Answer:
0.0177 L of nitrogen will be produced
Explanation:
The decomposition reaction of sodium azide will be:
[tex]2NaN_{3}(s)--->2Na(s)+3N_{2}(g)[/tex]
As per the balanced equation two moles of sodium azide will give three moles of nitrogen gas
The molecular weight of sodium azide = 65 g/mol
The mass of sodium azide used = 100 g
The moles of sodium azide used = [tex]\frac{mass}{molarmass}=\frac{100}{65}=1.54mol[/tex]
so 1.54 moles of sodium azide will give = [tex]\frac{3X1.54}{2}=2.31[/tex]mol
the volume will be calculated using ideal gas equation
PV=nRT
Where
P = Pressure = 1.00 atm
V = ?
n = moles = 2.31 mol
R = 0.0821 L atm / mol K
T = 25 °C = 298.15 K
Volume = [tex]\frac{P}{nRT}=\frac{1}{2.31X0.0821X298.15}=0.0177L[/tex]