Answer to Question #119517 in General Chemistry for Emma

Question

The production of a gas to fill an airbag rapidly is accomplished by the following reaction:

2NaN 3 (s) → 2Na(s) + 3N 2 (g)
How many moles of nitrogen gas will form from the reaction of 4.76 mol of NaN 3 (s)?

If 117 g of sodium azide (NaN 3 (s)) react, how many litres of nitrogen gas will form at
20.2 o C and 101.2 kPa?

How many grams of sodium metal will form as 10.5 L of nitrogen form at 25.0 o C and
132.0 kPa?

Answer 1

From 2 moles of NaN3, 3 moles of N2 gas formed

From 4.76 moles of NaN3, N2 gss will be formed = (3/2)× 4.76

= 7.14 moles

Molecular weight of NaN3 = 65 g/ mole

So from 130 gm of NaN3, 3 moles of N2 is formed

So from 117 g NaN3, N2 will be fromed = (3/130) × 117

=2.7 moles

Pressure, P = 101.2 kPa = 1 atm

Temperature, T = 20.2°C =293.2K

Gas Constant, R = 0.082

Mole no, n =2.7 moles

PV =nRT

So Volume, V = (nRT /P)

= (2.7×293.2× 0.082)/ 1

= 65 L

Here moles of N2, n = (PV/ RT)

P= 132kPa = 1.3 atm

T= 25°C =298 K

V = 10.5L

So n = (10.5× 1.3)/ (0.082× 298)

= 0.56 mole

For 3 moles of N2, 46 g of Na is formed

From 0.56 moles of N2, Na will be formed = (46/3)×0.56

= 8.6 g