Question

Automotive air bags inflate when sodium azide, NaN3, rapidly decomposes to its component elements: 2NaN3(s)→2Na(s)+3N2(g)

a. How many moles of N2 are produced by the decomposition of 1.70 mol of NaN3?

b. How many grams of NaN3 are required to form 13.0 g of nitrogen gas?

c. How many grams of NaN3 are required to produce 11.0 ft3 of nitrogen gas if the gas has a density of 1.25 g/L?

Answer 1

a)
From given chemical equation,
Mol of N2 formed = (3/2)*mol of NaN3 reacted
= (3/2)*1.70 mol
= 2.55 mol
Answer: 2.55 mol

b)

Molar mass of N2 = 28.02 g/mol

mass of N2 = 13 g
mol of N2 = (mass)/(molar mass)
= 13/28.02
= 0.464 mol


According to balanced equation
mol of NaN3 reacted = (2/3)* moles of N2
= (2/3)*0.464
= 0.3093 mol


Molar mass of NaN3,
MM = 1*MM(Na) + 3*MM(N)
= 1*22.99 + 3*14.01
= 65.02 g/mol


mass of NaN3 = number of mol * molar mass
= 0.3093*65.02
= 20.11 g
Answer: 20.1 g

c)
1 ft = 30.48 cm

So,
Volume = 11.0 ft^3
= 11.0 * (30.48 cm)^3
= 3.11*10^5 cm^3
= 3.11*10^5 mL
= 3.11*10^2 L

Now use:
Mass of N2 = density * volume
= 1.25 g/L * 3.11*10^2 L
= 389 g

Molar mass of N2 = 28.02 g/mol

mass of N2 = 3.89*10^2 g
mol of N2 = (mass)/(molar mass)
= 3.89*10^2/28.02
= 13.88 mol

According to balanced equation
mol of NaN3 required = (2/3)* moles of N2
= (2/3)*13.88
= 9.255 mol


Molar mass of NaN3,
MM = 1*MM(Na) + 3*MM(N)
= 1*22.99 + 3*14.01
= 65.02 g/mol


mass of NaN3 = number of mol * molar mass
= 9.255*65.02
= 6.018*10^2 g
Answer: 602 g