If 450.0 g of NaN3 react what is the volume of N2 gas if it is collected at 15.7 psi and 25 degrees Celsius

Question

2NaN3 yields 2Na + 3N2

If 450.0 g of NaN3 react what is the volume of N2 gas if it is collected at 15.7 psi and 25 degrees Celsius

Answer 1

2NaN3 ===> 2Na + 3N2

moles of NaN3 used = 450.0 g x 1 mole/65.01 g = 6.922 moles NaN3

Using the Ideal Gas Law PV = nRT and solving for V (volume)...

V = nRT/P

Let's convert 15.7 psi to atm so we can use R = 0.0821 L-atm/K-mole (the gas constant)

15.7 psi x 1 atm/14.70 psi = 1.068 atm

Convert 25ºC to K: 25 + 273 = 298K

Now solve the equation for V: V = (6.922 moles)(0.0821 Latm/Kmole)(298K)/1.068 atm

V = 158.6 liters = 159 liters (to 3 significant figures)