1. \"Air\" bags for automobiles are inflated during a collision by the explosion
ID: 891464 • Letter: 1
Question
1. "Air" bags for automobiles are inflated during a collision by the explosion of sodium azide, NaN3. The equation for the decomposition is
2NaN3(s) 2Na(s) + 3N2(g). What mass of sodium azide would be needed to inflate a 15.7-L bag to a pressure of 1.3 atm at 25°C?
2. Calculate the volume of nitrogen dioxide produced at 795.2 torr and 24.7°C by the reaction of 7.95 cm3 copper (density = 8.95 g/cm3) with 214.8 mL of concentrated nitric acid if the acid has a density of 1.42 g/cm3 and contains 68.0% HNO3 by mass).
Cu(s) + 4HNO3(aq) Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)
3. A mixture of helium and neon gases has a density of 0.2830 g/L at 30.8°C and 470 torr. What is the mole fraction of neon in this mixture?
Explanation / Answer
1)
2NaN3(s) ---------------> 2Na(s) + 3N2(g)
volume = 15.7-L
pressure = 1.3 atm
temperature = 25°C = 298 K
from ideal gas equation
PV = nRT
1.3 x 15.7 = n x 0.0821 x 298
n = 0.834
moles of sodium azide n = 0.834
molar mass of sodium azide = 65 g/mol
moles = mass / molar mass
0.834 = mass / 65
mass of NaN3 = 54.24 g
3)
T = 30.8 + 273 = 303.8 K
Pressure (P) = 470 torr = 470 / 760 = 0.618 atm
density (d) = 0.2830 g /L
Volume (V) = 1 Litre
R = universal gas constant = 0.0821 L-atm / mol K
PV = nRT
0.618 x 1 = n x 0.0821 x 303.8
n = 0.0248
mass =0.2830
let mass of He = x
mass of Ne = 0.2830 -x
moles of He = x / 4
moles of Ne = 0.2830 -x / 20.18
total moles = x /4 + 0.2830 -x / 20.18
x /4 + 0.2830 -x / 20.18 = 0.0248
20.18 x + 1.772 -4x = 2.571
16.18 x = 0.868
x = 0.0536
mass of He = 0.0536
mass of Ne = 0.2830 -x = 0.2830 - 0.0536 = 0.2294 g
moles of He = 0.0536 / 4 = 0.0134
moles of Ne = 0.2294 / 20.18 = 0.01136
mole fraction of Ne = moles of Ne / total moles
= 0.01136 / 0.0248
= 0.458
mole fraction of Ne = 0.458
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