What mass of NaN 3 (s) must be reacted to inflate an air bag to 62.0 L at STP?

Answer 1

At STP, 1 mole of any gas occupies 22.4 L. I assume the reaction is decomposition, 2 NaN₃ --> 2 Na + 3 N₂, so 62.0 L N₂ * (1 mol N₂ / 22.4 L N₂) * (2 mol NaN₃ / 3 mol N₂) * (65.0g NaN₃ / 1mol NaN₃) = 120. g NaN₃.