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

They are all decomposition reactions.