"air" bags for automobiles are inflated during a collision by the explosion of sodium azide, nan3. the equation for the decomposition is 2nan3 → 2na + 3n2. what mass of sodium azide would be needed to inflate a 14.9 l bag to a pressure of 1.1 atm at 25°c?

We need 29.3 grams of sodium azide

Explanation:

Step 1: The balanced equation

2NaN3(s) → 3N2(g)+2Na(s)

Step 2: Calculate moles of N2

In this situation, we can use the ideal gas equation to find the moles of nitrogen gas present:

P*V = n*R*T

with P = the pressure of the gas ( in this situation 1.1 atm)

with V = the volume of the gas ( in this situation 14.9L)

with n = the number of moles = TO BE DETERMINED

with R= the gas constant = 0.0821 L*atm / mol* K

with T = the temperature ( in this case 25°C = 298.15 Kelvin)

n = P*V / R*T

n = (1.1 * 14.9)/(0.0821 * 298.15) = 0.67 moles of N2

Calculate moles of NaN3

We need 2 moles of NaN3 to produce 3 moles of N2

So if there is produced 0.67 moles of N2, we need 0.45 moles of NaN3

Calculate mass of NaN3

Mass of NaN3 = moles of NaN3 * Molar mass of NaN3

Mass of NaN3 = 0.45 moles * 65.01 g/mole = 29.3 grams of NaN3

We need 29.3 grams of sodium azide

a

explanation: