Question

How many grams of sodium azide are needed to provide sufficient nitrogen gas to fill a 40.0 × 40.0 × 25.0 cm bag to a pressure of 1.17 atm at 15.0 °C?

Answer 1

Volume of bag = 40.0 * 40.0 * 25.0 cm^3

= 40,000 cm^3

= 40,000 mL

= 40.0 L

Given:

P = 1.17 atm

V = 40.0 L

T = 15.0 oC

= (15.0+273) K

= 288 K

find number of moles using:

P * V = n*R*T

1.17 atm * 40 L = n * 0.08206 atm.L/mol.K * 288 K

n = 1.98 mol

Thew reaction taking place is:

2 NaN3 -> 2 Na + 3 N2

From above reaction,

Mol of NaN3 required = (2/3)*mol of N2

= (2/3)*1.98 mol

= 1.32 mol

Molar mass of NaN3,

MM = 1*MM(Na) + 3*MM(N)

= 1*22.99 + 3*14.01

= 65.02 g/mol

use:

mass of NaN3,

m = number of mol * molar mass

= 1.32 mol * 65.02 g/mol

= 85.83 g

Answer: 85.8 g