An airbag is filled with 18L nitrogen gas at 1 atm and 27 degrees celsius, what mass of NaN2 must be decomposed?
Given:
P = 1.0 atm
V = 18.0 L
T = 27.0 oC
= (27.0+273) K
= 300 K
find number of moles using:
P * V = n*R*T
1 atm * 18 L = n * 0.08206 atm.L/mol.K * 300 K
n = 0.7312 mol
This is number of mol of N2 produced
The reaction taking place is:
NaN2 —> Na + N2
from reaction,
mol of NaN2 required = mol of N2 formed
= 0.7312 mol
Molar mass of NaN2,
MM = 1*MM(Na) + 2*MM(N)
= 1*22.99 + 2*14.01
= 51.01 g/mol
use:
mass of NaN2,
m = number of mol * molar mass
= 0.7312 mol * 51.01 g/mol
= 37.3 g
Answer: 37.3 g
Answer: 37.3 g
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