Question

Air bags are activated when a severe impact causes a steel ball to compress a spring and electrically ignite a detonator cap. This causes sodium azide (NaN3) to decompose explosively according to the following reaction: 2NaN3 (s) → 2Na(s) + 3N2(g) What mass of NaN3 must be reacted to inflate an air bag to 70.0 L at STP?

Answer 1

Answer:

135.2791 g

Explanation:

For STP conditions:

Pressure = 1.00 atm

Temperature = 273.15 K

Volume = 70.0 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1.0 atm × 70.0 L = n × 0.0821 L.atm/K.mol × 273.15 K  

⇒n = 3.1214 moles

Moles of $N_2$ = 3.1214 moles

From the reaction given,

$2NaN_3_{(s)}\rightarrow 2Na_{(s)}+3N_2_{(g)}$

3 moles of Nitrogen gas is formed from 2 moles of sodium azide

1 mole of Nitrogen gas is formed from 2/3 moles of sodium azide

3.1214 moles of Nitrogen gas is formed from (2/3)*3.1214 moles of sodium azide

Moles of sodium azide = 2.0809 moles

Molar mass of sodium azide = 65.0099 g/mol

Mass = Moles * Molar mass = 2.0809 moles * 65.0099 g/mol = 135.2791 g

Answer 2

136.5 g of azide must be reacted to inflate an air bag to 70.0 L at STP.

The equation of the reaction is; 2NaN3 (s) → 2Na(s) + 3N2(g)

We can obtain the number of moles of N2 which inflates air bag at STP as follows;

Pressure (P) = 1 atm (standard pressure)

Volume (V) = 70.0 L

T = 273 K (standard temperature)

n =?

R = 0.082 atm LK-1mol-1

From PV =nRT

n = PV/RT

n = 70.0 L × 1 atm/0.082 atm LK-1mol-1  × 273 K

n = 3.13 moles of N2

Back to the reaction equation;

2 moles of azide yields 3 moles of N2

x moles of azide yields 3.13 moles of N2

x = 2 moles  ×  3.13 moles/ 3 moles

x = 2.1 moles of azide

Mass of azide = 2.1 moles of azide × 65 g/mol

= 136.5 g of azide

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