A vehicle airbag uses a gas generator that contains a mixture of three compounds

Question

A vehicle airbag uses a gas generator that contains a mixture of three compounds: NaN_3, KNO_3, and SiO_2. A series of three reactions is used to convert the highly toxic sodium azide, NaN_3, into N_2 gas and harmless silicate glass. The nitrogen gas fills up the airbag. causing it to rapidly expand in a head-on collision. Balance the reactions involved in the deployment of an airbag by inserting the appropriate coefficients. NaN_3(s) rightarrow Na(s) + N_2(g) Na(s) + KNO_3(s) rightarrow K_2O(s) + Na_2O(s) + N_2(g) K_2O(s) + Na_2O(s) + SiO_2(s) rightarrow K_2O_3 Si(s) + Na_2O_3Si(s) The drive-side airbag deploys when a van is in a head-on collision. The airbag inflates to a volume of 67.8 L and a pressure of 1.13 atm at 25.0 degree C. Calculate the amount of NaN_3 in grams, that the manufacturer of the vehicle placed into the gas generator to produce this amount of nitrogen gas. Assume the nitrogen gas behaves as an ideal gas and the only source of the nitrogen gas is the first reaction shown above. A list of physical constants can be found here.

Explanation / Answer

N2 is the only gas present in the whole system. only N2 will expand the balloon so we need to balance the equation involving N2 only

Balancing the given equations

NaN3 (s) -----> Na (s) + 1.5 N2 (g)  

10 Na (s) + 2 KNO3 (s) ------> K2O (s) + 5 Na2O (s) + N2 (g)

Na (s) + 0.2 KNO3 (s) ------> 0.1 K2O (s) + 0.5 Na2O (s) + 0.1 N2 (g)

so for one mole of NaN3 1 mole of Na and 1.5 Mole of N2 are given

1 mole of Na which comes from NaN3 gives 0.1 mole of N2

so total of 1.6 moles of N2 are relased per mole of NaN3

Volume = 67.8 L

P = 1.13 atm

T = 25 C = 298 K

R = 0.08206 L.atm/mol.K

P*V = n* R * T

1.13 atm * 67.8 L = n * 0.08206 L.atm/mol.K * 298 K

n = 3.133 moles of N2 will be released as N2 is the only gas released in the reaction

1 mole of NaN3 releases 1.6 mole of N2

so 3.133 moles of N2 requires 3.133 / 1.6 = 1.96 mole of NaN3

Molar mass of NaN3 = 65 g/mol

Mass of NaN3 = 1.96 mol * 65 g/mol = 127.28 g Answer

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