Question

In the following reaction, how many liters of nitrogen, N2, measured at STP, would be produced from the decomposition of 371 g of sodium azide, NaN3?

Answer 1

Balanced chemical reaction is

2NaN₃    2Na + 3N₂

molar mass of NaN₃ = 65.0099 gm/mole then 2 mole of NaN₃ = 130.0198 gm

molar mass of N₂ = 28.013gm/mole then 3 mole of N₂ =84.039 gm

According to reaction 2 mole of NaN₃ produce 3 mole of N₂ that mean 130.0198 gm of NaN₃ produce 84.039 gm of N₂ then 371 gm of NaN₃ produce = 371 84.039 / 130.0198 = 239.798 gm of N₂

molar mass of N₂ = 28.013gm/mole then 239.798 gm of N₂ = 239.798 / 28.013 = 8.56 mole of N₂

At STP 1 mole of gas occupy volume 22.414 liter then 8.56 mole nitrogen occupy volume = 22.414 8.56 = 191.869 liter

191.869 liter N₂ at STP.