Sodium azide (NaN 3 ) can decompose at 300 o C to produce sodium metal (Na) and

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.

Explanation / Answer

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

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