"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

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.