Respuesta :
Answer : The mass of [tex]NaN_3[/tex] required is 71.175 grams.
Explanation :
To calculate the moles of nitrogen gas, we use the equation given by ideal gas :
PV = nRT
where,
P = Pressure of nitrogen gas = 763 torr
V = Volume of the nitrogen gas = 40.0 L
n = number of moles of gas = ?
R = Gas constant = [tex]62.364\text{ L.torr }mol^{-1}K^{-1}[/tex]
T = Temperature of helium gas = [tex]25^oC=273+25=298K[/tex]
Putting values in above equation, we get:
[tex]763torr\times 40.0L=n\times 62.364\text{ L.torr }mol^{-1}K^{-1}\times 298K\\\\n=1.642mol[/tex]
Now we have to calculate the moles of [tex]NaN_3[/tex].
The balanced chemical reaction is:
[tex]2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)[/tex]
From the balanced chemical reaction, we conclude that
As, 3 moles of [tex]N_2[/tex] produced from 2 moles [tex]NaN_3[/tex]
So, 1.642 moles of [tex]N_2[/tex] produced from [tex]\frac{2}{3}\times 1.642=1.095[/tex] moles [tex]NaN_3[/tex]
Now we have to calculate the mass of [tex]NaN_3[/tex].
Molar mass of [tex]NaN_3[/tex] = 65 g/mol
[tex]\text{Mass of }NaN_3=\text{Moles of }NaN_3\times \text{Molar mass of }NaN_3[/tex]
[tex]\text{Mass of }NaN_3=1.095mole\times 65g/mole=71.175g[/tex]
Therefore, the mass of [tex]NaN_3[/tex] required is 71.175 grams.
