Question

How many grams of NaN3 must be used to produce 85.5 L of N2 gas at 28 C and 95.0 psi of

pressure?

Answer 1

PV = nRT

R = 0.082057 L⋅atm/mol⋅K

Pressure = 95 psi = 6.5atm

Temp= 273 + 28= 301K

Volume = 85.5L

$6.5×85.5= n [0.082057×301]$

$n= 22.5 moles$

$Mass = 22.5×65=1462.5grams$