Automobile air bags are inflated with nitrogen gas, which is formed by the decomposition of solid sodium azide (NaN₃). The other product is sodium metal. Calculate the volume of nitrogen gas at 27 °C and 756 torr formed by the decomposition of 125 g of sodium azide

In the event of a collision, an airbag is an inflatable safety device meant to protect the occupants of a vehicle.

\[2{\rm{Na}}{{\rm{N}}_3}\to3\;{{\rm{N}}_2}+2{\rm{Na}}\]

Calculating molecular weight of\[{\rm{Na}}{{\rm{N}}_3}\]

Atomic weight of Sodium\[({\rm{Na}})=23\]grams.

Atomic weight of Nitrogen\[({\rm{N}})=14\]grams.

So, molecular weight of\[{\rm{Na}}{{\rm{N}}_3}=23+3\cdot (14)\]grams.

\[=23+42\]grams.

\[=65\]grams.

Calculating molecular weight of\[{{\bf{N}}2}\]

Atomic weight of Nitrogen\[({\rm{N}})=14\]grams.

So, molecular weight of\[{{\rm{N}}_2}=2 \cdot(14)\]grams.

\[=28\]grams.

Calculating molecular weight of Na

Atomic weight of Sodium\[({\rm{Na}})=23\]grams.

Given, 2 moles of\[{\rm{Na}}{{\rm{N}}_3}\to3\]moles of\[{{\rm{N}}_2}+2\]moles of\[{\rm{Na}}\]

\[2\cdot(65)\]grams\[\to3\cdot(28)\]grams\[+2\cdot(23)\]grams

So, 130 grams of\[{\rm{NaN}}\to84\]grams of\[{{\rm{N}}_2}+46\]grams of\[{\rm{Na}}\]

We have, for 130 grams of\[{\rm{NaN}} \to 84\]grams of\[{{\rm{N}}_2}\]

So, for 125 grams of\[{\rm{NaN}} \to \]? grams of\[{{\rm{N}}_2}\]

\begin{aligned}\to\frac{{(125)\cdot(84)}}{{130}}{\rm{gramsof}}{{\rm{n}}_2}\\\to\frac{{(10500)}}{{130}}{\rm{gramsof}}{{\rm{N}}_2}\\\to80.76{\rm{gramsof}}{{\rm{N}}_2}\end{aligned}

28 grams of\[{N_2} = 1\]mole of\[{N_2}\]

\[80.76\]grams of\[{{\rm{N}}_2} = \]moles of\[{{\rm{N}}_2}\]

\begin{aligned}=\frac{{(80.76)\cdot(1)}}{{28}}{\rm{moles}}\\=\frac{{80.76}}{{28}}{\rm{molesof}}{{\rm{N}}_2}\\= 2.88{\rm{ moles of }}{{\rm{N}}_2}\end{aligned}

We have \[n = 2.88\] moles, and given \[P = 756\]torr, \[T = {35.5R^\circ }\].

Converting the temperature in degrees Celsius to kelvins.

We have, \[{0R^\circ } = (0) + 273.15K\]

So,

\begin{aligned}27=(27)+273.15K\\=300.15K\end{aligned}

Converting pressure in torr into atm.

We have, 1 torr\[ = 0.001315\]atm.

\[{\rm{So}}\],

\begin{aligned}756{\rm{torr}}=(756)\cdot(0.001315){\rm{atm}}\\=0.99414{\rm{atm}}\end{aligned}

From Ideal gas law

\[PV = nRT\]

Where,

\begin{aligned}{\rm{P}}={\rm{pressureofthegas}}{\rm{.}}\\{\rm{V}}={\rm{volumeofthegas}}{\rm{.}}\\{\rm{n}}={\rm{numberofmolesofthegas}}{\rm{.}}\\{\rm{T}}={\rm{temperatureofthegas}}{\rm{.}}\end{aligned}

\begin{aligned}{\rm{R}}={\rm{Idealgasconstant}}\\=0.08206\;{\rm{Latmmo}}{{\rm{l}}^{-1}}{K^{-1}}\end{aligned}

If,

Amount of the gas\[(n)\]is in moles,

Temperature\[(T)\]is in Kelvin\[(K)\],

\[{\mathop{\rm pressure}\nolimits} (P)\]is in atm.

\begin{aligned}PV=nRT\\V=\frac{{nRT}}{P}\\=\frac{{(2.88)\cdot(0.08206)\cdot(300.15)}}{{0.99414}}\\=\frac{{70.9352}}{{0.99414}}\\= 71.35\;{\rm{L}}\end{aligned}

Hence, the required volume of nitrogen gas is 71.35L.