Answer:
V = 36.7L of [tex]N_{2}[/tex]
Explanation:
1. Write the chemical reaction for the decomposition of sodium azide:
[tex]_{2}NaN_{3}(l)=_{2}Na(s)+_{3}N_{2}(g)[/tex]
2. Find the number of moles of [tex]N_{2}[/tex] produced by the decomposition of 71.4g of [tex]NaN_{3}[/tex]:
[tex]71.4gNaN_{3}*\frac{1molNaN_{3}}{65gNaN_{3}}*\frac{3molesN_{2}}{2molesNaN_{3}}=1.65molesN_{2}[/tex]
3. Use the ideal gas equation to find the Volume of [tex]N_{2}[/tex] occupied by 1.65 moles of [tex]N_{2}[/tex], at the temperature and pressure given by the problem:
[tex]PV=nRT[/tex]
Solving for V:
[tex]V=\frac{nRT}{P}[/tex]
Converting the temperature from ◦C to K:
25◦C+273.15=298.15K
Replacing values:
[tex]V=\frac{1.65mol*(0.082\frac{atm.L}{mol.K})*298.15K}{1.10atm}[/tex]
V=36.7L of [tex]N_{2}[/tex]
