Question

How many grams of sodium azide are needed to provide sufficient nitrogen gas to fill a $50.0\times 50.0\times 25.0cm$ bag to pressure of $1.15atm$ at $25.0{}^\circ C$ ?

Answer 1

The ideal gas equation is applicable to the real gases which shows ideal behaviour at certain conditions of temperature and pressure, however the real gases tend to deviate from the ideal behaviour beyond those specific conditions.

Formula used: $PV=nRT$
Where, $P$ is the pressure, $V$ is the volume, $n$ is the number of moles of the gas, $R$ is the gas constant and $T$ is the temperature.

Complete step-by-step answer:
We are supposed to find out the mass of sodium azide in terms of grams, which is required to produce a specific amount of nitrogen. We know that the chemical formula of sodium azide is $Na{{N}_{3}}$ and nitrogen is ${{N}_{2}}$.
In order to solve this question we will use the ideal gas equation, in order to find out the number of moles of nitrogen gas. The pressure is given as $1.15atm$ and we will denote it by $P$. The volume of the gas would be the volume of the container as the gases are fluid and it will be evenly distributed in the total volume of the container. So the volume becomes $50.0\times 50.0\times 25.0cm=62500c{{m}^{3}}$, and we will denote it by $V$. The volume should be converted to litres so,
$62500c{{m}^{3}}\left( \dfrac{1mL}{1c{{m}^{3}}} \right)\left( \dfrac{1L}{{{10}^{3}}mL} \right)=62.5L$.
We know that the ideal gas equation is,
$PV=nRT$
Where, $P$ is the pressure, $V$ is the volume, $n$ is the number of moles of the gas, $R$ is the gas constant and $T$ is the temperature.

The value of temperature is given as $25.0{}^\circ C$, which should be converted to kelvin scale. So, we get,
$T=25.0{{~}^{o}}C+273=298~K$
For the value of gas constant we will use the value ${0}{.082057}Latm{{K}^{-1}}mo{{l}^{-1}}$.

Now substituting these values in the above equation we get,
$n=\dfrac{PV}{RT}$
$\left( (\dfrac{(1.15atm)(65.5L)}{{0}{.082057}Latm{{K}^{-1}}mo{{l}^{-1}})(298K)} \right)$

After solving the above equation we get,
$n=2.94mol$ of nitrogen.

Now, consider the chemical reaction which is taking place in the reaction, which is,
$2Na{{N}_{3}}\to 3{{N}_{2}}+2Na$

As we can see the coefficients of the sodium azide and nitrogen are two and three, we will use these to find out the moles of sodium azide which should react.
$2.94mol{{N}_{2}}(\dfrac{2mol{ }Na{{N}_{3}}}{3mol{ }{{N}_{2}}})=1.96~molNa{{N}_{3}}$

So, we can see that the number of moles of sodium azide is $1.96mol$, so in order to find out the mass of the sodium azide we will use this number of moles of azide.
$1.96molNa{{N}_{3}}\left( \dfrac{65.01gNa{{N}_{3}}}{1molNa{{N}_{3}}} \right)=127gNa{{N}_{3}}$
So the mass of sodium azide in grams is $127$.

Note:
i) The amount of any chemical consumed in a chemical reaction can be calculated in terms of grams, once we calculate the number of moles of that chemical which is present in the reaction.
ii) Then the coefficient of that compound will help us find the mass of the compound, as the number of moles of a substance is the given mass present per molar mass of the compound.