Answer: 71.65 L
Explanation:
Decomposition of sodium azide is shown by equation below:
[tex]2NaN_3\rightarrow 2Na+3N_2[/tex]
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
[tex]\text{Number of moles of sodium azide}=\frac{125g}{65g/mol}=1.92moles[/tex]
According to stoichiometry:
2 moles of [tex]NaN_3[/tex] produce 3 moles of [tex]N_2[/tex]
Thus 1.92 moles of [tex]NaN_3[/tex] will produce=[tex]\frac{3}{2}\times 1.92=2.88[/tex] moles of [tex]N_2[/tex]
According to the ideal gas equation:
[tex]PV=nRT[/tex]
P = Pressure of the gas = 756 torr = 0.99 atm (1 torr= 0.0013 atm)
V= Volume of the gas = ?
T= Temperature of the gas = 27°C = 300 K (0°C = 273 K)
R= Gas constant = 0.0821 atmL/K mol
n= moles of gas= 2.88
[tex]V=\frac{nRT}{P}=\frac{2.88\times 0.0821\times 300}{0.99}=71.65L[/tex]
Thus the volume of nitrogen gas at 27 °C and 756 torr formed by the decomposition of 125 g of sodium azide is 71.65 L
