Chemistry questions

Question

Sodium azide, NaN₃, undergoes a decomposition reaction to produce metallic sodium and nitrogen gas. If 43.2 g sodium azide decomposes in the reaction conducted at 792 torr pressure and 26.2 ^oC, calculate the volume of nitrogen gas produced under the same conditions of temperature and pressure.

Answer 1

molar mass NaN₃ = 65.0 g / mol

molar mass nitrogen gas (N2) = 28.0 g / mol

First, write the correctly balanced equation for the decomposition of sodium azide:

2NaN₃(s) ==> 2Na(s) + 3N₂(g) .. balanced equation for decomposition of sodium azide

Next, use the stoichiometry of the balanced equation to find the moles of N2 produced:

43.2 g NaN₃ x 1 mol NaN₃ / 65.0 g x 3 mol N₂ / 2 mol NaN₃ = 0.997 mol N₂

Finally, use the ideal gas law to convert mols of N2 to volume:

PV = nRT

P = pressure in atm = 792 torr x 1 atm / 760 torr = 1.04 atm

V = volume in L = ?

n = mols of N₂ = 0.997 mol

R = gas constant = 0.0821 Latm/Kmol

T = temperature in Kelvin = 26.2º + 273 = 299.2K

Solving for V we have ...

V = nRT / P = (0.997)(0.0821)(299.2) / 1.04

V = 23.5 L