Given:
P = 1.0 atm
V = 42.0 L
T = 28.3 oC
= (28.3+273) K
= 301.3 K
find number of moles using:
P * V = n*R*T
1 atm * 42 L = n * 0.08206 atm.L/mol.K * 301.3 K
n = 1.70 mol
From reaction,
Mol of NaN3 required = (2/3)*mol of N2
= (2/3)*1.70 mol
= 1.13 mol
Molar mass of NaN3,
MM = 1*MM(Na) + 3*MM(N)
= 1*22.99 + 3*14.01
= 65.02 g/mol
use:
mass of NaN3,
m = number of mol * molar mass
= 1.13 mol * 65.02 g/mol
= 73.47 g
Answer: 73.5 g
QUESTION 3 Automobile airbags use the decomposition of sodium azide, NaN3, to provide gas for rapid...
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