Question

Calculate how many grams of sodium azide (NaN3) are needed to inflate a 25.0 × 25.0 × 20.0 cm bag to a pressure of 1.35 atm at a temperature of 20.0°C. How much sodium azide is needed if the air bag must produce the same pressure at 10.0°C? The equation is 20NaN3+6SiO2+4KNO3=32N2+5Na4SiO4+K4SiO

Answer 1

Answer : The mass of
`NaN_3` at temperature
`20^oC` 28.47 g.

The mass of
`NaN_3` at temperature
`10^oC` 29.51 g.

Solution : Given,

Pressure of gas = 1.35 atm

Temperature of gas =
`20^oC=273+20=293K`
`(0^oC=273K)`

Volume of gas =
`25* 25* 20cm=12500cm^3=12.5L`
`(1L=1000cm^3)`

Molar mass of
`NaN_3` = 65 g/mole

Part 1 : First we have to calculate the moles of gas at temperature
`20^oC`. The gas produced in the given reaction is
`N_2`.

Using ideal gas equation,

`PV=nRT`

where,

P = pressure of the gas

V = volume of the gas

T = temperature of the gas

n = number of moles of gas

R = Gas constant = 0.0821 Latm/moleK

Now put all the given values in this formula, we get

`(1.35atm)* (12.5L)=n* (0.0821Latm/moleK)* (293K)`

By rearranging the terms, we get the value of 'n'

`n=0.7015moles`

The moles of
`N_2` = 0.7015 moles

The given balanced reaction is,

`20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)`

As, 32 moles of
`N_2` produced from 20 moles of
`NaN_3`

So, 0.7015 moles of
`N_2` produced from
`(20)/(32)* 0.7015=0.438` 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`

`\text{ Mass of }NaN_3=(0.438moles)* (65g/mole)=28.47g`

Therefore, the mass of
`NaN_3` needed are 28.47 g.

Part 2 : We have to calculate the moles of gas at temperature
`10^oC` and same volume & pressure.

Using ideal gas equation,

`PV=nRT`

Now put all the given values in this formula, we get

`(1.35atm)* (12.5L)=n* (0.0821Latm/moleK)* (283K)`

By rearranging the terms, we get the value of 'n'

`n=0.726moles`

The moles of
`N_2` = 0.726 moles

The given balanced reaction is,

`20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)`

As, 32 moles of
`N_2` produced from 20 moles of
`NaN_3`

So, 0.726 moles of
`N_2` produced from
`(20)/(32)* 0.726=0.454` 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`

`\text{ Mass of }NaN_3=(0.454moles)* (65g/mole)=29.51g`

Therefore, the mass of
`NaN_3` needed are 29.51 g.