What mass (g) of nan3 is required to provide 26.5 l of n2 at 22.0 °c and 1.10 atm?

NaN₃ (sodium azide) produces N₂ gas when it is heated. The balanced equation is,

2NaN₃ → 2Na + 3N₂

To find out mass of NaN₃, we have to find the moles of N₂ formed by using given data.

Let's assume that N₂ gas is an ideal gas. Then we can use ideal gas equation,

PV = nRT

Where,

P = Pressure of the gas (Pa)

V = volume of the gas (m³)

n = number of moles (mol)

R = Universal gas constant (8.314 J mol⁻¹ K⁻¹)

T = temperature in Kelvin (K)

The given data for N₂ gas is,

P = 1.10 atm = 111457.5 Pa

V = 26.5 L = 26.5 x 10⁻³ m³

T = (273 + 22) K = 295 K

R = 8.314 J mol⁻¹ K⁻¹

n = ?

By applying the formula,

111457.5 Pa x 26.5 x 10⁻³ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 295 K

n = 1.204 mol

Hence, produced N₂ gas = 1.204 mol

The stoichiometric ratio between NaN₃ and N₂ is 2 : 3

hence moles of NaN₃ = 1.204 mol x 2 / 3

= 0.803 mol

Molar mass of NaN₃ = 65 g/mol

Hence, required mass of NaN₃ = 0.803 mol x 65 g/mol

= 52.195 g