Starting with 400 mg of NaN3 what will be the volume of the products at the end of the reaction ?

The products are Na solid and N₂ gas. When comparing with the volume of a gas, the solid volume is negligible. Therefore, the volume of N₂ is approximately the total volume of the products.

Molar mass of NaN₃ = (23.0 + 14.0×3) g/mol = 65.0 g/mol
No. of moles of NaN₃ reacted = (400/1000 g) / (65.0 g/mol) = 0.006154 mol

2NaN₃(s)→ 2Na(s) + 3N₂(g)
Mole ratio NaN₃ : N₂ = 2 : 3
No. of moles of N₂ produced = (0.006154 mol) × (3/2) = 0.009231 mol

For the N₂ gas produced :
No. of moles, n = 0.009231 mol
Pressure, P = 99.3 kPa
Absolute temperature, T = (273 + 25) K = 298 K
Gas constant, R = 8.314 J / (mol K)

PV = nRT
V = nRT/P = 0.009231 × 8.314 × 298 / 99.3 L = 0.230 L = 230 mL