Answer:
0.0177 L of nitrogen will be produced
Step-by-step explanation:
The decomposition reaction of sodium azide will be:
![2NaN_(3)(s)--->2Na(s)+3N_(2)(g)](https://img.qammunity.org/2020/formulas/chemistry/high-school/tngmv87phdsp3rhthbaghl8d7srcecztuo.png)
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 =
![(mass)/(molarmass)=(100)/(65)=1.54mol](https://img.qammunity.org/2020/formulas/chemistry/high-school/vastdcmxjqsn6u7wva448q7x425th9pzf9.png)
so 1.54 moles of sodium azide will give =
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 =
![(P)/(nRT)=(1)/(2.31X0.0821X298.15)=0.0177L](https://img.qammunity.org/2020/formulas/chemistry/high-school/5bf7ao6k6mt8rdxll0d2ct0ljjw668fon5.png)