 47.055615 atm L mol^-1 = 0.0555 mol

Therefore, moles of N₂ in 2.62 L sample = 0.0555 mol

# According to the stoichiometry of balanced reaction, 2 mol NaN₃ produces 3 mol N₂ gas.

So,

            Required moles of NaN₃ = ( 2/ 3) x moles of N₂ = (2/ 3) x 0.0555 mol = 0.037 mol

            Required mass of NaN₃ = moles x molar mass

= 0.037 mol x (65.01 g// mol)

                                                = 2.41 g

# Part C: Given, Volume of N₂ = 1 quart = 0.946353 L

Putting the values in equation 1 to calculate moles of N₂ -

            1.00 atm x 1.0 L = n x (0.0821 atm L mol^-1K^-1) x 573.15 K

            Or, n = 1.0 atm L / 47.055615 atm L mol^-1 = 0.0555 mol

Therefore, moles of N₂ in 2.62 L sample = 0.02125 mol

# According to the stoichiometry of balanced reaction, 2 mol NaN₃ produces 3 mol N₂ gas.

So,

            Required moles of NaN₃ = (2/ 3) x 0.02125 mol = 0.01417 mol

            Required mass of NaN₃ = moles x molar mass

= 0.01417 mol x (65.01 g// mol)

                                                = 0.92 g