Question

The air bags in cars are inflated when a collision triggers the explosive, highly exothermic decomposition of sodium azide (NaN3): 2NaN3(s) → 2Na(s) + 3N2(g) The passenger-side air bag in a typical car must fill a space approximately four times as large as the driver-side bag to be effective. Calculate the mass of sodium azide required to fill a 113-L air bag. Assume the pressure in the car is 1.00 atm and the temperature of N2 produced is 85°C.

Answer 1

To calculate the mass of sodium azide required to fill a 113-L air bag, you need to convert the volume of nitrogen gas to a mass using the molar mass of sodium azide. Approximately 229 grams of sodium azide is required to fill a 113-L air bag.

Step-by-step explanation:

To calculate the mass of sodium azide required to fill a 113-L air bag, we need to convert the volume of nitrogen gas to a mass using the molar mass of sodium azide. The balanced equation tells us that 2 mol of NaN3 produces 3 mol of N2 gas. We can use the ideal gas law to convert the volume of N2 gas to moles, and then use the molar mass to convert moles to grams.

First, let's calculate the number of moles of N2 gas using the ideal gas law:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Plugging in the values, we get:

(1.00 atm)(113 L) = n(0.0821 L atm / mol K)(85 + 273 K)

Solving for n, we find that n = 5.29 moles of N2 gas.

Next, we need to find the moles of NaN3 that will produce 5.29 moles of N2 gas. From the balanced equation, we know that 2 mol of NaN3 produces 3 mol of N2 gas. So, the number of moles of NaN3 required can be calculated as follows:

(5.29 moles N2 gas)(2 moles NaN3 / 3 moles N2 gas) = 3.53 moles of NaN3.

Finally, we can use the molar mass of sodium azide to convert moles to grams:

(3.53 moles NaN3)(65.01 g / mol) = 229 g of NaN3.

Therefore, approximately 229 grams of sodium azide is required to fill a 113-L air bag.

Answer 2

Answer : The mass of
`NaN_3` required is, 166.4 grams.

Explanation :

First we have to calculate the moles of nitrogen gas.

Using ideal gas equation:

`PV=nRT`

where,

P = Pressure of
`N_2` gas = 1.00 atm

V = Volume of
`N_2` gas = 113 L

n = number of moles
`N_2` = ?

R = Gas constant =
`0.0821L.atm/mol.K`

T = Temperature of
`N_2` gas =
`85^oC=273+85=358K`

Putting values in above equation, we get:

`1.00atm* 113L=n* (0.0821L.atm/mol.K)* 358K`

`n=3.84mol`

Now we have to calculate the moles of sodium azide.

The balanced chemical reaction is,

`2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)`

From the balanced reaction we conclude that

As, 3 mole of
`N_2` produced from 2 mole of
`NaN_3`

So, 3.84 moles of
`N_2` produced from
`(2)/(3)* 3.84=2.56` moles of
`NaN_3`

Now we have to calculate the mass of
`NaN_3`

`\text{ Mass of }NaN_3=\text{ Moles of }NaN_3* \text{ Molar mass of }NaN_3`

Molar mass of
`NaN_3` = 65 g/mole

`\text{ Mass of }NaN_3=(2.56moles)* (65g/mole)=166.4g`

Therefore, the mass of
`NaN_3` required is, 166.4 grams.