automotive air bags inflate when sodium azide rapidly decomposes as shown in this equation: 2nan3(s) → 2na(s) + 3n2(g) what mass of nan3 is necessary to produce 70.0 l of n2 gas at 315 k and 1.20 atm? 141 g nan3 422 g nan3 723 g nan3 948 g nan3

141.0 g NaN₃.

Explanation:

It is a stichiometric problem.

Firstly, we should wright the reaction as a balanced equation:

2NaN₃ → 2Na + 3N₂

It is clear that 2.0 moles of NaN₃ decompose to 2.0 mole of Na and 3.0 moles of N₂.

Then we should convert the volume of N₂ (70.0 L) to moles via using the gas law of ideal gas: PV = nRT, n = PV / RT,

Where, P is the pressure of the gas in atm (P = 1.2 atm).

V is the volume of the gas in L(V = 70.0 L).

R is the general gas constant (R = 0.082 L.atm/mol.K).

T is the temperature in K(T = 315 K).

∴ n of N₂ = PV / RT = (1.2 atm) (70.0 L) / (0.082 L.atm/mol.K) (315 K) = 3.252 mole.

From the stichiometry:

2.0 moles of NaN₃ decomposes to → 3.0 moles of N₂

??? mole of NaN₃ decomposes to → 3.2520 moles of N₂

The number of moles of NaN₃ = (2.0 moles of NaN₃) (3.2520 moles of N₂) / (3.0 moles of N₂) = 2.168 mole.

Finally, we can convert the number of moles of NaN₃ to mass using the relation: m = n x molar mass.

Molar mass of NaN₃ = 65.0 g/mol.

the mass of NaN₃ = n x molar mass = (2.168 mole) (65.0 g/mol) = 140.92 g ≅ 141.0 g.

i believe it is-

d.) gravity pulled particles of dust and gas together to form planets.

there are two significant figures.